Paper IV Complex Analysis Block -1 Complex Analysis

Home , Picard theorem

Post Graduate Programme

M.Sc. Mathematics (Previous)

Distance Educations

Self Instructional Material

Block:1

Unit-1 Complex Integration Unit-2 Calculus of Residue

1 | P a g e

Paper IV Complex Analysis Block -1 Complex Analysis

Introduction : The fundamental idea of extending the real number system by the introduction of complex numbers was first necessitated by the solution of algebraical equations L. Euler (1707-1783) was the first mathematician who introduced the symbol I =√-1 C.F. Gaurs (1777-1855) studied that the algebraic equation with real coefficients has complex roots of the from a+ ib where a and b are real numbers . W.R. Hamilton (1805-1865) also made considerable contribution to the development of the theory of complex number. The BLOCK contains two units. In unit 1 we will discuss complex integration. Under this we will studies the proofs at various theorems based on complex integration like Cauchy’s integral formula. Morera’s theorem, Schwarz lemma etc. in unit 2 we will study about residues and evaluation of integrals using it. Complex Analysis: Unit-1 Complex integration Cauchy –Gousat theorem Cauchy integral formula. Higher order derivatives. Morera’s Theorem Cauchy’s inequality and Liouville’s theorem. The Fundamental theorem of algebra. Taylor’s theorem. Maximum modulus principle. Schwarz lemma. Laurent’s series. Lsoluted singularities Menomarphlc Punctions. The argument principle . Rouche’s theorem. Inverse function theorem. Unit–II Residues. Cauchy’s residue theorem. Evaluation of integrals. Branches of many valued functions with special n reference to argz. logzandz .

2 | P a g e

Unit -1 Complex Integration Objectives: After reading this unit, your should to able to,  understand the concept of complex integration  prove many results based on it  understand singularities STRUCTURE: 1.1 Complex integration 1.2 Basic Definitions 1.2.1 Partition 1.2.2 Jordanarc 1.2.3 Rectifiable Arcs 1.2.4 Contours 1.2.5 Simple and Multiply connected Regional. 1.2.6 Analytic functions 1.3 Cauchy’s –Goursat. Theorem 1.4 Cauchy’s Integral Derivalives 1.5 Higher order Inequality 1.6 Cauchy’s Inequality 1.7 Liouville’s theorem 1.8 Taylor’s theorem 1.9 Zero’s of Analysis functions 1.91 Singularities 1.92 Types of Singularity 1.10 Meromorphic functions 1.11 Argument Principle 1.12 Rouche’s Theorem 1.13 Fundamental theorem of algebra 1.14 Maximum modules principle 1.15 Schwarz lemma 1.16 summary 1.17 Self Assessment Test 1.18 Further Readings

3 | P a g e

1.1 Complex integration: In the domain of real variable there are two points of view from which integration may be considered. In the domain of complex variable use start with the definition of an integral as the limit of a sum and latter on establish the connection between differentiation and integration. 1.2 Basic Definitions: 1.2.1 Partitions: Let [a,b] be a closed interval where a,b are real number. Subdivide the interval [a,b] into n Sub- intervals

[t0,t1], [t1,t2], ______[tn,tn],

by inserting n-1 intermediate. points t1, t2, _ _ _ tn-1, Satisfying the inequality

a= t0 < t1 < t2 _ _ _ _ < tn = b

Then the set P={ t0, t1 _ _ _ _ tn } is called partition of the interval [a,b]. 1.2.2 Jordan Arc: Suppose a point Z lies on an ace L is defined by Z=Z(t)= x (t) +iy (t) Where t runs through the interval a ≤ t ≤ b and x (t), y(t) are continuous function of t Then L is called simple or Jordan are if Z(t1)= Z(t2) only when t1 = t2 1.2.3 Rectifiable Arcs: Suppose L denotes a Jordan arc defined by

Z=Z(t)= x (t) + iy (t), a ≤ t ≤ b

Zn-1 Z 1 Z 3 Z2

4 | P a g e

Let P={a=t0, t1 , t2 , _ _ _ _ _ tn = b } be any partition of [a,b] let Z0 ,Z1, …...Zn be the points of the are wr responding to their value. Then the length of this polygonal are is;

∑

∑ | |

If this sum tends to a unique value then it is called rectifiable.

1.2.4 Contour: A contour is a continuous chain of a finite number of regular arc (rectifiable arc) 1.2.5 Simply and Multiply Connected Regions: A region in the argand plane in which every closed curve can be shrunk to a point without passing out of the region is called a simply connected region otherwise it is multiply connected. 1.2.6 Analytic Functions: A function of Z which is one valued and differentiable at every point of a domain D. Save possibly for a finite number of exceptional points is called analytic in the domain D. Those exceptional points are called Singularities. 1.3 Cauchy-Goursat Theorem Statement: If a function f(Z) is analytic and single valued inside and on a simple closed contour C, then

∫ ( )

Proof: For the proof of the theorem use will use two lemmas which are as follows; Lemma1: if C is a closed contour, then the two integrals

∫ ∫

Lemma 2 : Let f(z) be analytic within and on a closed contour C. Then for any given >0, it is always possible to divide the region inside C into number of meshes either

5 | P a g e

complete square Cn or partial Squarer Dn Such that within each mesh there exists a point Z0 such that

| ( ) ( ) ( ) ( )| | | For every Z within or on the mesh. Now we divide the interior of C into complete squares C1, C2, _ _ _ _ Cm and partial Squares D1, D2, ______Dn. Now

∫ ( ) ∑ ∫ ( ) ∑ ∫ ( )

A D

B C

x 0

For the proof of the theorem. We will consider two cases. Case- I : Consider the mesh Cm Then

( ) ( ) ( ) ( ) ( ) ( ) ( ) Where | ( )| ̇

∫ ( ) ∫ ,( ) ( ) ( )-

∫ ( ) ( )

From lemma (1) we get

∫ ( ) ( )

as | ( )|

|∫ ( ) | √

Where denotes the length of diagonal of Cm 6 | P a g e

Case II – Consider the irregular sub regions Dn. let Sn denotes the length of the curved part of Dn Then

|∫ ( ) ( ) | √ ( )

Now (1) becomes

|∫ ( ) | ∑ |∫ ( ) | ∑ ∫ ( )

∑ |∫ ( ) ( ) |

|∫ ( ) | √ ∑( ) √ ∑

all the terms in the right hand side are constant

|∫ ( ) |

this is only true when

∫ ( )

1.4 Cauchy’s Integral Formula: Statement : let ( ) be analytic within and on a closed contour C and let any point within C . Then ;

( ) ( ) ∫

Proof : We define a circle by the equation | | = where and is the distance from to the neares point of C.

7 | P a g e

Consider a function ( ) ( )

( ) ( ) ∫ ∫

where C and are both traversed in the positive sense. As ( ) is continuous at so by defintion for given

| ( ) ( )| whenever | | ------(1) Now ( ) ∫

( ) ( ) ( ) ∫ ∫

------(2)

Put =

We have ( ) ∫

( ) ∫

( ) Hence (2) becomes

( ) ( ) ( ) ∫ ∫ ( )

( ) ( ) ( ) ∫ ( ) ∫

------(3) Since we may choose as small as wee please, we take . Thus the inequality (1) is satisfied for all points on .

8 | P a g e

Hence ( ) ( ) ( ) ( ) | ∫ | | ∫ )|

∫ | ( ) ( )| ∫

Thus ( ) ( ) | ∫ |

−------(4)

Then it follows from (3) and (4) that ( ) | ∫ ( )|

We now observe that is arbitrary and the left hand Side is independent of it. Thus implies that. ( ) ∫ ( )

( ) ∫ ( )

or ( ) ∫

1.5 Higher order derivatives: Statement: Let ( ) be analytic within and on the boundary C* of a simply connected region and let be any point within C. Then derivatives of all orders are analytic and given by

9 | P a g e

( )

( ) ∫ ( )

Proof: We first show that ( ) is analytic inside C. To prove this it is enough to show that ( ) has a differential coefficient at every point inside C. We have by Cauchy’s

formula for ( ) and ( ) ( ) ( )

( ) ∫ [ ] ( ) ( )

( ) ( ) ∫ ( ) ( ) ( )

follows that ( ) ( ) ( ) ∫ ( )

( ) ( ) ∫ ( ) [ ] ( ) ( ) ( )

( ) ∫ ( ) [ ] ( ) ( )

0 ( ) 1 ∫ ( ) ( ) ( )

Where is the circle | | lying entirely within C. Hence

( ) ( ) ( ) | ∫ | ( )

10 | P a g e

| | | | | | ∫ | ( )|| | ( ) ( )

| | | |

( ⁄ ) by means of arguments parallel to those used in the proof to Cauchy’s formula for ( )

| | . | |/

Where M is the upper bound of ( ) in D. Hence when | | the right hand side of (1) will tends to Zero. We then have

( ) ( ) ( ) ∫ ( ) or ( )

( ) ∫ ( ) Similarly ( ) ( ) ( ) ∫ ( ) ̇ ( ) ( )

( ) ( ) [∫ ∫ ] ( ) ( )

∫ 6 ( )

( )( ) 7 ( )

As we have ( ) ( ) ( ) ( ) ( ) ( ) ∫ ( )

11 | P a g e

or ( ) ( ) ( ) ( ) ∫ ( ) Replacing by ( )

( ) ∫ ( ) 1.6 Cauchy’s Inequality: Statement : let ( ) be analytic within and on a circle C ( ) defined by | | If | ( )| on C, then | ( )|

Proof: By Cauchy’s formula for the nth derivative of an analytic function at a point we have ( )

( ) ∫ ( ) ̇ | ( )| | | ( ) | ( )| ∫ | | ∫ | | | |

------(1)

Now we may write the equation of the circle | | as

So that

Thus | | Hence (1) will give

| ( )( )| ∫

Thus

| ( )( )|

Remark : If the function ( ) is analytic in every finite region of the Z plane then ( ) is called entire function. 1.7 Liouvilles’s Theorem: Statement: A bounded entire function is constant. Proof:- Let a be any arbitrary point of the Z plane. Now ( ) is by hypothesis analytic for | |

12 | P a g e

| ( )|

As we get ( ) But is an arbitrary point in the Z plane. Thus derivative of ( ) every where hence ( ) is constant. 1.8 Taylor’s Theorem: Statement: Let ( ) be analytic at all points within a circle C0 with center Z0 and radius R and Let Z be any point inside C. Then

( ) ( ) ( ) ∑ ( )

Proof: Let Z be any point inside the circle C0 with centre Z0 and radius R. Let | | and Let C be the Circle with centre Z0 and radius Such that then Z lies inside C. By Cauchy integral formula we have ( ) ( ) ∫

( ) ∫ ( ) ( )

( ) ∫ [ ] ( )

( ) ∫ 6 ( ) ( )

( ) 7

13 | P a g e

( ) ( ) ∫ ∫ ( ) ( )

( ) ( ) ∫ ( )( )

Now we will apply Cauchy’s formula and we get ( ) ( ) ( ) ( )

Where

( ) ( ) ∫ ( )( )

Now for the proof of the theorem it is sufficient to show that as

( ) ( ) | | | ∫ | ( )( )

| | | ( )| ∫ | | | | | || |

∫ | | ( )

( )

( ) ( ) ( ) ∫ ( )

14 | P a g e

Remark: when Z0=0 the series is called Maclaurin series Example: Prove that ( ) ( ) ( ) + -----+

Solution : Let ( ) Then by Tuylor’s theorem

( )( ) ( ) ∫( )

( )

( ) ( ) ( ) ----- (A)

Now replace Z by 1 in (a) ( )

( )

‘ ‘ ‘ ( ) ( ) ( )

( ) ( ) ( ) ( ) ) ( )

( ) ( ) ( )

1.9 Zero’s of Analytic functions: A Zero of an analytic function ( )is any value of Z for which ( )vanishes. 1.9.1 Singularities: These points of function ( ) at which the function ceases to be analytic is called. 1.9.2 Different types of singularities: (a) Isolated Singularity: A point a is said to be an isolated singularity of a function ( ) if ( ) is analytic at each point in some neighborhood of a excepting at the point a itself. 15 | P a g e

(b) Isolated essential Singularity: If there exists no finite value of in such that ( ) ( ) finite non zero constant then is called an isolated essential singularity Example: has isolated Singularity at

1.10 Meromorphic Function: To define meromorphic function firstly we will understand the concept of pole. If there exists a positive integer in such that ( ) ( )

Then is called the pole of order of ( ). Meromorphic function are those functions which has poles as its only singularities in the finite part of the plane. 1.11 Argument Principle : Statement: If ( ) is analytic within and on a closed contour C, having N zeroes inside C but no zeroes on C, then

( )

where ( ) denotes the variation in ( ) as z moves round C. Proof; Let ( ) is the analytic function with N zeroes inside C and denotes numbers of poles inside ( ) ∫ ( )

Put ( ) so that | ( )| and ( ). Then

( ) ( ) ( ) ( )

∫

Now

∫ , -

Retunes to its original value if z moves once round C. Also,

16 | P a g e

∫ , -

( ), as ( ) does not return to its original value us z moves round C and so ( ) is not necessary zero.

( )

As ( ) is analytic i.e. it has no poles inside C hence P=0

( )

1.12 Rouche’s Theorem: Statement: Let ( ) and ( ) be analytic inside and on a simple closed curve C and Let | ( )| | ( )| On C, then ( ) and ( ) ( ) have the same number of zeroes inside C. Proof: As | ( )| | ( )| on C and as | ( )| cannot to negative and hence we get ( )

( ) ( )

| ( )| | ( )|

But it is a contradiction to the fact that | ( )| | ( )| does not have a zero on C. Let N and M be the number of zeroes of ( ) and ( ) ( ) respectively inside C.

Then by argument principle ( )

( ) ( ) ( ) ( ) { } ( ) Now to prove the theorem it is Sufficient to show that

17 | P a g e

( ) 8 9 ( ) Since | ( )| | ( )| C, the transformation ( ) + givens point in the plane interior to the circle ( ) with center 1 and radius unity .Then We can write

( ) 0 1 must lie between –II/2 and II. This shows ( ) ( ) that 0 1 must return to it ‘s original value as Z ( ) describes C. ( ) Since 0 1 cannot increases or decrease by a ( ) multiple of 2 , we conclude that ( ) 6 7 ( ) Thus we have 2 => N=M 1.13 Fundamental Theorem of algebra: Statement: Every polynomial of degree n has roots. Proof: Consider the Polynomial in Z

( ) Where

Clearly has roots and they are all at origin if

Let ( )

( ) Now we define C as the circle with center at the origin and radius R>i then.

| ( )| | | | |

| | on C and similarly

| ( )| (| | | | | |)

| ( )| | ( )| on C if

| | (| | | | | |)

(| | | | | |) 〉 | |

18 | P a g e

Then by Rouche theorem we get that ( ) has same number of Zeroes as ( )

( ) has roots 1.14 Maximum Modulus Principle: Statement: let ( ) be analytic within and on a simple closed contour C. Then | ( )| reaches it’s maximum values on C and not inside C, unless ( ) is a constant. Proof: let us assume that ( ) is analytic within and on a simple closed contour C ( ) is continuous within and on C ( ) has a maximum value M Within and on C. Now for the proof of the theorem it is sufficient to show that | ( )| has a maximum value M on the boundary of C not inside C. On the contrary let ( ) has a maximum value at Z=a inside C. | ( )| | ( )| ------(1)

| ( )| ------(2) Now we describe a circle within C with Center at A. ( ) is Z=b inside S’-1 | ( )| | ( )| 〉 〉 || ( )| | ( )|| | | ( ) as | ( )| | ( )| || ( )| | ( )|| ( ) becoems. || ( )| | ( )|| || ( )| | ( )||

=M-

=M- | ( )| M- inside a circle with center b and radius Now draw another circle with center at a such that it passes through b. The are of the circle lies within s.t.

19 | P a g e

On we have z-a= ( ) ( ) ∫

∫ ( )

∫ ( ) ∫ ( )

| ( )| ∫ ( ) ∫ ( )

∫ ( ⁄ ) ∫

( ⁄ ) ( )

But

| ( )|

this is a contradiction to our assumption maximum value lies inside C. 1.15 Schwarz lemma Statement: If ( ) is analytic in a domain | | and satisfies the conditions | | ( ) Then | ( )| | | and | ( )| Equality holds only if ( ) is a linear transformation

20 | P a g e

( ) Where is a real constant. Proof: Let us consider that ( ) is analytic in the unit disc | | and hence by taylor’s expansion of ( ) about the origin gives.

( ) Again as ( )

( ) + We define ( ) ( )

( ) ------(1) This shows that ( ) defined in (1) has a Singularity at Z=0 which may be removed by defining

( ) ( ) Let z=a be an arbitrary point of the unit disc. Let us choose such that | | As | ( )| we get on the circle | | the inequality given by | ( )| | ( )| ( ) | | But by the maximum modulus principle the inequality (2) also holds in the disc | | and hence | ( )| | ( )| | | Letting we get ( ) | ( )| | |

| ( )| | |

or | ( )| | ( )|

as a is arbitrary we get

| ( )| | | | |

Then by maximum modulus principle we conclude that | ( )| will hold only when ( ) | ( )|

21 | P a g e

1.16 Summary: In this block we have understood the concept of complex integration and we have proved a | | of theorems based on it. Some important results are as follows; 1. Let ( ) be a analytic function on and single valued inside and on a simple closed contour C. then

∫ ( )

2. If C is closed contour then

∫ ∫

3. If ( ) be analytic within and on a closed contour C and let Z0 be any point within C then ( ) ( ) ∫

4. Every polynomial of degree has roots.

1.17 Self Assessment Test: (a) If ( ) be analytic within and on the boundary C of a simply connected region and let Z0 be any point within c. Then prove that

( ) ( ) ∫ ( ) (b) Expand and Sin Z in faylor’s series about z=0 (c) Let ( ) be analytic in the ring shaped region between two concentric circles C and C’ with center Z0 and radii R and R’ and Let z be any point of D. Then

( ) ∑ ( ) ∑ ( )

Where ( ) ∫ ( )

∫ ( ) ( )

(Laurent’s Theorem)

22 | P a g e

(d) Prove that if ( ) is a continuous in a simply connected domain and Let for every closed contour C in the domain D.

∫ ( )

(Morero’s Theorem)

(e) Let a function be analytic at a point z=z0 where

( ) and Let ( ) Then there exists a of the point in the plane in which the function ( ) has a Unique inverse ( ) and

( ) ( ) (Inverse Function theorem)

1.17 Further Readings:

(i) Complex Analysis by Dr. H.K. pathak.

(ii) Function of complex variable by Dr. J.C. Chatuvedi and Prof. S.J. Seth.

(iii) Complex Analysis by L.V.Anlfor.

(iv) Real and Complex Analysis by Walter Rudin.

(v) Foundations of complex analysis by S. Ponnusumy.

23 | P a g e

Unit-2 Calculus of Residue Objectives: After reading this unit you should able to;  Understand the concept of residue .  Solve typical problem of integration in a simpler way. STRUCTURE : 2.1 Concept of Residue 2.1.1 Residue at pole. 2.1.2 Residue at infinity. 2.2 Cauchy Residue theorem. 2.3 Calculations of residue at special cases. 2.3.1 Expression for the residue of ( ) at simple pole z=a 2.3.2 Expression for the residue of ( ) at pole of order 2.4 Problems based on Cauchy Residue theorem. 2.5 Jordan lemma 2.6 Problems based on Jordan lemma 2.7 Problems having branches of many function. 2.8 Summary 2.9 Self Assessment Test 2.10 Further Readings

24 | P a g e

1.1 Concept of Residue: Let us consider a single valued analytic function ( ) then in the neighborhood of an isolated singularity ( ) can be expanded with the help of Laurent’s series given by

( ) ∑ ( ) ∑ ( )

Then the coefficient of B, is called the residue at z=a and

∫ ( )

Where is a circle with center a 2.1.1 Residue of pole: The above stated definition is also known as residue at pole. 2.1.2 Residue at infinity: let ( ) has an isolated singularity at infinity and C is a closed curve such that ( ) is analytic except at singularity, then clearly C will be very

large and residue will become ∫ ( ) where

integration is taken around C in anti-clockwise direction. 1.2 Cauchy Residue Theorem: It is the most important theorem which is frequently used in solving the problem of Integration by using the concept of residue. Statement: Let ( ) be single valued and analytic within and on a closed contour C except at a finite number of poles and Let be respectively the residue of ( ) at these poles then

∫ ( ) ( )

Proof: Let be the circles with centers at respectively and radii so small that they lie entirely within the closed curve C and do not overlap. Then ( ) is analytic within the region enclosed by the curve C and these circles. Hence by Cauchy’s theorem for multi-connected region we have

25 | P a g e

∫ ( ) ∫ ( ) ∫ ( ) ∫ ( )

------(1)

But ∫ ( )

∫ ( )

Similarly

∫ ( )

‘ ‘ ‘ ‘ ‘

∫ ( )

Therefore (1) becomes

∫ ( )

( )

2.3 Calculation of Residue: In general we require residue two different situations. 2.3.1 Residue at Simple pole : ( ) ( )

Example :- Find Residue of at

Solution- we have

( )

( )( )

26 | P a g e

Clearly is a simple pole therefore residue at is

( ) ( )

( ) ( )( )

( )

2.3.2 Residue at pole of order : If ( ) is an analytic function having as a pole of order then ( ) ( ) ( ) then Residue is given by ( )

( ) 2.4 There are many types of problems which can be solved using the concept of Cauchy residue theorem. They are as follows. 2.4.1 Integration Round the Unit Circle. Example: Show that

∫

∫ ∫

| |

. / ∫

. /

∫ ( )

27 | P a g e

∫ ∫ ( )

( ) . /

Clearly ( ) has there pole, out of

which only lie within C

Therefore Sum of residues at o and will be

( ) ( ) [ ]

( ) ( ) . / . /

, -

Hence

∫ ( ) ∫

Try these questions

1) ∫

) ∫

2.4.2 Integration of the type ( ) ∫ Example: Apply the calculus of residues to prove

∫

Solution : Let us consider

∫ ∫ ( )

Where C is the contour consisting of a large semi circle CR by radius R. Then by Cauchy residue theorem

28 | P a g e

∫ ( ) ∫ ∫ ( )

( ) | |

( ) | | Hence we get

∫ ( ) ∫ ( )

Clearly are the simple poles of ( ) out of which lies inside C.

Therefore residue will be

( )

∫ ( ) ( )

∫

Try these questions

) ∫ ( )

29 | P a g e

2.5 Jordan lemma; This lemma is generally used in solving the integrals of the form

( ) ( ) ∫ ∫ ( ) ( )

Where (i) ( ) and ( ) are polynomials (ii) degree of ( ) exceeds that of ( ) (iii) the equation ( ) has no real roots. Statement: If (i) ( ) | | , uniformly for and (ii) ( ) is meromorphic in the upper half plane, then

∫ ( )

Where CR denotes Semi-circle | | ( ) 2.5.1 Problems based on Jordan’s lemma. Example: Apply the calculus of residue to prove that

∫

Solution:

∫ ∫ ( )

Where C is a semicircle of very large radius R. Above real . Then by Cauchy residue theorem.

∫ ( ) ∫ ( ) ∫ ( ) ( )

As R by Jordan’s lemma second integral will vanish and we get

∫ ( ) ( )

------(i) as

( )

So, residue at ( ) ( )

30 | P a g e

( )

Hence (1) becomes

∫ ( ) ( )

Example: Prove that ∫ using as a contour a ( ) large semicircle in the upper half plane indented at the origin. Solution:

∫ ∫ ( ) ( )

Where C is the closed contour consisting of the upper half of the large semi-circle by | | and real from –R to R indented at by a small circle of radius . Now poles of ( ) arc given by ( ) will not lie in C. also is the double pole Therefore Residue will be ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( )

. /

Now by Cauchy residue theorem

31 | P a g e

∫ ( ) ∫ ( )

∫ ( )

∫ ( ) ∫ ( ) ( )

4 . /5

( )

Also ( ) | |

4 5 | | ( ) Therefore

∫ ( )

Also

( ) ( )

( )

( ) ( )

32 | P a g e

Also

∫ ( )

Taking ( )

∫ ( ) ∫ ( ) ( )

( ) ∫ ∫ ( ) ( ) ( )

∫ ∫ ( ) ( ) ( )

Equating real parts

∫ ( )

Equating real parts

∫

2.6 Branches of Many- valued functions. If this section we will consider the integrals involving many-valued function. Such as where a is not an integer. In such cases we consider only those contours whose interiors do not contain any branch points and particulars branches should also be specified. In orders to avoid branch point at origin, we can take double circles contour indented at the center. Example:1: Prove that

∫ (

Solution: Consider the integral

∫ ∫ ( )

33 | P a g e

Contour consisting of a large Semi-circle defined by | | in the upper half plane and the real from –R to R indented at the origin by a small semi-circle of radius P which avoids the branch point o of the only simple pole of ( ) within C is at Residue at ( ) ( )

. /

Now we apply Cauchy residue theorem,

∫ ( ) ( )

. /

∫ ( ) ∫ ( ) ∫ ( ) ∫ ( ) .

------(1) Clearly

|∫ ( ) | ∫ | | | |

∫

As R

34 | P a g e

|∫ ( ) |

Also ( )

∫ ( )

Similarly when ( ) Becomes

∫ ( ) ∫ ( )( ) . /

Equating real parts we get

∫ ( )

∫

2.7 Try these question:

( ) ( ) ∫

( ) ∫

2.8 Summary: In this section we have studied about Poles. They arc of two types. We have studied about the application of Cauchy

35 | P a g e

residue theorem in solving the typical problems of complex integration. 2.9 Self assessment test: 1. Prove

∫

2. Prove that

∫ ( ) √

∫

4. If a>b>0, that prove that

∫ 4 5 ( )( )

2.10 Further Reading:

1. Complex Analysis by H.K. Pathak. 2. Complex Analysis from Shrivatava. 3. Functions of a complex variable by B.S. Tyagi.

36 | P a g e

FINANCIAL INFORMATION ANALYSIS FOR INTERNAL DECISION MAKERS ACCOUNTS FOR DECISION MAKING:

3.0 OBJECTIVES After studying this unit you will be able to: Understand the concept of marginal costing Develop an insight about different alternatives available for decision making Choose the best alternative among the alternatives available Give an overview on activity based costing Explain the concept of break even analysis Describe the concept of cost-volume profit for profit planning Explain the concept of margin of safety and profit volume ratio in decision making Understand and analyze the cause of variance between planned and actual results. 3.1 INTRODUCTION The element of costs can be divided into fixed and variable costs. There are certain costs which are a combination of fixed and variable costs. These costs are called semi variable costs. It is necessary to segregate the mixed costs into fixed and variable costs for managerial decisions. The analysis of cost behaviour is necessary for planning, control and decision making. It means analysis of variability of each cost element in relation to the level of output. The analysis of costs plays a vital role in selecting the alternatives available before the management. Costs could shape alternative opportunities and therefore, it influences and shapes future profits. Management is not only interested in the historical cost analysis but it is also interested to study those costs, which are influencing the future operations. After the standard costs have been set, the next step is to ascertain the actual cost of each element and compare them with the standard already set. The difference of actual from standard is variance. In this unit you will learn about the concept of break-even point, cost- volume- profit analysis, margin of safety, profit volume ratio and their role in decision making. You will also learn about the importance of variance for effective cost control and decision making. 3.2 MARGINAL COSTING AS A TOOL FOR DECISION MAKING

An important role of cost accounting is to assist in the process of managerial decisions. In this context profitability of two or more alternative option is compared and such option is selected which offers maximum profitability along with fulfillment of objectives of the enterprises. Though there are number of accounting techniques for such comparison and decision making, marginal costing has their significant role. Marginal costing plays a vital role in decision making

37 | P a g e

.It is a very useful technique in solving various managerial problems and contribution in various areas of decisions. Few important ones are explained here:  Make or Buy Decision  Lease or Buy  Change in Product Mix  Pricing Decision  Shut –down Decisions.  Exploring New Market

3.2.1 Make or Buy Decision: In this, management has to decide whether a certain product or a component should be made in the factory itself or bought from outside suppliers. It follows two types of decision:  Stopping the production of the part and buying it from the market  Stopping the purchase of the component and to produce it in own factory. Stopping the production of the part and buying it from the market : A business concern is already making a part or a component which is used in the business. Now due to some reason, a decision has to be taken whether to bring the part from the market or should be made in the factory. In case of a decision like the stopping the production of the part or component and buying it from the market, a comparison of marginal cost of such production with that of buying price should be made. But in this fixed cost remain fixed. Marginal Cost < Buying Price ------Making of a part inside the business is a relevant decision. Marginal Cost> Buying Cost------Purchasing of a part from the market is a relevant decision. However opportunity cost may also be taken into consideration while taking such decisions. Example: Suppose a component is being manufactured with the help of a machine and 10,000 units at a cost of Rs 10 per unit.( Rs. 9 variable cost ,Rs1 fixed cost).are produced. The same component can be bought from the market @ Rs.9.50 per unit. What will be beneficial to go either for producing the component or buying a component? What if the machine is released from production and can be hired at an annual rent of Rs 6,000? If the item to be made: Rs. Marginal Cost of 10,000 units @ Rs.9 90,000 Buying price of 10,000 units @ Rs. 9.50 95,000

38 | P a g e

Savings if „Made‟ 5,000 If the item is bought and machine is let out on hire: Buying Price 10,000 units @ Rs. 9.50 95,000 Less: Hire of Machine 6,000 89,000 Marginal Cost of Rs. 10,000 @ Rs. 9 90,000 Saving if „bought‟ and „machine is hired‟ 1,000 If we take decision without considering opportunity cost then we can say that it is beneficial for the company to produce the component. But with the consideration of opportunity cost decision will totally change then at this it is beneficial for the company to buy the component from the market. Stopping the purchase of a component and to produce it in own factory: Another point to be considered while making make or buy decision may be that a component thus far being purchased from the market should be purchased or made in the factory or not. In this case generally some extra arrangements regarding space, labour , machines, raw material ,rent etc will be required which involves huge investment. Some extra space, staff requirement is also required which leads to additional cost. Prices paid to the outsiders for purchasing the component should be compared with the additional cost which will be incurred in the form of raw material, rent, salaries etc. Rule Decision Additional Cost < Buying Price ----- Manufacture the component Additional Cost> Buying Price------Purchase the component No compromise in respect of quality .Reliability of regular supply, reputation, financial position of supplier should be given due consideration. There are large fluctuation in demands it is better to purchase from outside .But if there is possibility of increasing demand in future then own production may be preferred because it will lead to low cost. Example ABC Ltd. Purchases 20,000 Pens per annum from an outside supplier at Rs. 5 each. The management feels that these be manufactured and not purchased. The following relevant information are available: Material Cost per unit Rs.2 Labour Cost per unit Re.1 Fixed Cost per Pen Re.1 Variable Overheads per Pen 100% of Labour Cost 39 | P a g e

You are required to advise whether: The company should continue to purchase the pens from the outside supplier or should make them in the factory, and whether the company should accept an order to supply 10,000 pens to the market at the selling price of Rs. 5 per unit. Solution: Calculation of Manufacturing Cost Particulars Per Pen 20,000 Pen Rs. Rs. Variable Cost: Material 2 40,000 Labour 1 20,000 Variable Overheads 1 20,000 4 80,000 Fixed Cost 1 20,000 5 1,00,000

Decisions: the cost of manufacturing as well as of purchasing per pen is equal. If there is assurance of regular supply of quality product from the suppliers, the company can continue to purchase the pens. If there are some problems in this context, the company can decide to make them in the factory. (II) Yes, the company should accept the order because the marginal cost is Rs. 4 only, where the order will be at Rs. 5 per pen. 3.2.2 Lease or Buy The fast changing technology and innovations require replacement of old machineries so as to modernize the plant and make it up-to-date for the production which is in consonance with the customers‟ tastes and requirements. Moreover, cost reduction is always the demand of the industry, particularly in the wake of all time rising prices going to the extent of run-way inflation. Leasing as an alternative to purchasing the plant, machinery and other equipments, besides land and building is considered by managements these days. The factors which influence the decision as to own or lease are: Cost of alternatives, Long-term stability of earnings, Financing, and Return on Investment Flexibility Example:

40 | P a g e

In view of increasing cost of operating own fleet of cars, your company is presently considering two proposals, viz: To hire cars with drivers from an agency @ Rs. 800 per car per month. The company will bear the cost of petrol, oil and tyres.The executives will be given Rs. 25,000 interest –free loan repayable (in 5 years) to buy their own car. The company will, however, provide them with free petrol and Rs. 500 per month for maintenance and drivers wages. If the present cost of a car is Rs. 50,000 and monthly average running is 2,000 kilometers, find out the most economic way with the help of the following data: Paisa per km. Petrol 65 Oil 8 Tyre 7 Repairs 10 Tax and Insurance Rs.560 per year Drivers‟ wages and bonus Rs. 720 per month Life of a car 5 years Resale value at the end of the fifth year @18% per annum Rs. 10,000 Solution:

Comparative Cost of operation Cost per car per month Company's own car Hired car Executive car Rs. Rs. Rs. Petrol 1,300 1,300 1,300 Oil 160 160 …… Tyre 140 140 …. Repairs 200 Tax and insurance 46.67 Wages and Bonus 720 Depreciation 666.67 Hire Charges 800 Maintenance Allowance 500 Interest on Loan 375

Total Cost 3,233.34 2,400 2,175 Cost per km. 1.62 1.2 1.09

* The interest of Rs. 375 relates only to the first year of operation. When annual instalments of loan are paid back by the executives,the interest in the subsequent year of operations will gradually decrease and thus the cost per km. of operations of Executive cars will be reduced.

41 | P a g e

The cost of opeartion of the Executive cars,being the lowest ,this alternative is recommended.

3.2.3 Change in Product Mix: Introducing a New Product Line or Department: It involves two decisions:  Whether a new product or line should be added to the existing production or not.  If it should be introduced, then what should be the model or design or shape of new product? For this, the marginal cost of new product in all its possible model should be considered. And also some additional investment of plant and machinery will likely increase the fixed overheads, which should also be considered along with marginal costs. Selecting Optimum Product Mix: When a company is engaged in a number of lines then the problem arise of selecting the most optimum product mix which would maximize the earnings. This problem becomes complicated, when one of the factors happens to be the limiting or key factors. Under such a situation, profitability will be improved only by economizing the scarce resources .Thus, while deciding a profitable mix of products, Contribution per unit of Key Factor should be considered. he product giving highest contribution per unit of key factor should be considered as most desirable product and in this way all products may be assigned ranks ion order of priority. Selection of products in this way will offer an optimum product-mix at which the profit will be maximum. Principles for taking a decision in respect of product-mix are:- (a) Calculate contribution per unit of key factor (b) Assign ranks on the basis of highest contribution per unit of key factor. (c) Available key factor should be utilized in the manufacture of that product which has been assigned first rank then in the production of product having second rank and so on.

Example: The following data are available in respect of Product Soap by ABC Co. Ltd: Rs. Sales 50,000 Direct Materials 20,000 Direct Labour 10,000 Variable Overheads 5,000

42 | P a g e

Fixed Overheads 10,000 The Company now proposes to introduce new product Shampoo so that sales may be increased by Rs.10, 000.There will be no rise in fixed costs and the estimated variable costs of product Shampoo are: Rs. Materials 4,800 Labour 2,200 Overheads 1,400 Advise whether product Shampoo will be profitable or not. Solution: Statement of Profit if Shampoo is introduced

Product Product Particulars Soap Shampoo Total Rs. Rs. Rs. Direct Materials 20,000 4,800 24,800 Direct Labour 10,000 2,200 12,200 Variable Overheads 5,000 1,400 6,400 Marginal Costs 35,000 8,400 43,400 Sales 50,000 10,000 60,000 Contribution 15,000 1,600 16,600 Less: Total Fixed Costs 10,000 Net Profit 6,600 Advise: By introducing Shampoo profit has been enhanced by Rs.6,600 . So, it is profitable to introduce this product.

Illustration: A company produces two products A and B. The following facts are given regarding them: A B Profit contribution per unit Rs 2 Rs.3 Required production hours per unit 1 2 Sales Potential in units 1,500 1,800 Available Production Hours 2,000 Determine the optimum product mix.

Solution:

Particulars A B Contribution per unit Rs.2 Rs.3 Production hours per unit 1 2

43 | P a g e

Contribution per hour Rs.2 Rs.1.50 Because of the limited production hours, product A will be preferred over product B as product A has greater contribution per hour. The sales potential of A is 1,500 units which will require 1,500 hours. Total available hours are 2,000. Hence, the balance hours can be used in the production of product B. Product of B will be 250 units @ 2 hours per unit. Thus, optimum Product Mix=A:B, 1,500units: 250 units. 3.2.4 Pricing Decisions: In long Run prices should be such as to cover total cost which includes marginal cost and fixed cost as well as desired profit. And secondly, in competitive markets prices are determined by market forces. Thus in both the cases Marginal Costing will not play any significant role. Marginal Cost is helpful in price determination only in short run and monopoly conditions. Various Aspects to be considered under Pricing Decision are as follows:  Normal Price  Minimum Price  Special Price including dumping  Price changes Normal Price: The normal price should be such as to maximize the contribution, so that maximum profit is assured. Sale Price = Marginal Cost + Contribution (Fixed Cost + Profit) Minimum Price Sale Price = Variable Price + Fixed Cost This price is used:  When there is tough competition.  Price –cut is on war.  When new product is to be introduced in the market for the first time. In these situations normal price may not be useful .In these situation manufacture faces a problem of what should be the minimum price. It is obvious that if a manufacturer is not in a position to earn profit, he will also not incur any loss. As , such ,he would fix a price ,which must cover total cost at least. Such price is known as Minimum Price. Special Price: Sometimes the big concern may face a problem of accepting or rejecting a special offer at a price which is below at existing price. Similarly, sometimes there may be a possibility of capturing a

44 | P a g e new market, attracting new customers, patronizing special customers, if goods are sold at a price below cost. Normally, such offer should be rejected, because total cost will not be recovered. For this rule should be followed: Price> Marginal Cost------Accept the proposal Price< Marginal Cost -----Reject the proposal. Price Change: This is needed in order to captalise the market situation. It involves hike in price as well as reduction in price. In both cases, quantity solids affected depending on the degree of elasticity of production and in turn affect profit position. A careful analysis is required through the technique of Marginal Cost. If change is necessary then that course is adopted which leads to maximum profit or present level profit. If price –reduction is forced by government ,in such cases adverse affect have to be neutralize by making suitable adjustments like quantity sold, fixed cost, variable cost, production method and techniques etc. Example: The annual operating capacity of a manufacturing company is 10 lakh direct labour hours and it is currently operating at 75% of its capacity. The company has recently been approached by a distribution to buy 1,00,000 units of its product at a special price of Rs. 10.50 per unit .The standard Cost Sheet is as under: Rs. Material 6 kg @ Re. 0.40 per kg 2.40 Direct Labour 2 hours @ Rs. 2.50 p.h. 5.00 Overheads: Variable 2 hours @ Re. 0.75 1.50 Fixed 2 hours @ Rs. 1.20 2.40 11.30

The management of the company is confronted with the following questions:  Would it be profitable to accept the offer?  What would be the minimum price for the offer, if company‟s target is Rs.2,00,000 as profit on this offer?

Solution:

45 | P a g e

 Total production capacity is (10,00,000/2) 5,00,000 units and present production is 3,75,000 units( 5,00,000*75/100)  As present selling price is unknown, decision will be taken on the basis of contribution of special order. Statement of contribution on Special Order

Rs. Sales: 1,00,000units @ rs. 10.50 10,50,000 Less: Marginal Costs: Materia;ls @ Rs.2.40 p.u. 2,40,000 Direct Labour @ Rs.5.00 p.u. 5,00,000 Variable Overhead @ Rs.1.50 p.u. 1.,50,000 8,90,000 Contribution 1,60,000

Hence, the acceptance of order is profitable. If company‟s target is Rs. 2,00,000 as profit, the minimum price for the offer will be: Marginal costs 8,90,000 Desired Profit 2,00,000 Sales desired 10,90,000 Minimum Selling Price =10,90,000/1,00,000 = Rs. 10.90

3.2.5 Shut –Down Decisions: Shut down decisions may be of two types: Permanent Closure of entire business, and Temporary Closure of entire business Permanent Closing Down or Continue: If the project undertaken does not yield the minimum rate of return expected by the investor it will have to be given up. The decision to continue or otherwise shut –down will depend upon the cost – benefit analysis of the two alternatives. Costs items to be considered are:  Setting up Cost;  Effect of Fixed overhead costs;  Packing and storing of plant and equipment costs;  Loss of goodwill and or market;  Retrenchment or lay off compensation to workers; Items of benefits to be quantified are:  Avoiding operating losses;

46 | P a g e

 Saving in Fixed costs;  Saving in repairs and maintenance costs;  Savings in indirect labour costs;  Savings in heat and light costs;  Saving in other indirect costs. Temporary Closure of the business: When trading activity particularly plant operation is suspended for a short period; it is known as temporary closure. Such closure is necessitated either due to depression/recession or due to ensuring off-season. In the former case, the period of closure will run over the period of recession/depression, while in the latter case, it will cover the period of off-season. For the period of recession/depression, the trading activity should not be suspended for a short period till there is contribution on that level of trading activity. But there is need for re-examination and deep analysis of fixed costs. There may be few items of fixed costs which can be eliminated or saved by suspending the trading activities. Such fixed costs are known as „escapable‟ or „avoidable‟ fixed costs. But there are certain items of fixed costs, which cannot be avoided; there will be incurred after the suspension of activities. These are known as „inescapable „or „unavoidable‟ fixed costs. Again, additional expenses would have to be incurred in setting up the plant again after shut-down period. Such fixed costs are known as‟ Special costs‟. It is significant point to note that inescapable fixed costs are all the time „loss factor‟. Thus, escapable fixed costs minus special costs, known as net escapable fixed costs is a relevant factor and amount of contribution should be compared with net escapable fixed costs. This is done by preparing a statement of marginal costs and contribution for zero level of production and for possible demanded quantity of production. As regards the level of production to close down the plant or point of shut –down in terms of quantity, the following formulas may be used: Shut –down point = Net escapable fixed costs Contribution per unit It is also significant to note that a decision regarding temporary closure should not only be based on cost data; some other economic and social factors may also be considered. Example: Mr. Singhania has a sum of Rs.3,00,000 which is invested in a business. He wishes 15% return on his fund. It is revealed from the present cost data analysis that variable costs of operation are 60% of sales and fixed costs are Rs. 1,50,000 p.a. On the basis of this information, you are required to find out: a) Sales Volume to earn 15% return.

47 | P a g e b) Shut- down point of the business, if he would spend RS. 50,000 even if business has to be closed. Solution: (a) Return= 3,00,000 *15/100 = Rs. 45,000 Fixed Cost = Rs. 1,50,000 Total Contribution = Rs. 1,95,000 Contribution percentage or P/V ratio = 100% -60%= 40 % Required Sales Volume = 1,95,000 *100 40 = Rs. 4,87,500 (b) Net Escapable Fixed Cost = 1,50,000-50,000= Rs. 1,00,000 Shut –Down Point= Net Escapable Fixed Cost P/V Ratio= = 1,00,000 * 100 40 = Rs. 2,50,000

3.2.6 EXPLORING NEW MARKETS

Whether Indian or foreign, but if new markets is to be explored, the decision rests on a consideration of the incremental gain resulting there from. However, certain other factors are also to be considered:  Availability of surplus capacity, If the surplus capacity is not available, the question of creating additional capacity will have to be separately studied, since it will involve additional investment.  Maintenance of present sales at current prices, The already existing market should not be affected at all by the tapping of the fresh market. If in the new market, the selling price is lower than that prevailing in existing market, it should be fully ensured that the new reduced selling price outside does not influence the current higher prices in the present market.

Example: A manufacturer has planned his level of production at 50% of his total plant capacity of 30,000 units. At 50% of the capacity his expenses are as follows: a. Direct labour 11,160 b. Direct material 8,280 c. Variable and other manufacturing expenses 3,960 d. Total fixed expenses regardless of production 6,000 The home selling price Rs. 2.00 per unit. Now, the manufacturers receive a trade enquiry from overseas for 6,000 units at a price of Rs. 1.45 per unit. If you were the manufacturer,

48 | P a g e would you accept or reject the offer? Support your statement with suitable cost and profit details. Solution: To decide whether it would or would not benefit financially by accepting the offer, the incremental cost is to be compared with the incremental revenue. The reason is, the fixed has “sunk” and there is idle capacity to the extent of 50 per cent capacity. As such, only incremental cost has to be considered and if the incremental revenue exceeds the incremental cost, the offer is worth accepting. The selling price in overseas market is not going to affect home selling price. For calculating incremental cost, we are concerned with variable cost per unit. Assuming linearity, the variable cost as per question is: Variable Cost per unit = 23,400 15,000 The above calculation is based on the fact that, at 50% capacity utilization, production is 15,000 units and as per data given in the question, total variable cost would be equal to Rs. 23,400. Incremental cost per unit = Rs. 1.56 Incremental revenue per unit = Rs. 1.45 Incremental loss per unit = Re. 0.11 Therefore, the offer should not be accepted

3.3 STANDARD COSTING & VARIANCE ANALYSIS

Standard costing is a very important technique of cost control. Every organization wants to minimize the cost of production and maximize the profits. Standard costing is such a system which seeks to control the cost of each unit and then it is compared with actual cost. The difference between the actual and pre-determined costs is known as variance which need to be analyzed through carefully planned accounting procedures and then the results are promptly reported to managers. The term „Standard‟ refers to „a specific measurement‟ or „pre-determined scale or measurement‟ Webster‟s Dictionary has given several meanings of the word „Standard‟ but we can prefer to understand it as “ something established as a rule or basis of comparison in measuring or judging quantity, quality, value, etc.” This something is not vague but “established as a rule or basis of comparison.” Thus, in the context of management accounting „standard‟ may be defined as measurable quantity of material, labor and other elements of cost required in the production of pre-determined quality, level or technical characteristics. According to I.C.M.A., LONDON,” Standard cost is defined as a pre-determined cost, which is calculated from management‟s standards of efficient operation and the relevant necessary expenditure. It may be used as a basis for price fixing and for cost control through variance analysis.”

Characteristics or Salient Features of Standard Costing The salient features of characteristics of standard costing may be summarized as follows:-  Determination of Standards:- First of all, standards and standard cost of various elements of cost are ascertained separately.  Computation of Actual:-After the completion of production actual cost incurred in computed.

49 | P a g e

 Comparison of Standard and Actual Costs:-The most important feature of standard costing in the comparison of actual cost incurred with standard cost specified.  Computation of Variances:-The most important aspect of comparison and analysis of actual cost and standard cost is to find out variances. These variances are calculated separately for element of cost.  Ascertainment of Reasons of Variances:-If there is variance between standard cost and actual cost, then causes of these variances are identified. These causes may be controllable as well as non-controllable.  Study of Options:-After identifying various causes of variances. Different options to overcome these variances are studied and points of incidence of these variances are determined.  Presentation of Report:- After analysis of variances and various options to deal with a report is prepared for management so that necessary corrective actions may be taken and variances may be minimized.

Application of Standard Costing The application of standard costing required the following conditions to be fulfilled:-  A sufficient volume of standard products or components should be produced.  Methods, procedures and materials should be capable of being standardized.  A sufficient number of costs should be capable of being controlled.

OBJECTIVES OF STANDARD COSTING The following are the main objectives of standard costing:-  Increase in Efficiency and Productivity:-The first objective of standard costing is to improve the quality and minimize the cost so as to face competition effectively. In fact, standard costing is a tool of management control with the help of which efficiency and productivity can be improved and these objectives can be achieved.  Cost Control:-The purpose of determining standard cost and then to compare it with actual cost is to make effective.  Determination of Responsibility:- One important objective of standard costing is to identify the persons or centers responsibility for variances, so that they may be controlled properly.  Supplement to Budgetary Control:- Standard costing is also adopted to make budgetary control a success. In fact, management control becomes more effective if budgetary control and standard costing are introduced simultaneously.  Information to the Management:-To provide important information to management is also an objectives of standard costing. This costing provides all such information due to which the production work could not be completed as per pre- determined plan and standards so that necessary corrective action may be taken at appropriate time.

50 | P a g e

 Progressiveness of Management:- One objective of standard costing is to develop the feeling of looking forward among management personnel and to make management dynamic and progressive continuously.

ADVANTAGES OF STANDARD COSTING The various advantages of standard costing are as follows:-  Simplification of Cost Bookkeeping: - It is very simple in comparison to historical costing. Once the standards are fixed for the product, the records can be simplified through uniformity which saves the time and money.  Basis for Measuring Operating Performances:-Standards work as yardsticks for measuring the operating efficiency or inefficiency. For it, the comparisons are made between standard cost and actual cost.  Cost Reduction and Control:-Standard costing is very useful in cost reduction and control by eliminating or limiting lost time, spoilage, material, wastage and lost machine hours.  Helpful In Budgeting:-Standards costing is linked with budgetary control. It helps in making budgets and planning though budgeting.  Management by Exception:-Standard costing is helpful in applying the principle of management by exception. Variance analysis brings the inefficient operations in light and management can focus its attention towards those matters only.  Prompt Reporting:-In standard costing, reporting is very prompt. Reports get ready as soon as the operations come to a halt. Moreover, the reports prepared in it are found to be very simple and improved in nature.  Formulation of Production and Price Policies:-Standard costs represent long-term estimates, cost and prices. It helps the management in the formulation of ideal production policy.  Implementing Incentive Schemes:-Standard costing promotes the implementation of incentive schemes in the organization because every incentive scheme is based on certain standards which are determined under this system.  Facilities Comparison:-Cost comparison between different products and departments can be done under standard costing. It also makes possible the comparison of costs of one period with another.  Promotes Cost Consciousness and Efficiency:- It also promotes cost consciousness as the employees know that their performances shall be in assessing manufacturing inefficiencies and fixing responsibilities. This improves the efficiency of the organization.

LIMITATIONS OF STANDARD COSTING Through standard costing is an important tool of cost control, it has certain practical limitations also. These limitations are as follows:-

51 | P a g e

 Not Appropriate for Small Concerns:-It is not appropriate for small concerns as the installation of standard costing requires high degree of skill and the small concerns may not have expert staff for handling or operating the system.  Division of Variances is Difficult:-The exact division of variances into controllable and uncontrollable variances is a difficult task.  Not suitable for Certain Industries:-This system is not suitable for industries which produce non-standardized products and for job works which change according to customer requirements.  Problem of Updating:-There is also a problem of keeping standards up –to- date. The circumstances keep changing but the standards cannot be changed frequently.  Unsuitable for Concerns Dealing in Non-Standardized Products:-Standard costing is not much useful in those concerns where non-standardized products are produced or production is undertaken according to customer‟s specification. In such a case it becomes difficult to set up standard for each job separately.  Difficulties in Setting Up Standards:-The process of setting up standards is a complicated task as it requires technical skill and expertise competence. The time and motion studies are required to be undertaken for this purpose, which require a lot of time and money. Moreover, if wrong standards are determined, the whole purpose of the system would fail.  Not Suitable for Small Firms:-The system of standard costing may not be suitable for small concerns keeping in view the cost of high degree of skill required in it.  Difficulty in Fixing Responsibility:-The responsibility for variances is fixed in the process of standard costing but it is not an easy task. Under many circumstances variances may arise due to various reasons when it becomes difficult to fix the specific responsibility. For example, in the case of unfavorable labor efficiency variance, it is not necessary that workers are inefficient. This situation may also arise due to inferior quality of raw material or defect in machine.  Changing Business Conditions:-In standard costing standards are fixed under specified conditions. In case these condition changes, then either standard loose their suitability or they have to be modified. Thus, standard costing is not suitable in those concerns where frequent technological changes take place or prices of material fluctuate frequently.  Need of Budgetary Control:-This system can be effective only when budgetary control is also adopted in the concern.  Feeing of Dissatisfaction among Employees:-If standards related to labor are fixed at a high level, it may develop a feeling of dissatisfaction and psychological frustration among employees.

52 | P a g e

VARIANCE ANALYSIS Variance analysis is the distinctive feature of standard costing. A variance is the deviation of the „actual‟ from the standard. In standard costing, it means the differences between the standard cost and the actual cost. Variances depict the extent to which standard have been achieved or not have been achieved. Variances are computed to know the deviations and fix the responsibility. It is a very important managerial tool to interpreting operating results and spotting situations requiring corrective measures.

MANAGERIAL USES OR VARIANCE ANALYSIS  Measurement of Operational Efficiency:-Efficiency of business activities can easily be measured on the basis of variances. If variances are unfavorable, the activity is considered efficient. If it is otherwise the activity is considered inefficient.  Technique of Cost Control:-Variances analysis is a useful managerial technique from the view of cost control because necessary corrective measures may be taken after analyzing the cause of unfavorable variances.  Knowledge of Variance Centre:-Variances analysis helps in identifying the level, activity or departments where variances are occurring so that special attention may be focused to these activities or departments.  Determination of Responsibility:-On the basis of variance analysis the persons or departments responsible for variances can be identified which helps in taking corrections measures.  Measurement of Accuracy of Standards:-If there are no variances or variance are very limited. It is an indication of accuracy of standards. On the contrary, permanent existence of variances indicated the inaccuracy of standards.  Relative Measurement of Performances:-The performance of difference of different department of an enterprise can easily be evaluated in relative terms on the basis of variance analysis.  Basic of Future Action and Planning:-Analysis of causes of variances, identification of responsible persons and suggestions of corrective measures provide a useful base for future action and planning by management.  Cost Consciousness:-Variance analysis and its reporting develops the feeling of cost consciousness among employees and executives, which proves an effective motivational tool from the view of management.

CAUSES OF VARIANCES

(a) MATERIAL COST VARIANCE:-Material Cost Variance may arise due to variance in price or quantity or in both:- Material Price Variance: - Material price variance may arise due to following causes:-  Changes in market price of materials  Failure to secure expected discount on purchases  Failure to purchase materials at proper time  Changes in prices of materials due to changes in tax by government

53 | P a g e

 Purchasing non-standard lots and the consequent reduction in quantity discount  Increase in transport cost due to small lot or additional transport cost for quick delivery.  Excessive shrinkage or loss in transit.  Failure to purchase the standard quality of materials.

A material price variance is normally the responsibility of the purchase manager. However, price variances due to frequent market fluctuations or changes in tax fall in the category of uncontrollable variances.

Material Usage Variance:-The variance arising due to difference in quantity of material is called as Material Usage Variance or Material Quantity Variance. This variance may arise due to the following causes:-  Negligence in the handling and use of materials.  Increase in wastage either due to untrained workers or defective method of production  Increase in the usage rate of material due to difference in quality  Changes in technique or method of production  More or less yield from materials  Change in material mix, etc. (b) LABOR COST VARIANCE: - The variances in labor cost may arise due to the following causes:- Labor Rate Variance:-  Changes in basic wage rates, it may be either due to change in demand and supply of worker or due to new wage settlement  Different rates being paid to workers employed to meet seasonal demands or to get urgent work done  Employment of workers of different grades and wage rates  New workers being paid lower rates from the standard rates  More payment of overtime wages etc. It may be mentioned that major portion of labor rate variance is beyond the direct control of management. Labor Efficiency or Labor Time Variance:-If the actual hours worked are less or more than standard hours, the difference is known as labor efficiency variances. This variance may be unfavorable on account of following causes:-  Use of sub-standard materials requiring extra labor  Defective working conditions  Lack of proper supervision of work  Use of defective machinery and equipment  Inadequate training of employees  Discontentment among workers due to unsatisfactory working conditions, etc. Idle Time Variance:-It will always be unfavorable and the main causes of this variances are:-  Non-availability of material at proper time

54 | P a g e

 Gap in work due to defect in machinery  Stoppage of work due to power failure  Closure of work due to strike or lock out Labor Mix Variance:-If different types or grades of employees are engaged in a work and practically the actual composition differs from the standard composition, the variance due to such differences is called as labor mix variance.

3.4 ACTIVITY BASED COSTING This technique has been of recent origin and primarily concerned with absorption of overheads (Indirect Costs) in an organization having products that differ in volume and complexity of production. The crux of activity based costing is in accurately assigning the overhead cost to the end product. The traditional costing system does not serve effective purposes of product costing and pricing decision. Activity based costing is a method of cost attribution to cost units on the basis of benefits received from indirect activities. Their performance of particular activities and demands made by these activities on the quantity of resources of organization are linked together so that the cost of product is arrived at as per the quantum of actual activities performed to produce a product or service. The reason for such a basis is that products themselves do not consume resources directly rather several activities are required to be performed for them, and these activities consume the resources of organization as driven by cost drivers. Cost centre pay for these resources, depending upon the number of activities required for a product. Activity based costing may be defined as a technique which involves identification of costs with each cost driving activity and making it as the basis for absorption of costs over different products or jobs. The Chartered Institute of Management Accountants (CIMA), London, defines it as “a technique of cost attribution to cost units on the basis of benefits received from indirect activities, e.g., ordering, setting up, assuring quality.”

3.4.1 Characteristics of ABC  The characteristics of activity based costing can be summarized as follows:-  It increases the number of cost pools used to accumulate overhead costs. The number of pools depends upon the cost driving activities. Thus, in stead of accumulated overheads costs in a single company-wise pool or departmental pools, the costs are accumulated by activities.  It charges overhead costs to different jobs or products in proportion to the costs driving activities in place of a blanket rate based on direct labor cost or direct hours or machine hours.  It improves the traceability of the overhead costs which results in more accurate unit cost data for management.  Identification of cost during activities and their causes not only help in computation of more accurate cost of a product or a job but also eliminate non- value added activities. The elimination of non-value added activities would drive down the cost of the product. This, in fact, is the essence of activity based costing. 3.4.2 Elements Involved in ABC

55 | P a g e

 Activity:-An activity can consist of one or more of the tasks associated with one another to attain an objectives. For example, customer order processing includes receiving orders from customers, as to its capacity of producing and interaction with customers regarding delivery time etc.  Activity cost centre (Pool):-The result of identifying the overhead cost to an activity is called an activity cost pool.  Cost drivers:-These are used to assign costs to products by using an appropriate measure of resources consumed by each activity.  Process:- When related activities are grouped together, it is known as a process.  Cost Objectives:- are links to the whole of the enterprise.  Non-value Adding Activities:- Certain activities do not contribute anything to the value of a product, but which are required to be carried out in the organization because of reasons beyond the control of management.

3.4.3 Steps Involved in ABC  Evaluation of prevalent costing system:- The already existing costing system should be able to adopt the activity based costing system so that product costs can be accurately determined and correct pricing decisions can be taken in a competitive business environment.  Identification activities:- A physical plan of the work place and listing of pay-rolls can be examined, to begin with, supplemented by holding interviews with staff. Such an activity analysis can throw light on how the work spaces have been utilized and how the staff members have spent their time. After chalking out the different tasks in detail, the prime activity can be identified. For it, a cost-benefit analysis is required to be performed. An activity can be a very small job or a combination of several small jobs.  Selection of cost basis:- The immediate past actual cost or average cost of a specified period may be used under the system.  Determined cost pools:- Cost centre on cost pools should have a similarity under financial accounting and cost accounting systems in order to have a comparative utility.  Assigning costs pools:- Cost of resources consumed are to be associated or allocated (apportioned) to each activity, in order to find out the amount spent by the enterprise on each of its activity. Methods of direct attribution , apportionment on a reasonable basis in case of joint costs and the methods of estimation of certain costs may all be appropriately utilized assignment of costs to activity cost centre.  Determined activity hierarchies:-The activities may be classified according to:-  Unit-level activities  Batch-level activities  Product-level activities  Facility-level activities

56 | P a g e

 Identification outputs:- Outputs are required to be categorically identified because without it, The basic purpose of applying the system cannot be meaningful.  Selecting suitable cost drivers:- Interviews should be conducted with concerned employees to determine the cost drivers for each activity. The cost drivers should be easily measureable. Past data should be made available in order to ascertain potential cost drivers. The management can finally choose a single cost driver or multiple cost drivers for the activities.  Computing Cost Driver Rates:- The cost driver rates are required to be calculated on a suitable basis.  Identifying cost to products:- The cost driver rates are applied to products. The aggregate cost can be computed by multiplying the rate of consumption of resources (cost driver rate) with the number of activities.  Test Run:- The activity cost an all the relevant data used for consumption purposes should be tested to evaluate and judge the reliability of activity based costing system adopted.

3.4.4 Activity based Costing versus Traditional Costing Following are the main differences between activity based costing system and traditional costing system:-  Effectiveness of purposes:-The objectivities of product costing and pricing decisions are more effectively served by adopting ABC systems.  Basis of Identification:- Under ABC system, overhead costs are identified to each major activity in stead of the department as under traditional costing system. It results in greater number of cost centre under ABC system.  Terms used:-The term „ Cost Drivers‟ is not used under traditional costing system. Popular terms are basis of allocation or apportionment. Under ABC system, cost drivers are fewer in number for the purpose of charging overheads to products.  Methodology:-ABC system uses separate rates for support centre and there is no reallocation to production centre, as is the case under traditional costing system.  Thus, ABC is refinement over traditional costing system. This is primarily because cause and effect relationship is considered under ABC system to identify support cost objects, which is not the case under traditional costing system.

Requirement of ABC System: ABC system requires the following:-  Setting up of an information system which could help trace all the costs to cost objects.  Support from top to bottom because the system involves people at all levels.  Integration of system into financial system. For it, computerization may be required.

57 | P a g e

3.4.5 Reasons for Adoption of ABC System Manufacturing as well as non-manufacturing industries adopt Activity Based Costing system because of the following reasons:  More accurate cost estimation:-ABC helps in estimating costs of individual product or service more accurately, which, in turn, assists in formulating appropriate marketing and corporate strategy.  More accurate accounting:-ABC improves accuracy of accounting for support service costs, which occupy an important place in total cost of delivering value of customers.  More accurate profitable measurement:- Profitability of the product and even customer-wise profitability can be more accurately measured by tracing the consumption of resources to each individual product and customer. It is essential as demand on resources by products and customers differ among products and customers respectively.  Reduction in cost of data processing:-ABC helps in reducing the cost of data processing, with consequently brings down the cost of tracking consumption of resources by variety of activities.  Reduction in cost of wrong decision-making:-ABC induces reduction in costs associated with poor decisions made to a substantial extent because of rational apportionment of costs. The final words of comment over ABC system are that the adoption, implementation and operation of the system is not an end in itself. The benefits can be derived by translating the system design and its operation into action-oriented managerial performance. Ultimately, it amounts to effective cost management for the success of the system.

3.4.6 Problems with the ABC Approach  Cost of change will be high as everything will have to be worked out from scratch.  It would be difficult to correlate the marginal increase in cost with a particular cost driver.  Over a period of time, the ABC will tend to standardize the cost of activities related to a particular product or process. But in practice there will be differences in set-up time, production run, and meeting a delivery order for the product or process, as well as for different product.  The ABC system will require a change due to changes associated with new products and new technology. This will put strain on the costing system and resources due to certain degree of inbuilt standardization. There exists catch 22 situation in the implementation of the ABC. Measure of activity performance will change again and again. A trade-off will be required between the accuracy and time spent on replacing the existing system with the ABC.  The ABC is at the stage of evolution. Literature on the ABC concept at present is primarily restricted to the manufacturing environment. The Activity Based Costing (ABC) has been successfully adopted by many Japanese Corporations. As a matter of fact, the elimination of “ non-value added activities”

58 | P a g e was the secret of the Japanese capturing a significant market share with limited products and being considered serious threat to the entire US automobile and electronics. As a result, now many US Corporations are also increasingly adopting Activity Based Costing

3.5 COST–VOLUME-PROFIT ANALYSIS Cost –Volume- Profit Analysis is a logical extension of the concept of marginal costing, in which cost of production is divided into two parts, i.e., fixed cost and variable cost. These two cost leads to decrease in cost per unit and increase in profit per unit with the increase in volume of production. In short, there is negative relation between cost of production and amount of profit; volume of production and cost of production and positive relation between volume of production and amount of profit. The concept of cost –volume-profit analysis is used in the narrower as well as in the broader sense. In narrower concept, it is concerned with finding out „break –even point‟, i.e., the point at which there is no profit or no loss. In its broader sense, it is a technique of management accounting which determines profit, cost and sales value at different levels of production. It also establishes relationship among these three factors.

IMPORTANCE OF COST–VOLUME-PROFIT ANALYSIS It is an important tool in the process of managerial decision and it is extensively helpful to management in a variety of problems involving planning and control. The main objectives of such analysis are as follows: Setting up Flexible Budget: This analysis is helpful in setting up flexible budget which indicates that what trend of amount of sales and cost of production at different level of activity will be. Determination of B.E.P.: The most important objective of cost –volume-profit analysis is to find out break –even point. Profit Planning: This analysis is useful in profit planning also because whereas, on the one hand, we can determine the amount of profits at different levels of activity we can also determine the volume of sales or production to earn desired profit on the other hand. Decision relating to Selection of Alternatives: This analysis helps the management in taking decision in respect of various alternative proposals, like: 1. Which of the product is more profitable? 2. Whether the firm should accept the proposal of supply of additional products at a particular price? 3. What is the optimum mix of sales or production, so that profit may become maximum?

Performance Evaluation for Control: This analysis assists in evaluation of performance for the purpose of control. On the basis of profit achieved and costs incurred it can be analyzed that what the role of volume of production and others factors was in effecting the amount of profit. Helpful in Price Fixation: This analysis is also helpful in price fixation by studying the effect of different price structures on cost and profit.

59 | P a g e

Allocation of Overhead Costs: This analysis assists in finding out the amount of overhead costs to be charged to the products at various levels of production because predetermined overheads rates are related to a selected volume of production. Analysis of Effect of Changes in Cost: It assists in analyzing the impact of the fluctuations in fixed cost, material cost, labour cost and overheads etc.

3.5.1 BREAK EVEN ANALYSIS Break-Even Point (B.E.P.) is that point of sales where the total revenues are equal to total costs. In other words, it is a point where there is no profit no loss. It is also known as „No Profit Point „or Zero loss Point.‟ At this point, contribution is equal to fixed costs. If the sales are more than this level, there will be profit to the firm and if sales are less than this level, there will be a loss. It is also called as „Equilibrium Point‟, „Balancing Point‟ or „Critical point‟. In broader sense, the term break-even analysis refers to the study of relationship between costs, volume and profit at different levels of sales or production. DEFINITION: “ The Break –Even Point is that point of sales volume where total revenues and total expenses are equal, it is also said as the point of zero profit or zero loss.” ------Charles T. Horngren “ The Break-Even Point of a company or unit of a company is that level of sales income which will equal the sum of its fixed cost and its variable costs.”------Keller and Ferrara ASSUMPTIONS OF BREAK EVEN ANALYSIS 1. All elements of costs are segregated into fixed and variable components. 2. There will be no change in general price level. 3. Production and sales are in synchronization 4. Fixed cost remains fixed at all volumes of output. 5. Selling price per unit remains constant. 6. Variable cost remains constant per unit of output.

ADVANTAGES /USES OF BREAK EVEN ANALYSIS Break –Even Analysis helps to evaluate the profitability of new order. 1. It also helps in determining the sales to earn desired profit. 2. It also helps in estimating margin of safety. 3. Impact of increase or decrease in fixed or variable costs on profit can be ascertained with break –even analysis. 4. It also helps in make or buy decision. 5. It also helps in determination of optimum sales-mix. 6. It is useful at the time of changing capacity 7. It helps in the calculation of comparative profitability of projects, product lines, etc.

LIMITATION OF BREAK EVEN ANALYSIS 1. The assumption that variable cost remains constant is not valid assumption since the prices of material, labour and other expenses do not remain constant. 2. The assumption that the fixed cost remains constant is not valid because the fixed cost increases after a certain level of production.

60 | P a g e

3. The assumption that the selling price remains constant is not valid because it varies with change in demand and supply of all the products. 4. It may be difficult to divide all costs into fixed and variable costs. 5. The break even analysis assumes a static situation that cannot exist for long periods of time . For example, it assumes no change in general price level, selling price, production technology etc. but in practice there is constant change in these factors as management wants to improve production system and efficiency.

COMPUTATION OF BREAK EVEN POINT There are three methods for the computation of B.E.P. 1. Equation Method 2.Contribution Method 3. Graphic Method (a) EQUATION METHOD: A simple equation is used to calculate the break even point which also expresses the relationship of the items of income statement. Sales = Variable Expenses + Fixed Expenses + Profit For example, selling price is Rs.10 p.u, variable cost is Rs. 6 p.u and fixed cost Rs.10,000. Assuming x unit is to be sold to break –even, then by applying formula: 10x=6x+10,000 10x-6x=10,000 4x=10,000 X=10,000/4 = 2,500 units.

(b) CONTRIBUTION METHOD: This is the second method to calculate the break even point. This method involves two basic tools, i.e. „Contribution‟ and „Profit – Volume Ratio‟. Contribution: It is the difference of sales and variable cost. It may also be defined as the excess of sales over variable cost. It contributes towards fixed expenses and profit. The amount of contribution may be computed as follows: (A) Total Contribution:

Contribution = Sales-Variable Cost or S-V Contribution = Fixed Cost + Profit (-Loss) Contribution= Sales x P/V Ratio (B) Per Unit Contribution :

Contribution per unit = Sales per unit- Variable Cost per unit

Profit Volume Ratio or P/V Ratio It is the relationship between contribution and sales and is generally expressed in terms of percentage. It is also known as „Margin Ratio‟ or „Contribution Ratio‟. It can be expressed as under: P/V Ratio = Contribution *100 or C*100 Sales S = Sales- Variable Cost × 100 Sales = Fixed Cost + Profit ×100 Sales = Contribution per unit ×100 OR Spu –Vpu ×100 Sales per unit Spu

61 | P a g e

= Change in Profit ×100 Change in Sales

Calculation of Break Even Point B.E.P. ( Rs.) : It is also known as Sales B.E.P. = Fixed Cost ×Sales Contribution = Fixed Cost × Selling Price p.u. Contribution p.u = Fixed Cost P/V Ratio = Sales- Margin of Safety B.E.P.(in units): It is also known as „Break-even Point in Quantity‟ or „output B.E.P.‟ = Fixed Cost Contribution Per unit = B.E.P. (Rs.) Spu

(c ) GRAPHIC METHOD OR BREAK EVEN CHART

Break-even chart is a graphic representation of marginal costing. It presents the information relating to cost, sales, revenue and profit, etc. It also indicates the break – even point and the estimated profit and loss. There are three methods of drawing a break- even chart.

Methods of Drawing a Break- even Chart

First Method: The following steps are followed in it: 1. Volume of production or output or sales is shown on X-axis. 2. Costs and sales revenue are plotted on Y-axis. 3. Fixed cost line is drawn parallel to X-axis. This shows that fixed cost remain constant at all levels of output. 4. The variable cost and sales value at different levels of output are plotted and lines are drawn to join their respective points. The variable cost starts above the fixed cost line. 5. The point where the two lines intersect each other, that point is known as break –even point. 6. The number of the units that should be produced to reach break-even can be known by drawing a perpendicular to the X-axis from the break-even point. 7. Similarly sales revenue can be known by drawing a perpendicular from this point to Y-axis. 8. The area below the break-even point is known as loss area and above it is known as profit area.

62 | P a g e

Second Method: In this method, variable cost is plotted \first and then the fixed cost line is plotted above the variable cost line. It is drawn parallel to variable cost line. It represents the total cost of the product. In this method, contribution is represented by the difference between total revenue line and variable cost line.

3.5.2 Margin of Safety It is the difference between actual total sales and B.E.P. sales and may be calculated in rupees, units or even in percentage form as explained below: M.O.S. in Rupees:

63 | P a g e

M.O.S.(Rs.)= Sales (Rs.) – B.E.P (Rs.) M.O.S.( Rs.)= Profit P/V Ratio

M.O.S. in Units:

M.O.S. (Units) = Sales (Units) - B.E.P (Units)

M.O.S. (Units) = Profit Contribution per Unit

M.O.S in Percentage: = M.O.S × 100 Total Actual Sales

Example: The following information is obtained from XY and Co. for 2013 Rs. Sales 20,000 Variable Cost 10,000 Fixed Cost 6,000 1. Find the P/V ratio, break-even point and margin of safety at the level. 2. Calculate the effect of: 20% decrease in fixed costs; 10% increase in fixed costs; 10% decrease in variable costs; 10%increase in selling price; 10% increase in selling price together with an increase of fixed overheads by Rs.1,200 10% decrease in sales price; 10% decrease in sales price accompanied by 10% decrease in variable costs. Solution: Profit Volume Ratio = Contribution ×100 Sales = Sales – Variable Cost × 100 Sales

Break- Even Point (B.E.P) = Fixed Cost × 100 P/V Ratio

Margin of Safety = Profit P/V Ratio Or Actual Sales- Sales at B.E.P

(a) At Existing Level : P/V Ratio = Rs. 20,000 – Rs. 10,000 × 100 Rs. 20,000 = Rs. 10,000 × 100

64 | P a g e

Rs. 20,000 = 50%

B.E.P = Rs. 6,000 50% = Rs. 12,000. M.O.S = Rs. 20,000- Rs. 12,000 =Rs. 8,000 (b) (i) 20 % decrease in fixed cost P/V Ratio = 50% B.E.P. = Rs. 4,800 50% = Rs. 9,600. M.O.S = Rs. 5,200 50% = Rs. 10,400 Or (Rs. 20,000- Rs. 9,600)

(ii) 10 % Increase in Fixed Costs P/V Ratio = 50% B.E.P = Rs. 6,600 50% = Rs. 13,200 M.O.S = Rs. 3,400 50% = Rs. 6,800 Or (Rs. 20,000 – Rs. 13,200)

(iii) 10% Decrease in Variable Costs P/V Ratio = Rs 20,000 – Rs 9,000 Rs 20,000 = Rs 11,000 ×100 Rs 20,000 = 55% B.E.P = Rs. 6,000 55% = Rs. 10,909.09 M.O.S = Rs. 5,000 55% = Rs. 9090.91

(iv) 10% Increase in Selling Price P/V Ratio = Rs. 22,000- Rs. 10,000 ×100 Rs. 22,000 = Rs. 12,000 ×100 Rs. 22,000 = 54.55 % (Approx) B.E.P = Rs. 6,000 54.55%

65 | P a g e

= Rs. 11,000 M.O.S = Rs. 6,000 54.55% = Rs.11,000

(v) 10% Increase in Selling Price with Increase in Fixed Overhead by Rs. 1,200 P/V Ratio = Rs. 22,000 – Rs 10,000 Rs 22,000 = 54.55% B.E.P = Rs. 7,200 54.55% = Rs. 13,200 M.O.S = Rs. 4,800 54.55% = Rs. 8,800

(vi) 10% Decrease in Sales Price P/V Ratio = Rs. 18,000 – Rs 10,000 × 100 Rs 18,000 = 44.44 % B.E.P = Rs. 6,000 44.44 % = Rs. 13,500 M.O.S = Rs 2,000 44.44% = Rs. 4,500 (vii) 10% Decrease in Sales with 10% Decrease in Variable Costs P/V Ratio = Rs. 18,000 – Rs. 9,000 × 100 Rs. 18,000 = 50% B.E.P = Rs.6,000 50% = Rs. 12,000 M.O.S = Rs. 3,000 50% = Rs. 6,000

3.6 CHECK YOUR PROGRESS

1. The break-even point is the point at which: a. There is no profit no loss; b. Contribution margin is equal to total fixed cost; c. Total revenue is equal to total cost; d. All of the above

2. Cost- Volume profit analysis is based on several assumptions. Which of the following is not one of these assumptions? a. The sales-mix of the product is constant; b. Inventory quantities change during the year; c. The behavior of both revenues and cost is linear throughout the relevant range; 66 | P a g e

d. Factor prices e.g. material prices and wage rate remain unchanged.

3. Standard Costing involves the :

a. Fixation of estimated cost

b. Determination of standard cost

c. Setting of budgeted cost

d. None of these

4. The difference between standard cost and actual cost is known as

a. Variance

b. Profit

c. Differential Cost

d. Marginal Cost

5. An activity-based costing system is one that:

a. traces costs to activities and then to products. b. traces costs to resources and then to activities. c. traces activities to costs and then to resources. d. traces products to activities and then to resources. 6. In Make or Buy decision: a. Only variable costs are relevant b. Fixed Cost that can be avoided in future are relevant c. Fixed Cost that will continue regardless of the decision are relevant d. Both a and b

7. Relevant Costs are: a. Standard Costs b. Controllable Costs c. Future Costs d. Historical Costs

8. In deciding whether to manufacture a part or buy it from outside supplier, a cost that is not relevant to short run decision is: a. Direct Labour b. Variable Overheads c. Fixed overheads that will be avoided if the part is bought from an outside supplier.

67 | P a g e

d. Fixed overhead that will continue even if the part is bought from an outside supplier. 3.7 LET US SUM UP

Marginal costing plays a vital role in decision making .It is a very useful technique in solving various managerial problems and contribution in various areas of decisions. Few important ones are explained here: Make or Buy Decision, Lease or Buy, Change in Product Mix, Pricing Decision, Shut –down Decisions and, Exploring New Market. Standard costing is a very important technique of cost control. Every organization wants to minimize the cost of production and maximize the profits. Standard costing is such a system which seeks to control the cost of each unit and then it is compared with actual cost. The difference between the actual and pre- determined costs is known as variance which need to be analyzed through carefully planned accounting procedures and then the results are promptly reported to managers. A variance is the deviation of the „actual‟ from the standard. In standard costing, it means the differences between the standard cost and the actual cost. Variances depict the extent to which standard have been achieved or not have been achieved. Variances are computed to know the deviations and fix the responsibility. It is a very important managerial tool to interpreting operating results and spotting situations requiring corrective measures. The traditional costing system does not serve effective purposes of product costing and pricing decision. Activity based costing is a method of cost attribution to cost units on the basis of benefits received from indirect activities. Activity based costing may be defined as a technique which involves identification of costs with each cost driving activity and making it as the basis for absorption of costs over different products or jobs. Cost –Volume- Profit Analysis is a logical extension of the concept of marginal costing, in which cost of production is divided into two parts, i.e., fixed cost and variable cost. These two cost leads to decrease in cost per unit and increase in profit per unit with the increase in volume of production. In short, there is negative relation between cost of production and amount of profit; volume of production and cost of production and positive relation between volume of production and amount of profit. The concept of cost –volume- profit analysis is used in the narrower as well as in the broader sense. In narrower concept, it is concerned with finding out „break –even point‟, i.e., the point at which there is no profit or no loss. In its broader sense, it is a technique of management accounting which determines profit, cost and sales value at different levels of production. It also establishes relationship among these three factors. Break-Even Point (B.E.P.) is that point of sales where the total revenues are equal to total costs. In other words, it is a point where there is no profit no loss. It is also known as „No Profit Point „or Zero loss Point.

3.8 ANSERS TO CHECK YOUR PROGRESS

1 ( a), 2 (b ), 3 (b ), 4 (a ), 5 (a ), 6 (d ), 7 (c ), 8 (c )

Suggested Reading/References

 Management Accounting by Aggarwal, Aggarwal and Jain, Ramesh Book Depot, Jaipur

68 | P a g e

 Management Accounting by I M Pandey, Vikas publicationj, Delhi  Management Accounting and Financial Management by D K Goyal, Avichal Publishing Company, Kala Amb, Himachel Pradesh.  Accounting for Managerial decision by BP Aggarwal and BK Mehta, Sahitya Bhawan Publication, Agara  Accounting for Managers by Prof. Jawaharlal, Himalaya Publication, Delhi  Accounting for Managers by S. N. Maheshwari, Vikas Publication, Delhi  Financial Accounting by Dr. Mohinder Kumar Gupta, Mahaveer Prakashan, Delhi

69 | P a g e

Block II - UNIT-IV SPECTRAL THEORY OF BOUNDED SELF-ADJOINT LINEAR OPERATORS Introduction In mathematics, on a finite-dimensional inner product space, a self-adjoint operator is an operator that is its own adjoint, or, equivalently, one whose matrix is Hermitian, where a Hermitian matrix is one which is equal to its own conjugate transpose. By the finite dimensional spectral theorem, such operators can be associated with an orthonormal basis of the underlying space in which the operator is represented as a diagonal matrix with entries in the real numbers. In this Chaptere, we consider generalizations of this concept to operators on Hilbert spaces of arbitrary dimension. Self-adjoint operators are used in functional analysis and quantum mechanics. In quantum mechanics their importance lies in the Dirac–von Neumann formulation of quantum mechanics, in which physical observables such as position, momentum, angular momentum and spin are represented by self-adjoint operators on a Hilbert space. Of particular significance is the Hamiltonian

2 HV    2  2m which as an observable corresponds to the total energy of a particle of mass m in a real potential field V. Differential operators are an important class of unbounded operators. Objectives The structure of self-adjoint operators on infinite-dimensional Hilbert spaces essentially resembles the finite-dimensional case, that is to say, operators are self- adjoint if and only if they are unitarily equivalent to real-valued multiplication operators. With suitable modifications, this result can be extended to possibly unbounded operators on infinite-dimensional spaces. Since an everywhere defined self-adjoint operator is necessarily bounded, one needs be more attentive to the domain issue in the unbounded case. This is explained below in more detail. For simplicity we will always assume that the Hilbert spaces considered in this chapter are separable and complex (although most results extend to nonseparable complex Hilbert spaces).Let H1, H2 be separable Hilbert spaces and A be a linear operator A : D(A) ⊂ H1 → H2. We denote by B(H1,H2) the set of all bounded linear operators from H1into H2 and write B(H,H) = B(H) for simplicity.We recall that A = B if D(A) = D(B) = D and Ax = Bx for all x D. Next, let H1 = H2 = H. In this chapter we introduce and explain spectral properties of Bounded self adjoint Linear operator and and spectral family of a Bounded self adjoint linear operator with properties. Extension of spectral theorem to continuous function is main aim. 4.1 Spectral Properties of Bounded Self-Adjoint Linear Operators Definition 4.1.1. (i) Let T be densely defined in H. Then T* is called the adjoint of T if,

70 | P a g e

Dom(T*) = {g  H | there exists an hg  H such that (hg, f) = (g, Tf) for all f  Dom(T)}, T*g = hg. (ii) An operator A in H is called symmetric if A is densely defined and A  A*. (iii) A densely defined operator B in H is called self-adjoint if B = B*. (iv) A densely defined operator S in H is called normal if SS* = S*S. We note that for every self-adjoint operator A in H one has D(A) = H. For every bounded operator A we will assume D(A) = H unless explicitly stated otherwise. Definition 4.1.2. (i) z  C lies in the resolvent set of A if (A − zI)−1 exists and is bounded. The resolvent set of A is denoted by ρ(A). (ii) If z  ρ(A), then (A − zI)−1 is called the resolvent of A at the point z. (iii) σ(A) = C \ ρ(A) is called the spectrum of A. We will use the notation, R(z,A) = (A − zI)−1, z  ρ(A). Remark 1 - (A) = σ(A). 2 - A = A*  σ(A)  R.

3 - If A is a bounded operator, then σ(A) is a bounded subset of C. 4 - If A is a bounded self-adjoint operator, then σ(A)  R is compact. 5 - If A is a bounded self-adjoint operator, then ||A|| = supλσ(A) |λ|. 6 - If A is a self-adjoint operator, then R(z,A) is a normal operator for all z  ρ(A). 4.2 Further Spectral Properties of Bounded Self-Adjoint Linear Operators The spectrum (T) of a bounded self-adjoint linear operator T is real. This important fact was proved in the last section. We shall now see that the spectrum of such an operator can be characterized in more detail since it has a number of general properties which are mathematically interesting and practically important. It is clear that (T) must be compact, but in the present case we can say more: Theorem 4.2-1: (Spectrum). The spectrum (T) of a bounded self-adjoint linear operator T: H  H on a complex Hilbert space H lies in the closed interval [m, M] on the real axis, where (1) m inf Tx,x , M sup Tx, x ||x|| 1 ||x|| 1 Proof: (T) lies on the real axis (by 9.1-3). We show that any real  = M + c with c > 0 belongs to the resolvent set (T). For every x  0 and v = ||x||-1x we have x = ||x||v and Tx, x = ||x||2 Tv, v  ||x||2 = sup Tv, v x, xM. ||v|| 1 Hence -Tx, x  -x, xM, and by the Schwarz inequality we obtain

||Tx|| ||x||  -Tx, x = -(Tx, x) + (x,x)  (-M + ) x, x = c||x||2, where c =  - M > 0 by assumption. Division by ||x|| yields the inequality ||Tx||c||x||. Hence   (T) by 9.1-2. For a real  < m the idea of proof is the same. m and M in (1) are related to the norm of T in an interesting fashion: Theorem 4.2-2: (Norm). For any bounded self-adjoint linear operator T on a complex Hilbert space H we have [cf. (1)] (2) ||T|| = max(|m|, |M|)= sup |Tx, x|. ||x|| 1

71 | P a g e

Proof: By the Schwarz inequality, sup |Tx, x|  sup ||Tx|| ||x|| = ||T||, ||x|| 1 ||x|| 1 that is, K  ||T||, where K denotes the expression on the left. We show that ||T||  K. If Tz = 0 for all z of norm 1, then T = 0 (why?) and we are done. Otherwise for any z of norm 1 such that Tz  0, we set v = ||Tz||1/22 and w = ||Tz||-1/2Tz. Then ||v||2 = ||w||2 = ||Tz||. We now set y1 = v + w and y2 = v - w. Then by straightforward calculation, since a number of terms drop out and T is self-adjoint, Ty1, y, - Ty2, y2 = 2(Tu, w + Tw, v) (3) =2(Tz,Tz + T2z, z) = 4||Tz||2. Now for every y  0 and x = ||y||-1 y we have y = ||y|| x and |Ty, y| = ||y||2 |Tx, x|  ||y||2 sup |T x , x | = K||y||2, ||x|| 1 so that by the triangle inequality and straightforward calculation we obtain |Ty1, y1 - Ty2, y2  |Ty1, y1| + |Ty2, y2| 2 2  K(||y1|| + ||y2|| ) = 2K(||v||2 + ||w||2) = 4K||Tz||. From this and (3) we see that 4||Tz||2  4K||Tz||. Hence ||Tz||  K. Taking the supremum over all z of norm 1, we obtain ||T||  K. Together with K  ||T|| this yields (2). Actually, the bounds for (T) in Theorem 9.2-1 cannot be tightened. This may be seen from Theorem 4.2-3: (m and M as spectral values). Let H and T be as in Theorem 4.2-1 and H  {0}. then m and M defined in (1) are spectral values of T. Proof: We show that M  (T). By the spectral mapping theorem 7.4-2 the spectrum of T + kI (k a real constant) is obtained from that of T by a translation, and M  (T)  M + k  (T + kI). Hence we may assume 0  m  M without loss of generality. Then by the previous theorem we have M sup Tx, x = ||T|| ||x|| 1

By the definition of a suprimum there is a sequence (xn) such that

||xn|| = 1, Txn, xn = M - n, n  0, n  0.

Then ||Txn||  ||T|| ||xn|| = ||T|| = M, and since T is self-adjoint, 2 ||Txn – Mxn|| = Txn - Mxn, Txn - Mxn 2 2 = ||Txn|| - 2M Txn, xn + M ||xn|| 2 2  M - 2M(M - n) + M = 2Mn  0. Hence there is no positive c such that ||TMxn|| = ||Txn - Mxn||  c = c ||xn|| (||xn||= 1). Theorem 4.1-2 now shows that  = M cannot belong to the resolvent set of T. Hence M  (T). For  = m the proof is similar. 4.3 Positive Operators 4.3.1 Aspects of Positivity In this subchapter we extend some of the ideas earliar discussion to a more general context and describe some of the special spectral properties of positive operators. These were first discovered for n × n matrices with non-negative entries by

72 | P a g e

Perron and Frobenius, but many aspects of the theory can be extended to much more general level. When we write B := Lp(X, dx) in this chapter, we usually refer to the space of real- valued functions. We assume throughout that the measure space satisfies the assumptions. Sometimes we will consider the corresponding complex space, and when we need to distinguish between these we do so by adding subscripts, as in BR and BC. If X is a countable set and dx is the counting measure we write lp(X) in place of Lp(X, dx). A number of the theorems have slightly less technical proofs in the discrete case, because one does not have to worry about null sets and can use pointwise evaluation of functions. Later in the chapter we assume that X is a compact metric space, and consider certain positive one-parameter semigroups acting on C(X). If f  B, the positive and negative parts of f are defined by 1 f+ := max{f, 0} = (|f| + f), 2 1 f− := max{−f, 0} = (|f| − f). 2 Note that |f| ≤ |g| implies ||f|| ≤ ||g||. The set B+ of all non-negative f  B is a convex cone, and is closed with respect to the norm and weak topologies of B. An operator A : B → B is said to be positive, symbolically A ≥ 0, if Af ≥ 0 for all f ≥ 0. We say that Tt is a positive one-parameter semigroup on B if Tt ≥ 0 for all t ≥ 0. p Lemma 4.3.1 If A is a positive operator acting on B = LR (X, dx), then A is bounded and ||A|| = sup{||Af||/||f|| : f ≥ 0 and f  0}. + Proof: Suppose first that for all n  Z there exists fn ≥ 0 such that ||fn|| = 1 and ||Afn|| ≥ 4n. If we put  n f : 2 fn n1 −n −n then f ≥ 0, ||f|| ≤ 1 and 0 ≤ 2 fn ≤ f for all n. Hence 0 ≤ 2 Afn ≤ Af, and n −n 2 ≤ 2 ||Afn|| ≤ ||Af|| for all n. The contradiction implies that there exists c such that ||Af|| ≤ c whenever f ≥ 0 and ||f|| = 1. If c is the smallest such constant then c ≤ ||A|| ≤ +∞. Given f  B, the inequality −|f| ≤ f ≤ |f| implies −A|f| ≤ Af ≤ A|f| and hence |Af| ≤ A|f|. Therefore ||Af|| = || |Af| || ≤ ||A|f| || ≤ c|| |f| || = c||f||. This implies that ||A|| ≤ c. p In order to study the spectrum of an operator AR acting on BR = LR (X, dx), one must p pass to the complexification BC = LC (X, dx). The complex-linear operator AC is defined in the natural way by AC(f + ig) := ARf + iARg. The proof in Theorem 13.1.2 that ||AC|| = ||AR|| is only valid for positive operators. One may also adapt the proof of Theorem 12.1.1 to Lp(X, dx); this does not require A to be positive, but Problem 13.1.3 shows that it does require p = q. p q Theorem 4.3.2 Let 1 ≤ p, q ≤ ∞ and let AR : LR (X, dx) → LR (X, dx) be a positive linear operator. Then |AC(f + ig)| ≤ AR(|f + ig|) p for all f, g  LR (X, dx). Hence ||AC|| = ||AR||.

73 | P a g e

Proof:. Given   R we have |(ARf) cos () + (ARg) sin()| = |AR(f cos() + g sin())| ≤ AR(|f cos() + g sin()|) ≤ AR(|f + ig|). Let u, v, w : X → R be functions in the classes of ARf, ARg, AR(|f + ig|). Then we have shown that |u(x) cos() + v(x) sin()| ≤ w(x)  for all x not in some null set N(). If {}n n 1 is a countable dense subset of [−, ] then |u(x) + iv(x)| = sup |u(x) cos(n) + v(x) sin(n)| ≤ w(x) 1n   for all x not in the null set N() . This implies the first statement of the n1 n theorem, from which the second follows immediately. Problem - 1 The following shows that the positivity condition in Theorem 13.1.2 is necessary if p  q. Consider the matrix 11 A:  11 ∞ 1 3/2 as a bounded operator from l to l . Show that ||AR|| = 2 but ||AC|| = 2 . Problem – 2 Let A be a positive linear operator on Lp(X, dx) where 1 ≤ p < ∞, and let 1/p + 1/q = 1. Use following to prove that A ,   A |  |,|  | for all complex-valued   Lp((X, dx) and   Lq((X, dx). Also give the much more elementary proof available when A has a non-negative integral kernel. Our next lemma might be regarded as an operator version of the Schwarz inequality. An operator version of the H older inequality may be proved by the same method. p q Lemma 4.3.3 Let 1 ≤ p, q ≤ ∞ and let AR : LR (X, dx) → LR (X, dx) be a positive linear operator. Then 2 2 2 |AC(fg)(x)| ≤ {AR(|f| )(x)}{AR(|g| )(x)} 2p almost everywhere, for all f, g  LC (X, dx). Proof: If A has a non-negative integral kernel K then 2 | A (f )(x) |2 {K(x, y) 1/ 2 f (y)}{K(x, y) 1/ 2 g(y)}dy Cg X  K(x, y) | f (y) |22 dy K(x, y) | g(y) | dy XX 22 {ARR (| f | )(x)}{A (| g | )(x)} This finishes the proof if X is finite or countable. We deal with the general case by using an approximation procedure. If  : = {E1, ...,En} is a sequence of disjoint Borel sets with finite measures |Er|, we define the orthogonal projection P by n P f : | E |1 X f,X  r Err E r1

We then note that the operator PA has the non-negative integral kernel n K(x, y) : | E |1 X (x){A*(X )(y)}  r Err E r1 Hence

74 | P a g e

|P(p)(x)|2 ≤ {P(q)(x)}{P(r)(x)} 2 2 almost everywhere, where p := AC(fg), q := AR(|f| ) and r := AR(|g| ). The proof is completed by choosing a sequence of increasingly fine partitions (n) for which P(n)(p), P(n)(q) and P(n)(r) converge to p, q and r respectively not only in norm but also almost everywhere. The following theorem has a wider scope than is apparent at first sight, because it is not required that   L2(X, dx). Theorem 4.3.4 Let A be a positive linear operator on L2(X, dx) and let  be a measurable function on X. If (x) > 0 almost everywhere, 0 ≤ A ≤  and 0 ≤ A* ≤ μ then || A || (  )1/ 2 Proof:Assume first that   L2(X, dx), so that (x)2dx is a finite measure and L∞(X)  L2(X, 2dx). We define the unitary operator U : L2(X, 2dx) → L2(X, dx) by Uf := f. We then observe that B : = U−1AU is positive and satisfies 0 ≤ B*1 ≤ 1 and 0 ≤ B*1 ≤ μ1. If f  L∞(X) then |(Bf)(x)|2 ≤ B(|f|2)(x)B(1)(x) ≤ B(|f|2)(x) almost everywhere by Lemma 13.1.5. Therefore 2 2 2 2 2 ||Bf||22 B(|f|),1   |f|,B*(1)   |f|,1   ||f|| Since L∞(X) is dense in L2(X, 2dx) we deduce that ||A|| = ||B|| ≤ (μ)1/2. If   L2(X, dx) then the assumptions of the theorem have to be interpreted appropriately. We assume that 0 ≤ A ≤  for all   L2(X, dx) that satisfy 0 ≤  ≤ , and similarly for A*. We than define B as before, and observe that 0 ≤ Bf ≤ 1 and 0 ≤ B*f ≤ 1 for all f  L∞(X) ∩ L2(X, 2dx) such that 0 ≤ f ≤ 1. From this point on we work in the weighted L2 space. Let D denote the set of all bounded functions on X whose supports have finite measure with respect to the measure (x)2dx. If f  D and supp(f) = E then 2 2 2 2 2 |(Bf)(x)| = |(B(fXE))(x)| ≤ B(|f| )(x)B( XE )(x) ≤ B(|f| )(x) almost everywhere, If the set F has finite measure then | (Bf ) |2 2 dx   B(| f | 2 )  2 dx FF 2 = B(|f| ), XF 2 = |f| , B*XF ≤ μ|f|2, 1 = μ||f||2 Since F is arbitrary subject to having finite measure we deduce that 22 || Bf ||22  || f || for all f  D, and since D is a dense subspace of L2 we obtain the same bound for all f  L2. Therefore ||A|| = ||B|| ≤ (μ)1/2. 2 2 Corollary 4.3.5 Let A be a positivity preserving self-adjoint linear operator on LR (X, dx) and let   L2(X, dx). If (x) > 0 almost everywhere and A =  then ||A|| = . Our next lemma states that the singularity closest to the origin of certain operatorvalued analytic functions lies on the positive real axis. Lemma 4.3.6 Suppose that An are positive operators on B := Lp(X, dx) and that for all z such that |z| < R the series

75 | P a g e

 n A(z) :  An z ….(1) n0 converges in norm to an operator A(z). Suppose also that A(z) may be analytically continued to the region {z : |z − R| < S}. Then the series (1) is convergent for all z such that |z| < R + S. Proof: If 0 ≤ f  B and 0 ≤ g  B* then the function F(z) := A(z)f, g is analytic in D := {z : |z| < R}  {z : |z − R| < S}. We have F(n) (R) limF (n) (r) rR  m! mn  Am f,g R mn (m n)! by a monotone convergence argument that uses the non-negativity of the coefficients. Moreover  0 F(n) (R)x n / n!   n0 for all x such that 0 ≤ x < S by the analyticity of F in {z : |z − R| < S}. Therefore the series    mm! m n n Af,g(Rx)mm  Af,gRx m 0 n  0 m  n (m n)!n! of non-negative terms is convergent for 0 ≤ x < S, and the series  m  Am f,g z m0 has radius of convergence at least (R + S). The same holds for all f  B (resp. g  B*) since every element of B (resp. B*) is a linear combination of four elements of B+ (resp. B*+). The proof that (1) is norm convergent for all z such that |z| < R + S is similar Theorem 4.3.7 Let A be a positive operator on B and let r := max{|z| : z  Spec(A)} be its spectral radius. Then r  Spec(A). Proof: If |z| > r then the series  (zI A)1 z  n  1 A n n0 is norm convergent. Since the analytic function z → (zI −A)−1 cannot be analytically continued to any set {z : |z| > r − }, the function must have a singularity at z = r by Lemma 4.3.6 Therefore r  Spec(A). All of the above ideas can be adapted to the context of one-parameter semigroups. Semigroups of the following type occur in population growth models and the neutron diffusion equation. These models are unstable if ||Tt|| increases indefinitely with t. In such cases one either has a population explosion, or it is prevented by some non-linear effect. Lemma 4.3.8 Let the operator Z := −M +A act in B := Lp(X, dx), where M denotes the operator of multiplication by a measurable function m that is bounded below and

76 | P a g e

Zt A is a bounded, positive operator on B. Then Tt := e is a positive one-parameter semigroup. −Mt Proof: Putting St := e , we note that Z is a bounded perturbation of M, so Tt is a positive operator for all t ≥ 0 because every term in the perturbation expansion is positive. Given f  B+ one might interpret f(t, y) := (Ttf)(y) as the local density of some entities at a site y, which can increase (if m(y) < 0) or decrease (if m(y) > 0) as time passes without moving from y. Entities at the position y can cause new entities to appear at the position x at the rate A(x, y) if A has an integral kernel A(x, y). The nth term on the right-hand side of (11.10) describes the part of the state at time t for which exactly (n − 1) such creations have taken place. 4.4 Square Roots of a Positive Operator If T is self-adjoint, then T2 is positive since T2x, x = Tx, Tx  0. We consider the converse problem: given a positive operator T, find, a self-adjoint A such that A2 = T. This suggests the following concept, which will be basic in connection with spectral representations. Definition 4.4.1 (Positive square root). Let T : H  H be a positive bounded self-adjoint linear operator on a complex Hilbert space H. Then a bounded self-adjoint linear operator A is called a square root of T if (1) A2 = T. If, in addition, A  O, then A is called a positive square root of T and is denoted by T1/2 exists and is unique: Theorem 4.4.2: (Positive square root). Every positive bounded self-adjoint linear operator T: H H on a complex Hilbert space H has a positive square root A, which is unique. This operator A commutes with every bounded linear operator on H which commutes with T. Proof: We proceed in three steps: (a) We show that if the theorem holds under the additional assumption T  I, it also holds without that assumption. 1/2 (b) We obtain the existence of the operator A = T from Anx Ax, where A0 = 0 and 1 2 (2) An+1 = An+ (T – An ), n = 0, 1,…., 2 and we also prove the commutativity stated in the theorem. (c) We prove uniqueness of the positive square root. The details are as follows. (a) If T = 0, we can take A = T1/2 = 0. Let T  0. By the Schwarz inequality, Tx, x)  ||Tx|| ||x||  ||T|| ||x||2 Dividing by ||T||  0 and setting Q = (1/||T||)T, we obtain Ox, x  ||x||2 = Ix, x; that is, O  I. Assuming that Q has a unique positive square root B = Q1/2, we have B2= Q and we see that a square root of T = ||T||Q is ||T||1/2 B because (||T||1/2B)2 = ||T||B2 = ||T||O = T. Also, it is not difficult to see that the uniqueness of Q1/2 implies uniqueness of the positive square root of T. Hence if we can prove the theorem under the additional assumption T  I, we are done.

77 | P a g e

1 1 2 (b) Existence. We consider (2). Since A0 = 0, we have A1 = T. A2 = T - T , etc. 2 8 Each An is a polynomial in T. Hence the An's are self-adjoint and all commute, and they also commute with every operator that T commutes with. We now prove (3) An  I n = 0, 1,….;

(4) An  An+1 n = 0, 1,….; 1/2 (5) Anx Ax, A = T ; (6) ST=TS  AS = SA where S is a bounded linear operator on H. Proof of (3): 2 We have A0  I. Let n > 0. Since I - An-1 is self-adjoint, (I - An-1) s  0. Also T  I implies I - T  0. From this and (2) weobtain (3):

1 2 1 0  (I – An-1) + (I – T) 2 2 1 2 = I – An-1 - (T – A n-1) 2 = I - An. Proof of (4): 1 We use induction. (2) gives 0 = A0  A1 = T. We show that An-1  An for any fixed n 2 implies An  An+1. From (2) we calculate directly

1 2 1 2 An+1 – An = An + (T – An ) – An-1 - (T – A n-1) 2 2 1 = (An – An-1)[I - (An + An-1)] 2 Here An - An-1  0 by hypothesis and [….]  0 by (3). Hence An+1 - An  0 by 9.3-1. Proof of (5): (An) is monotone by (4) and An  I by (3). Hence Theorem 9.3-3. implies the existence of a bounded self-adjoint linear operator A such that Anx Ax for all x  H. Siace (Anx) converges, (2) gives 1 2 An+1x – Anx = (Tx – An x)  0 2 as n  . Hence Tx - A2x = 0 for all x, that is, T = A2. Also A  0 because 0 =

A0  An by (4), that is, Anx, x  0 for every x  H, which implies Ax, x  0 for every x  H, by the continuity of the inner product (cf. 3.2-2). Proof of (6): From the line of text before (3) we know that ST = TS implies AnS = SAn, that is,

AnSx = SAnx for all x  H. Letting n  , we obtain (6). (c) Uniqueness. Let both A and B be positive square roots of T. Then A2 = B2 = T. Also BT=BB2 = B2B = TB, so that AB = BA by (6). Let x  H be arbitrary and y = (A - B)x. Then (Ay, y)  0 and (By, y)  0 because A  0 and B  0. Using AB = BA and A2 = B2, we obtain Ay, y + By, y = (A+B)y, y = (A2-B2)x, y = 0. Hence Ay, y = By, y = 0. Since A  0 and A is self-adjoint, it has itself a positive square root C, that is, C2 = A and C is self-adjoint. We thus obtain 0 = Ay, y = C2y, y = Cy, Cy = ||Cy||2

78 | P a g e and Cy = 0. Also Ay = C2y = C(Cy) = 0. Similarly, By = 0. Hence (A - B)y = 0. Using y = (A - B)x, we thus have for all x  H ||Ax - Bx||2 = (A - B)2x, x = (A - B)y, x = 0. This shows that Ax - Bx = 0 for all x  H and proves that A = B. Applications of square roots will be considered in Sec. 9.8. la-deed, square roots will play a basic role in connection with the spectral representation of bounded self-adjoint linear operators. 4.5 Projection Operators The concept of a projection operator P or, briefly, projection P, was ^defined in Sec. 3.3, where a Hilbert space H was represented as the direct sum of a closed subspace Y and its orthogonal complement Y; thus H = Y Y x = y + z (y  Y, z  Y). Since the sum is direct, y is unique for any given x  H. Hence (1) defines a linear operator P : H H (2) x  y = Px. P is called an orthogonal projection or projection on H. More specifically, P is called the projection of H onto Y. Hence a linear operator P : H  H is a projection on H if there is a closed subspace Y of H such that Y is the range of P and Y is the null space of P and P|Y is the identity operator on Y. Note in passing that in (1) we can now write x = y + z = Px + (1 - P)x. This shows that the projection of H onto Y is I - P. There is another characterization of a projection on H, which is sometimes used as a definition:

The results in this section are extension to projection operator In the following we will assume H+ to be a self-adjoint operator. 2 By δk  l (Z), k  Z we will denote a vector, such that (δk)j = δk,j, j  Z. Theorem 4.5.1 There is a family of projection operators {E+(λ)}λ R corresponding to the operator H+ and the following representations are valid, I dE (  ) and H   dE (  ) RR   Theorem 4.5.2 The following formula is valid,

 P (  )P (  )d(  ,E (  )  ) (4.1) k,j  ,K  ,j 0  0 R In particular, the polynomials P +,j(λ) are orthonormal with respect to the measure d(δ0, E + (λ)δ0) on R. Proof: First, note that because of (H + δj, u) = (δj, H + u) = (H + u)j = aj − 1uj−1 + ajuj+1 + bjuj 2 = (aj−1δj−1 + ajδj+1 + bjδj, u), u  l0 (N0), + H acts on each δj as H+δj = aj−1δj−1 + ajδj+1 + bjδj, j  N0, n where we assume δ−1 = 0. Therefore, δj belongs to the domain of any H +, n  N, and analogously to (3.1) we find that

79 | P a g e

δj = P + j(H+)δ0. Now it is easy to establish (4.1) using δk,j = (δk, δj) = (P+,k (H+)δ0, P+,j(H+)δ0) = (δ0, P+,j(H+)P+,k(H+)δ0)

P (  )P (  )d(  ,E (  )  )  ,j  ,k 0  0 R Definition 4.5.1 Let σ+(λ) = (δ0, E+(λ)δ0), then dσ+(λ) is called the spectral measure associated with H+. Remark 4.5.1 Following the usual conventions we also call dσ+(λ) the spectral measure of H+ even though this terminology is usually reserved for the operatorvalued spectral measure dE+(λ). (This slight abuse of notation should hardly cause any confusion.) Lemma 4.5.3 The set of points of increase of the function σ+(λ) is infinite, that is, for 2 all polynomials P(λ)  L (R, dσ+(λ)), | P( ) |2 d  (  )  0 if and only if P() = 0   R 4.6 Further Properties of Projections Let us start, as indicated, by discussing basic properties of projections. In the first place we show that projections are always positive operators: Theorem 4.6-1: (Positivity, norm). For any projection P on a Hilbert space H, (3) Px, x = ||Px||2 (4) ; P  0 (5) T ||P||  1 ||P|| = 1 if P(H) {0}. Proof. (3) and (4) follow from Px, x = P2x, x = Px, Px = ||Px||2  0. By the Schwarz inequality, ||Px||2 = Px, x  ||Px|| ||x||, so that ||Px||/||x||  1 for every x  0, and ||P||  1. Also ||Px||/||x|| = 1 if x  P(H) and x  0. This proves (5). The product of projections need not be a projection, but we have the basic Theorem 4.6-2: (Product of projections). In connection with products (composites) of projections on a Hilbert space H, the following two statements hold. (a) P = P1P2 is a projection on H if and only if the projections P1 and P2 commute, that is, P1P2 = P2P1. Then P projects H onto Y = Y1  Y2, where Yj = Pj(H). (b) Two closed subspaces Y and V of H are orthogonal if and only if the corresponding projections satisfy PYPv = 0. Proof: (a) Suppose that P1P2 = P2Pl. Then P is self-adjoint, by Theorem 3.10-4. P is idempotent since 2 2 2 P = (P1P2)(P1P2) = P1 P1 = P1P2 = P. Hence P is a projection by 9.5-1, and for every x  H we have Px = P1(P2x) = P2(P1x). Since P1 projects H onto Y1, we must have P1(P2x) Y1. Similarly, P2(P1x) Y2. Together, Px  Y1  Y2 Since x  H was arbitrary, this shows that P projects H into Y = Y1  Y2. Actually, P projects H onto Y. Indeed, if y  Y, then y Y1, y  Y2, and Py = P1P2y = P1y = y.

80 | P a g e

Conversely, if P = P1P2 is a projection denned on H, then P is self-adjoint by 9.5-1, and P1P2 = P2P1 follows from Theorem 3.10-4. (b) If Y V, then Y  V = {0} and PYPvx = 0 for all x  H by part (a), so that PYPv = 0. Conversely, if PYPv = 0, then for every y  Y and v  V we obtain y, v = PYy, Pvv = y, PYPvv = y, 0 = 0. Hence Y  V. Spectral projections Let BR denote the set of all Borel subsets of R.

Definition 4.6.1 The family {P }  BR of bounded operators in H is called a projection-valued measure (p.v.m.) of bounded support if the following conditions (i)–(iv) hold: (i) PΩ is an orthogonal projection for all Ω  BR. (ii) P = 0, there exist a, b  R, a < b such that P(a,b) = I (the bounded support property).  N (iii) If Ω =  Ωi ∩ Ωj =  for i  j, then PΩ = s − limN→∞ P . k1 k k1 k (iv) PPP 1  2  1  2

Next, let A = A*  B(H), Ω  BR. Definition 4.6.2 PΩ(A) = χΩ(A) are called the spectral projections of A. We note that the family {PΩ(A) = χΩ(A)}Ω satisfies conditions (i)–(iv) of BR Definition 3.1. Next, consider a p.v.m. {PΩ}Ω . Then for any h  H, (h, PΩh) is a positive (scalar) measure since properties (i)–(iv) imply all the necessary properties of a positive measure. We will use the symbol d(h, Pλh) to denote the integration with respect to this measure. By construction, the support of every (h, PΩ(A)h) is a subset of σ(A). Hence, if we integrate with respect to the measure (h, PΩh), we integrate over σ(A). If we are dealing with an arbitrary p.v.m. we will denote the support of the corresponding measure by supp(PΩ). Theorem 4.6.3 If {PΩ}Ω is a p.v.m. and f is a bounded Borel function on supp(PΩ), then there is a unique operator B, which we will denote by supp(PΩ) f(λ) dPλ, such that (h,(Bh) f ( )d(h,P h), h  H supp(P )   Proof: A standard Riesz argument. Next, we will show that if PΩ(A) is a p.v.m. associated with A, then f (A) f ( )dP (A) (A) 

First, assume f(λ) = χΩ(λ). Then X ( )d(h,P(A)h)  d(h,P(A)h)  (h,P (A)h) (A)(A)     

= (h, X(A) h) Hence, holds for all simple functions. Next, approximate any measurable function f(λ) by a sequence of simple functions to obtain for bounded Borel functions on σ(A).

81 | P a g e

The inverse statement also holds: If we start from any bounded p.v.m. {PΩ}Ω and BR form A = dP , then χΩ(A) = PΩ(A) = PΩ. This follows from the fact that for supp(P )   such an A, the mapping f f ( )dP forms a functional calculus for A. By supp(P )   uniqueness of the functional calculus one then gets P (A) X (A)  X (  )dP  P  supp(P )     Summarizing, one obtains the following result: Theorem 4.6.4 There is a one-to-one correspondence between bounded self-adjoint operators A and projection-valued measures {PΩ}Ω in H of bounded support given BR by A → {PΩ(A)}Ω = {χΩ(A)}Ω , BR BR {P } A   dP  BR supp(P )   4.7 Spectral Family : We recall from projection operal or(4.5) that our present aim is a representation of bounded self-adjoint linear operators on a Hilbert space in terms of very simple operators (projections) whose properties we can readily investigate in order to obtain information about those more complicated operators. Such a representation will be called a spectral representation of the operator concerned. A bounded self-ad joint linear operator T: H  H being given, we shall obtain a spectral representation of T by the use of a suitable family of projections which is called the spectral family associated with T. In this section we motivate and define the concept of a spectral family in general, that is, without reference to a given operator T. The association of a suitable spectral family with a given operator T will be considered separately, in the next section. 4.8 The spectral family of a bounded selfadjoint Linear operators The construction of the spectral decomposition for unbounded self-adjoint operators will be based on the following theorem. 4.8.1 Definition (Spectral family or decomposition of unity). A real spectral family (or real decomposition of unity) is a one-parameter family  = (E)R of projections E defined on a Hilbert space H (of any dimension) which depends on a real parameter A and is such that (1) E  E, hence EE = EE = E ( < )

(2) lim E x 0 

(3) lim E x 0 

(4) E0 x lim E  x E  x (x  H). 0 We see from this definition that a real spectral family can be regarded as a mapping R  B(H, H)

  E; to each A  R 'there corresponds a projection E  B(H, H), where B(H, H) is the space of all bounded linear operators from H into H.  is called a spectral family on an interval [a, b] if (5) E = 0 for  < a, E = I for   b.

82 | P a g e

Such families will be of particular interest to us since the spectrum of a bounded self- adjoint linear operator lies in a finite interval on the real line. Theorem 4.8.1 Assume A = A*. Then there is a measure space (MA, dμA) with μA a finite measure, a 2 unitary operator UA : H → L (MA, dμA), and a real-valued function fA on MA which is finite a.e., such that 2 (i) ψ  D(A)  fA(・)(UAψ)(・)  L (MA, dμA). −1 (ii) If   U[D(A)], then (UAAUA )(m) = fA(m)(m). To prove this theorem we need some additional constructions. First we will prove a similar result for bounded normal operators. Definition 4.8.1 Let A be a bounded normal operator in H. Then ψ  H is a star- cyclic vector for A if Lin.span{Anm (A*) } H . n,m N0 Lemma 4.8.2 Let A be a bounded normal operator in H with a star-cyclic vector ψ  H. Then there is a measure μA on σ(A), and a unitary operator UA, such that UA : H → 2 L (σ(A), dμA) with -1 (UAAUA f)(λ) = λf(λ). 2 This equality holds in the sense of equality of elements of L (σ(A), dμA). n Proof: Introduce P c ij  ,c  C,n  N and take any p(・)  P . i,j 0 ij ij 

Define UA by UAp(A)ψ = p. (1) One can prove that for all x, y  H there exists a measure μx,y,A on σ(A) such that (p(A)x, y) p(  )d  , p  P (A) x,y,A Then || p(A) ||2  (p(A)*p(A)  ,  )  ((pp)(A)  ,  )  p(  )p(  )d  (A) ,,A 2 = || p || 2 L ( (A),d, ,A )

Next we choose μA = μψ,ψ,A. Since ψ is star-cyclic, UA is densely defined and equation (1) implies that UA is bounded. Thus, UA can be extended to an isometry 2 UA : H→ L (σ(A), dμA). 2 2 Since P (σ(A)) is dense in L (σ(A), dμA), Ran(UA) = L (σ(A), dμA) and UA is invertible. Thus, UA is unitary. Finally, if p  P (σ(A)), then −1 (UAAUA p)(λ) = (UAAp(A)ψ)(λ) = (UA(λ ・ p)(A)ψ)(λ) = λp(λ). 2 By continuity, this can be extended from P (σ(A)) to L (σ(A), dμA). Lemma 4.8.3 Let A be a bounded normal operator on a separable Hilbert space H. N Then there is an orthogonal direct sum decomposition H =  j1 Hj (N ≤ ∞) such that:

(i) For all j: AHj  Hj . (ii) For all j there exists an xj  Hj such that xj is star-cyclic for A| . H j

Proof: Take any h1  0  H. If {p(A)h1 ,p(.) P} = H, then h1 is starcyclic and we are done. Otherwise, denote H1 = {p(A)h1 ,p(.) P}, take any h2  H1, consider H2 =

83 | P a g e

{p(A)h2 ,p(.) P}, etc. Then (i) and (ii) are obvious. To show that {Hj} are orthogonal one computes (p(A)hj, q(A)hk) = (q(A)*p(A)hj, hk) = (( qp )(A)hj, hk) = 0, if j  k. Theorem 4.8.4 Let A be a bounded normal operator on a separable Hilbert space H. Then there is a measure space (MA, dμA) with μA a finite measure, a unitary operator 2 UA H → L (MA, dμA), and a bounded continuous function fA on MA, such that −1 (UAAUA )(λ) = fA(λ)(λ). Proof: Based on Lemmas 4.8.3 and 4.8.4 Now we return to the principal objective of this section: Proof of Theorem 4.6.1: Since R(λ, A) is a bounded normal operator, we can apply −1 −1 −1 Theorem 4.8.4 to (A+i) and get (UA(A+i) UA )(m) = gA(m)(m) for some gA. −1 −1 Since Ker(A + i) = {0}, then gA  0 μA-a.e., so gA is finite μA-a.e. Define fA(m) = −1 gA(m) − i. First, we prove that (i) holds: () Let ψ  D(A). Then there exists a   H such that −1 2 ψ = (A + i)  and UAψ = gAUA. Since fg is bounded, one obtains fA(UAψ)  L (MA, dμA). 2 () Let fA(UAψ)  L (MA, dμA). Then UA = (fA + i)UAψ for some  H. −1 Thus, gAUA = gA(fA + i)UAψ and hence ψ = (A + i)   D(A). Next, we show that (ii) holds: Take any ψ  D(A). Then ψ = (A + i)−1 for some   H and Aψ =  − iψ. Therefore, −1 (UAAψ)(m) = (UA)(m) − i(UAψ)(m) = (gA(m) − i)(UAψ)(m) = fA(m)(UAψ)(m). It remains to show that f is real-valued. We will prove this by contradiction. W.l.o.g. we suppose that Im(f) > 0 on a set of nonzero measure. Then there exists a bounded set B  {z  C | Im(z) > 0} with S = {x  R | f(x)  B}, μA(S) = 0. Hence, Im((χS, fχS)) > 0, implying that multiplication by f is not self-adjoint. Next, we can define functions of an operator A. Let h  Bor(R). Then −1 h(A) = UA Th(fA)UA, where 22 L(M,dAAAA  L(M,d  ) T:h(f )  A  T   h(f(m))(m)   h(fA ) A Using (2), the next theorem follows from the previous facts.

Theorem 4.8.5 Assume A = A*. Then there is a unique map A : Bor(R) → B(H) such that for all f, g  Bor(R) the following statements hold:

(i) A is an algebraic *-homomorphism.

(ii) A (f + g) = A (f) + A (g) (linearity).

(iii) || A (f)||B(H) ≤ ||f|| (continuity)

(iv) If {fn(x)}nN  Bor(R), fn(x)  x for all x R, and |fn(x)| ≤ |x| for all n  N, then n for any ψ  D(A), lim (fn)ψ = Aψ. n

(v) If fn(x)  f(x) for all x  R and fn(x) are uniformly bounded w.r.t. n

(x, n), then (fn) (f) strongly. Moreover, has the following additional properties: (vi) If Aψ = λψ, then (f)ψ = f(λ)ψ.

84 | P a g e

(vii) If f ≥ 0, then (f) ≥ 0. Again, formally, (f) = f(A). 4.9 Spectral representation of Bounded self adjaoint Linear operator :

Definition 4.9.1. The family {PΩ}Ω BR of bounded operators in H is called a projection-valued measure (p.v.m.) if the following conditions (i)–(iv) hold: (i) PΩ is an orthogonal projection for all ΩBR. (ii) P = 0, P(−∞,∞) = I.  N (iii) If Ω = k,  i  j   for i  j, then PΩ = s lim P k1 N k1 k (iv) PPP . 1  2  1  2 It is easy to see that {χΩ(A)} is a p.v.m. From now on {PΩ(A)} will always denote {χΩ(A)}. In analogy to the case of bounded operators we then define g(A) for any g  Bor(R) by  (h,g(A)h) g( )d(h,P (A)h), h  H   where d(h, Pλ(A)h) in (4.3) denotes integration with respect to the measure (h,

PΩ(A)h). One can show that the map g → g(A) coincides with the map g → A (g) in Theorem 4.6. At this point we are ready to define g(A) for unbounded functions g. First we introduce the domain of the operator g(A) as follows:  D(g(A)) h  H |g(  )|2 d(h,P(A)h      One observes that D(g(A)) = H. Then g(A) is defined by  (h,g(A)h) g( )d(h,P(A)h) , h  D(g(A))   We write symbolically, g(A) g( )dP (A) (A)  Summarizing, one has the following result: Theorem 4.9.1 There is a one-to-one correspondence between self-adjoint operators A and projection-valued measures {PΩ}Ω in H given by BR  A dP   If g is a real-valued Borel function on R, then  g(A) g( )dP (A) ,    D(g(A)) h  H |g(  )|2 d(h,P(A)h)       is self-adjoint. If g is bounded, g(A) coincides with (g) 4.9.2 Spectral Theorem for Bounded Self-Adjoint Linear Operators. Let T : H H be a bounded self-adjoint linear operator on a complex Hilbert space H. Then : (a) T has the spectral representation M (1) T dE m0  where = (E) is the spectral family associated with T (cf. 9.8-3); the integral is to be

85 | P a g e understood in the sense of uniform operator convergence [convergence in the norm on B(H, H)], and for all x, y  H, M (1*) Tx, x = dw( ) w() = Ex, y m0 where the integral is an ordinary Riemann-Stieltjes integral (sec. 4.4). (b) More generally, if p is a polynomial in  with real coefficients, say, n n-1 p() = n + n-1 + …. + 0, then the operator p(T) defined by n n-1 p(T) = nT + n-1T + ……. + 0I has the spectral representation M (2) p(T) p( )dE m0  and for all x, y  H, M (2*) p(T)x, y = p( )dw( ), w() = Ex, y m0 Theorem 4.9.3: (Properties of p(T)). Let T be as in the previous theorem, and let p, p1 and p2 be polynomials with real coefficients. Then : (a) p(T) is self-adjoint. (b) If p() = p1() + p2(), the p(T) = p1(T) + p2(T). (c) If p() = p1()p2(), then p(T) = p1(T) = p1(T)p2(T). (d) If p()  0 for all   [m, M], then p(T)  0

(e) If p1()  p2() for all   [m, M], then p1(T)  p2(T). (f) ||p(T)||  max |p()|, where J = [m, M] J (g) If a bounded linear operator commutes with T, it also commutes with p(T). 4.10 Extension of the Spectral Theorem to continuous Functions Theorem 4.9.1 holds for p(T), where T is bounded self-adjoint linear operator and p is a polynomial with real coefficients. We want to extend theorem to operators f(T), where T is as before and f is a continuous real-valued function. Clearly, we must first define what we mean by f(T). Theorem 4.10.1 Let T : H H be a bounded self-adjoint linear operator on a complex Hilbert space H. Let f be a continuous real-valued function on [m, M], where (1) m inf Tx,x , ||x|| 1 as before. Then by the Weierstrass approximation theorem 4.11-5 there is a sequence of polynomials (pn) with real coefficients such that

(2) pn() f() Uniformly on [m, M]. Corresponding to it we have a sequence of bounded self-adjoint linear operators pn(T). By Theorem 9.9-2 (f), ||pn(T) – pr(T)||  max |pn() – pr()|, J where J = [m, M]. Since pn() f(), given any  > 0, there is an N such that the right-hand side is smaller than  for all n, r > N. Hence (pn(T)) is Cauchy and has a limit in B(H, H) since B(H, H) is complete (cf. 2.10-2). We define f(T) to be that limit; thus

(3) pn(T) f(T). Or course, to justify this definition of f(T), we must prove that f(T) depends only on f (and T, of course) but not on the particular choice of a sequence of polynomials converging to f uniformly.

86 | P a g e

Proof1 Let ( pn ) be another sequence of polynomials with real coefficients such that

pn () f() uniformly on [m, M]. Then pn (T)  f (T) by the previous argument, and we must show that f (T) = f(T). Clearly,

pn () – pn() 0, hence pn (T) – pn(T) 0, again by 4.9-2(f). Consequently, given  > 0, there is an N such that for n < N,  || f (T) - p (T)|| < n 3  || p (T) – pn(T)||< n 3  ||pn(T) – f(T)||< 3 By the triangle inequality it flows that

|| f (T) – f(T)||  || f (T) - pn (T)|| + || pn (T) – pn(T)|| + ||pn(T) – f(T)|| < . Since  > 0 was arbitrary, f (T) – f(T) = 0 and f (T) = f(T). 4.11 Properties of the Spectral Family of a Bounded Self-Adjoint Linear Operator It is interesting that the spectral family = (E) of a bounded self-adjoint linear operator T on a Hilbert space H reflects properties of the spectrum in a striking and simple fashion. We shall derive results of that kind form the definition of (cf. Sec. 4.8) in combination with the spectral representation in Sec. 4.9 From sec. 4.7 we know that if H is finite dimensional, the spectral family = (E) has “points of growth” (discontinuities, jumps) precisely at the eigenvalues of T. In fact E E 0if and only if 0 is an eigenvlaue of T. It is remarkable, although 00  o perhaps not unexpected, that this property carries over to the infinite dimensional case : Theorem 4.11-1: (Eigenvalues). Let T : H H be a bounded self-adjoint linear operator on a complex Hilbert space H and  = (E) the corresponding spectral family. Then   E has a discontinuity at and  = 0 (that is, EE ) if and only if 00  o

0 is an eigenvalue of T. In this case, the corresponding eigenspace is (1) N(T - 0I) = ( EE ) (H). 00  o Proof: 0 is an eigenvalue of T if and only if N(T - 0I)  {0}, so that the first statement of the theorem follows immediately from (1). Hence it sufficies to prove (1). We write simply F0 = EE 00  o and prove (1) by first showing that (2) F0(H)  N(T - 0I) and then (3) F0(H)  N(T - 0I). Proof of (2) : 1 Inequality in Sec. 4.8 with  = 0 - and  = 0 is n 1 (4) 0 E(  0 )  TE(  0 )   0 E(  0 ) n

87 | P a g e where 0 = (0 – 1/n, 0]. We let n . Then E(0) F0, so that (4) yields 0F0  TF0  0F0

Hence TF0 = 0F0, that is, (T - 0I)F0 = 0. This proves (2). Proof of (3) : Let x  N(T - 0I). We show that then x  F0(H), that is, F0x = x since F0 is a projection. If 0  [m, M], then 0  (T) by 9.2-1. Hence in this case N(T - 0I) = {0}  F0(H) since F0(H) is a vector space. Let 0  [m, M] By assumption, (T - 0I)x = 0. This 2 implies (T - 0I) x = 0, that is by 9.9-1. b (   )dw(  )  0 w() = Ex, x a 0 2 where a < m and b > M. Here ( - 0)  0 and  Ex, x is monotone increasing by 4.7-1. Hence the integral over any subinterval of positive length must be zero. In particular, for every  > 0 we must have     000 (   )dw()2    2 dw()    2 E x,x aa0 0  and bb 0 (  )dw()2  2 dw()  2 Ix,x  2 E x,x  0   0  00 Since  > 0, for this and 4.5-2 we obtain  E x,x  = 0 hence Ex= 0 0  0  and  E x,x  = 0 hence x E x = 0 0  0  We may thus write x (E E )x 00   

If we let  0, we obtain x = F0x because   E is a continuous form the right. This implies (3), as was noted before.

4.12 Assignment Q 1. It was mentioned in the text that for a self-adjoint linear operator T the inner product Tx, x is real. What does this imply for matrices ? Q 2. Let T : H H and W : H  H be bounded linear operators on a complex Hilbert space H. If T is self-adjoint, show that S = W*TW is self- adjoint. 2 2 Q 3. Let T : l l be defined by (1, 2,….) (0, 0, 1, 2,….). Is T bounded ? Self-adjoint ? Find S : l2  l2 such that T = S2. Q 4. What theorem about the eigenvlaues of a Hermitian matrix A = (jk) do we obtain from Theorem 9.2-1 ? Q 5. Show that the spectrum of a bounded self-adjoint linear operator on a complex Hilbert space H  {0} is not empty, using one of of the theorems of the present section. Q 6. Show that a compact self-adjoint linear operator T : H  H on a complex Hilbert space H  {0} has at lest one eigenvalue. Q 7. Show that a bounded self-adjoint linear operator T on a complex Hilbert space H is positive if and only if its spectrum consists of nonnegative real values only. What does this imply for a matrix ?

88 | P a g e

Q 8. Let T : H  H and W : H H be bounded linear operators on a complex Hilbert space H and S W*TW. Show that if T is self-adjoint and positive, so is S. Q 9. Let S and T be bounded self-adjoint linear operators on a Hilbert space H. If S  0, show that TST  0. Q 10. Show that if T  0, then (I + T)-1 exists. Q 11. Let T be any bounded linear operator on a complex Hilbert space. Show that the inverse of I + T*T exists. Q 12. If T : H H and S : H  H are bounded linear operators and T is compact and S*S  T*T, show that S is compact. Q 13. Let T : H H be a bounded linear operator on an infinite dimensional complex Hilbert space H. If there is a c > 0 such that we have ||Tx||  c||x|| for all x  H, show that T is not compact. Q 14. Find operators T : R2 R2 such that T2 = I, the identity operator. Indicate which of the square roots is the positive square root of I. 4.13 Check your progress

Q 1. Let T : L2[0, 1] L2[0, 1] be defined by (Tx)(t) = tx(t). (Cf. 3.1-5) Show that T is self-adjoint and positive and find its positive square root. Q 2. Let T : H H be a bounded positive self-adjoint linear operator on a complex Hilbert space. Using the positive square root of T, show that for all x, y  H, |Tx, y  Tx, x1/2Tx, y1/2. Q 3. If S and T are positive bounded self-adjoint linear operators on a complex Hilbert space H and S2 = T2, show that S = T. Q 4. Show that a projection P on a Hilbert space H satisfies 0  P  1. Under what conditions will (i) P = 0, (ii) P = I? Q 5. Let Q = S-1PS : H H, where S and P are bounded and linear. If P is a projection and S is unitary, show that Q is a projection.

Q 6. If a sum P1 + …. Pk of projections Pj : H H(H a Hilbert space) is a projection, show that 2 2 2 ||P1x|| + … + ||Pkx||  ||x|| . Q 7. Show that the difference P = P2 – P1 of two projections on a Hilbert space H is a projection on H if and only if P1  P2. Q 8. (Limit of Projections) If (Pn) is a sequence of projections defined on a Hilbert

space H and Pn P, show that P is a projection defined on H. Q 9. (Invariant subspace) Let T : H H be a bounded linear operator. Then a subspace Y  H is said to be invariant under T if T(Y)  Y. Show that a closed subspace Y of H is invariant under T if and only if Y is invariant under T*. Q 10. (Reduction of an operator) A closed subspace Y of a Hilbert space H is said reduce of linear operator T : H  H if T(Y)  Y and T(Y)  (Y), that is, if both Y and Y are invariant under T. (The point is that then the investigation of T can be facilitated by considering T| and T| separately.) Y Y If P1 is the projection of H onto Y and P1T = TP1, show that Y reduces T.

89 | P a g e

Q 11. Prove that T-T = TT-. Q 12. Find T+, T-, (T2)1/2 and the other square roots of T2 if 20 T   03 Q 13. If in the finite dimensional case a linear operator T is represented by a real diagonal matrix T , what is the spectrum of T? How do we obtain from T the matrix (a) T + (representing T+), (b) T - (representing T-), (c) B (representing B) ? Q 14. Verify for the zero operator T = 0; H H. Q 15. If an operator T : R3  R3 is represented, with respect to an orthonormal basis, by a matrix 0 1 0  1 0 0 , 0 0 1 What is corresponding spectral family ? Using the result, verify (1) for this operator. 2 2 Q 16. Find the spectral family of the operator T : l l defined by (1, 2,

3,….,) (1/1, 2/2, 3/3….). Find and orthonormal set of eigenvectors. What from does (1) take in this case ?

4.14 Point for discussion/clarification

90 | P a g e

4.15 Suggested Study material 1. Kelley J. L. (1955) General topology. New Yark Van Nostrand. 2. Kreyszing E (1970) Introductory Mathematical statistics. New York wieley. 3. Edward R. E. (1965) Functional Analysis. New York.

91 | P a g e

Unit -5

OBJECTIVES: After reading this unit you should be above to;  Define canonical product and prove many results of it  Understand the concept of order of an entire function.  Understand the meaning of exponent of convergence and prove a number of important results.  Define univalent function.

STRUCTURE: 5.1 Canonical Product 5.1.1 Jensen’s formula 5.1.2 Poisson Jensen formula 5.1.3 Handmard’s there circle theorem 5.2 Order of an entire function 5.2.1 Definition 5.2.2 Theorem 5.2.3 Problems 5.3 Exponent of convergence 5.3.1 Definition 5.3.2 Proposition 5.4 Borel’s Theorem 5.5 Hadamard’s Factorization Theorem. 5.6 The Range of an analytic function 5.6.1 Bloch theorem 5.6.2 Little picard theorem 5.6.3 Montel Caratheodory 5.6.4 Great picrd Theorem 5.7 Univalent Function 5.7.1 Definition 5.7.2 Theorems 5.7.3 Meaning of 5.7.4 Statement of Bieberbach’s Conjecture 5.7.5 theorem

5.8 Summary 5.9 Self assessment problems 5.10 Points of Discussion 5.11 Further Readings

92 | P a g e

5.1 Canonical Product: This product is defined for entire function. Its concept is derived from weiestrass factorization theorem.

Definition: The product of the form

∏ ( ) 8 ( ) ( ) 9

Is called canonical product if it is convergent and represents an entire function with the condition that.

∑ ( )

is converges for all R. Here is called the genus of the canonical product. Example: Canonical product of sin z is

∏ 4 5

5.1.1 Jensen’s Formula: Let ( ) be analytic in the closed disc | | . Assume that ( ) and no zeros of ( ) lie on| | . If are the zeros of ( ) in the open disc | | each repeated as often as its multiplicity then

| ( )| ∑ ( ) ∫ | ( )| | |

Proof; let us define a function

̅ ( ) ( ) ∏ ( ) ( )

as ( ) is given to be analytic, hence ( ) is also analytic and hence

( ) on | | Also ( ) is never zero on an open disc | | for | ( )| | ( )| | |

Let , Then

93 | P a g e

̅ ̅ |∏ | |∏ | ( )

( ̅ ) ∏ ( ̅ )

̅ ∏ | | | | ̅

Now we can apply Gauss mean value theorem for | ( )|

| ( )| ∫ | ( )| ( )

From (1)

( ) ( ) ∏ ( )

| ( )| | ( )| ∑ | |

Also | ( )| | ( )| | |

| ( )| | ( )| Putting these values in eq(2), ewe get

| ( )| ∑ ∫ | ( )| | |

5.1.2 Poisson – Jensen Formula : Let ( ) be analytic in the closed disc | | . Assume that ( ) and no zeros of ( ) lies on | | . If are the zeros of

94 | P a g e

( ) in the open disc | | , each repeated as often as it’s multiplicity and ( ) then | ( )|

̅ ∑ | | ( )

( ) | ( )| ∫ ( )

Proof : Let us consider a function

̅ ( ) ( ) ∏ ( ) ( )

( ) is analytic in any domain in which f(z) is analytic and ( ) for | |

( ) is analytic and never 0 on an open disc | | for some

Hence | ( )| | ( )| | | Let

and hence

̅ |∏ | ( )

Again as F (z) is analytic and non zero on an open disc |z| < P, P > R.

log F (z) is analytic in |z| < P Real part of log |F (z)| is harmonic so by poisson’s formula for log |F (z)|,

( ) | ( )| Log |F (z)| = ∫ d ….. (2) ( )

as log |F (R )| = log |f (R )| on |z| = R

95 | P a g e

again from eq (1),

| ( )| | ( )| ∑ | | ( )

putting these value in eq (2) we get

| ( )|

∑ | | ( )

( ) | ( )| ∫ ( )

5.1.3 Hadamard’s Three circles Theorem: Theorem Let f (z) be analytic in the annular region r1 |z| r3, and r1, r2, r3 if M1, M2, M3 be the max |f (z)| on the three circles |z| = r1, r2, r3 respectively, then

( ) ( ) ( )

Proof. Let F (z) = zk f (z), where k is a ral constant to be determined later. Then F (z) is analytic in the annulus r1 |z| r3.

If k is not an integer, then F (z) is a multivalued function, and for convenience, one choose the principal branch. For this we cut the annulus along the negative real axis and we obtain a domain in which the principal branch of this function is analytic. By the maximum modulus principle, the maximum value of |F (z)| is attained on the boundary of the cut annulus. Now consider branch of this function which is

analytic in the part of the annulus for which arg z

, we see that the principal value cannot attain its

maximum modulus on the cut. Hence the max |F (z)| occurs

96 | P a g e

on one o f the bounding circles. Thus it is shown that when r1 |z| r3, k k |F (z)| max {r 1 M1, r 3 M3}

Hence, on |z| = r2, are must have k k k r 2 M2 max { r 1 M1, r 3 M3}

k k r 2 M2 r 1 M1 or M2 . / M1 …………. (1)

Since k is at our choice, we choose k to our best advantage by making the expressions in the parentheses on the right equal to each other. Thus k is defined by the equation

k k r 1 M1 = r 3 M3

whence we have k log r1 + log M1 = k log r3 + log M3

k [log r1 - log r3] = log M3 - log M1

k = - {log }/{log } ………………. (2)

By (1), we have

. / . /

- k log

Now substituting the value of k in the above inequality, we obtain

. / . / . /

. / .

. / . /

= . / .

97 | P a g e

log b [a = ( ) = = ( ) = ]

. / . /

5.2 Order of an entire function : we have already understood the concept of entire function in the previous unit now we define order of an entire function and for this we must know what is M (r)

M (r) = max {|f (z)| : |z| = r}

5.2.1 Definition : An entire function f (z) is said to be of finite order is M (r) exp ( ) where

= in f { : M (r) exp ( ) for all larger}

Similarly if M (r) > exp ( ) then the order of f (z) is considered to be finite

5.2.2 Theorem : Let f be a non constant function Define

= in f { ( ) exp ( ) for sufficiently large r} and

( )

Then =

Proof : For the proof of the theorem we consider two cases Case I : If is finite and > 0 be given then R ( ) > 0 such that

M (r) < exp ( ), r R ( ) …….. (1)

Thus for all r > R ( ), in eq (1) we take logar than twice Log log M (r) < ( ) log r

98 | P a g e

( )

Hence ( )

As is arbitrary so as 0

……….. (2)

Case II : If is finite and > 0 be given then

( )

log log M (r) < ( ) log r =

log M (r) < exp ( ( ))

= ( )

log M (r) < exp ( ( ))

in f ( ( ) ( ) for larger)

Again as us arbitray so an We get

………… (3)

From (2) and (3) we get

99 | P a g e

5.2.3 Problems : Find the order of ,

Solution let M (r) = | |

So ( )

(| | )

Then we apply L-hospital rule

| | | |

Hence order of

5.3 Exponent of Convergence : This is a very common term associated with the sequence of complex numbers.

5.3.1 Definnition : Let { } be a sequence of non-zero complex numbers such that |z,| |z2| ……….. |zn| as n .

The exponent of convergence of the sequence is define by

= inf { > : ∑ < } | |

5.3.2 Proposition : The convergence of exponent of a sequence * + is given by

| |

Proof : Let is consider as a finite quantity and hence we define a series

100 | P a g e

∑ | |

Clearly for < the series is convergent

| |

Let such that

| | | |

log n - log | | < 0

| |

……………. (1) | |

Now we consider be any other arbitrary number exceeding the

R.H.S. of (1)

( ) such that

| |

Thus | |

∑ is convergent for | |

Now from the definitin of exponent of convergence

| | ……………. (2)

101 | P a g e

From (1) and (2) we get

| |

5.4 Barel’s Theorem : The order of a canonical product is equal to the convergence exponent of its zeros.

Proof : Let us consider and as the order and convergence exponent of a conical product P (z). Then we know that

So to prove the theorem it is sufficient to show that

So let us consider

( ) ( ) 4 5

If we consider P > 0

a, b > 0 Such that | ( )| ( | | ) = ( | | ) If c = a + b, and |w| | and | ( )| ,( ) - ( | | ) ( | | )

On the other hand if |w| and P > 0

Then |E (w, p)| exp ( | | ) and if

102 | P a g e

| ( )| ( | | ) again if ½ | | | and p > 0 we get | ( ) ( | | )| = | ( ) ( | | )|

Now let us consider {z1 , z2 , z3 ……}

Be the sequence of zeros of the canonical product P (z). Then by definition h will be the genus and

also by definition of exponent of convergence, we have

{ ∑| | }

∑| |

( ) ∏ . /

We get

| ( )| ∑ . | | /

( | | ) Or |P (z)| ( | | ) Or

103 | P a g e

Hence we get

5.5 Hadamard’s factorization Theorem : Let f is an entire function of finite order then f has finite genus

Let ( ) be an entire function of finite order . Then ( ) ( ) ( ) Where is the order of the (possible) zeros of ( ) at is a polynomial of degree not exceeding and ( ) is the canonical product associated with the sequence of non- zero zeros of ( )

Proof: By Weierstrass’s factorization theorem, the entire function ( ) can be represented in the form given by ( ) ( ) ( ) Where ( ) is an entire function. Here we have the additional hypothesis that ( ) is of finite order . This additional hypothesis is used to show that ( ) is a polynomial. Its is evident that the division of ( )by does not affect either the hypothesis or the conclusion of the theorem and so it is sufficient to consider the repsrentation ( ) ( ) ( ) ( ) ( ) ( ) ( ) Therefore | ( )| | | ( )

( ) ( ) ( ) | | | | ( ) ( )

( ) | ( )| | ( )|

By the definition of order, it follows that

( ) ( )

For sufficiently large | | and all Therefore

104 | P a g e

| ( )| Suppose is the convergence exponent of the non-zero zeros of ( ), then

Also by Borel’s theorem, is the order of the cannonical product ( ), and so it follows from theorem II of the proceeding section . That | ( )| | | Thus, | ( )|

Form (2) and (3), we obtain | ( )| | ( )|

Hence by (1). We get ( ( )

Since is large, we conclude from Theorem Ii of the section . That ( ) is a polynomial of degree not exceeding . So the genus of . This completes the proof.

5.6 Range of an analytic function: Analytic functions can have a variety of range . Under this article we will prove many important results which given an idea about the range of an analytic function

5.6.1 Bloch’s Theorem: Let be an analytic function in a region containing the closure of the disc * | | + and satisfying ( ) ( ) . Then there is a disc S⊂D in which is one-one and such that ( ) contains a disc of

radius

Proof: Let us consider ( ) *| ( )| | | +

( ) ( ) ( ) , - is continuous and ( ) ( ) and ( ) ( )

105 | P a g e

Let * ( ) + ( ) ( ) Also maximum value of ( ) is 1

Now we choose as | | and | ( )| ( ) ( ) ( ) ( )

( ) ( ) | ( )| ( ) ( )

Also

| ( )| 4 ( )5 | |

[ ( ) ] ( ) | |

( )

| ( )| | | ( )

Form (1) and (2)

| ( )| | ( )| | ( )| | ( )|

Now we define ( ) ( ) ( )

Then by Schwarz’s lemma | ( )| | | * | | +

( ) ( ) | | ( )

| |

| ( ) ( )|

Again if . /

| |

| ( )| | ( )| | ( )|

106 | P a g e

So we get that is one-one as we have a result “ Let be am analytic function in the disc ( ) such that

| ( )| | ( )| | ( )| ( )

Now to prove the theorem , it remains to show that ( ) contains a disc of radius .

So we define

. / ( ) ( ) ( )

( ) ( ) ( )

| ( )|

Also

| ( )| . /

then we can apply “Let be analytic in ( ) ( ) | ( )| | ( ) |

( ( )) 4 5

and we get

4 . /5 ( )

. / . /

( ⁄ ) Hence

( ( ) )

{ | ( )| }

{ ( ) | ( | ( ) }

107 | P a g e

{ ( ) | ( ) | ( )}

* | | + ( )

4 . /5 ( )

( ) ( ( ) )

5.6.2 The little Picard Theorem: To prove this theorem firstly we should know the concept of branch of logarithm. If G is an open connected set in c an is a continuous function such that ( ) Called a branch of logarithm.

Statement: Let be an function that omits two values. Then is a constant. Proof: As it is given that omits two values so Let ( ) ( ) ( ) ( )

then the function ( ) omits 0 and 1

So we can assume that the given entire function omits two value 0 and 1 ( ) ( ) we get a function such that ( ) contains no disc of radius 1.

Now our aim is to prove that is constant, so on the contrary let is not constant is not constant

a point such that ( )

So without the loss of generality we may suppose that ( )

108 | P a g e

Then we use the result “ let be an analytic function in the region that contains ( ( )) contains a disc of radius | ( )| ( ( ))contains a disc of radius | ( )| If we choose R sufficiently large then ( ) does not contain a contradiction, hence is constant.

5.6.3 Montel Caratheodary Theorem: Statement: Let F be the family of all analytic functions defined in a region G that do not assume the value 0 and 1, then is normal in ( ).

Proof: Choose a point in G and keep it fix. Define the families and by *( | ( )| + And *( | ( )| + So

So U H. We now show that is normal. in H(G) and that is normal in ( ) Note that by considering the sequence of constant functions * + in defined by ( ) we see that .

To show that * | ( )| +11 is normal is H(G) we shall invoke Montel's theorem; that is, it is just sufficient to show that is locally bonded.

Let be arbitrary and let be a curve in from

z0 to a. Suppose Do, D1...... Dn be discs in G with centre z0,z1...... z, = a, respectively on * + The discs are

so constructed that zk - 1 and zk are in Dk - 1 Dk for

. Assume that ̅ ⊂ .

Now Applying Schottky's Theorem to D0, there is a

constant CO such that

| ( )|

We notice that if D0 = B(z0; r) and R > r is such that

109 | P a g e

B(zo; R) C G then by Corollary of Theorem 1 of §124, we have

| ( )| ( )

whenever is choosen in such a way that .

In particular, f ( ) so that Schottky's Theorem gives

that y is uniformly bounded by a constant C1 on D1. Continuing this process we have that is uniformly bounded on D. Since was arbitrary, it follows that is locally bounded. Hence by Montel's Theorem, is normal in H(G).

Next, we consider * | ( )| + Then 1/f is analytic on G because f never vanishes. Also, 11f never assumes the value 1; moreover | ( )| 1. Hence, if we define ̅ * +

then we see that ̅ ⊂ and ̅ is normal in H(G). It

follows that if * + is a sequence in then there is a

subsequence * + and an analytic function h defined in G

such that in H(G). Hence, either or never

vanishes.

( ) If it is easy to show that uniformly on compact subsets of G.

In case never vanishes, we see that is analytic and it follows that ( ) ( ) uniformly on compact subsets of G.

5.6.4 The Great Picard theorem: Let be an analytic function that has an essential singularity at Then in each neighbourhood of assumes each

110 | P a g e

complex numbers, with one possible exception, an infinite number of times. Proof. Without loss of generality we may suppose that has an essential singularity at z = 0. To effect this we may consider ( ) if necessary. Suppose, if possible, there is an such that there are two numbers not in * ( ) + we will obtain a contradiction. If ( ) ( ) for all in * + then the function defined by ( ) ( )

omits the values 0 and 1. So we may suppos that ( ) amd ( ) for | | .

Let ( ) * + and define by

( ) . /

Then each is analytic in , since is analytic and no assumes the value 0 or 1, Now , by Montel-Caratheodary Theorem, * + is a normal family in ( ).

Let * + be a subsequence of * + such that

( ) uniformly on . / 2 | | 3, where

is either analytic in G or . If is analytic, let

2| ( )| 3 then for e = M>0, such that for

all k , we have

( ) ( ) ( ) | ( )| | ( )| | | | |

( ) ( ) ( ) | | | |

. / |Z| =

Thus

| ( )|

111 | P a g e

For sufficiently large and |Z| = ; hence | ( )|

for sufficiently large and |Z| =

Now, by Maximum Modulus Principle, f is uniformly bounded on concentric annuli about zero. It follows that f is bounded by 2M on a deleted neighbouhood of zero. This shows that z = 0 must be a removable singularity. Therefore, cannot be analytic and so . Further, we see that if . Then f must have a ple at zero.

Hence we conclude that there is at most one complex number that is never assumed by f.

However, if there is a complex number w which is assumed only a finite number of times then by taking a sufficiently small disc, we again arrive at a punctured disc in which f fails to assume two values.

5.7 Univalent Functions : In this section we will study a special type of function

5.7.1 Definition : A function f (z) is said to be univalent in a domain if it is analytic in and assumes no value more than once in .

The condition is , implies ( ) ( ). Points where ( ) vanishes, ( ) is not univalent.

5.7.2 Theorem 1 : Prove that if w = f (z) is univalent in a domain D then the inverse mapping z = g (w) is univalent in f (D) Proof : As D and f (D) have one-to-one correspondence z is a function of w i.e. z = g (w)

g (w) assume no value in D more than one. Then by inverse function theorem, g (w) is analytic at each point of f (D)

g (w) is analytic in f (D).

112 | P a g e

Hence if is an essential singularity for f, so by weierstrass theorem, f (z) comes arbitrary close to in every neighbourhood of .

But this is a contradiction as f is simple is a pole of f is a polynomial and clearly the polynomial must have degres 1 otherwise ( ) would have atleast one zero. And hence a univalent function which maps | | onto | | must be linear.

5.7.3 Meaning of : Let us consider U as unit disc and f is

a complex function defined in U if and if ( ) =

( ) ( )

Exists for every Then f is holomorphic in U. Then the class of all holomorphic function in U is denoted by H (U). The class of all f H (U) which are in one to one in unit disc U and satisfy ( )

5.7.4 Bieberbach Conjecture: This was defined by Bieberback in 1916 and proved by L de Branges in 1984 If and

( ) ∑

then | | for all and ( ) ( )

113 | P a g e

5.7.5 ¼ Theorem: Statement: Let and

( ) ∑

then

( )| | ( ) ( ) ( )

Proof (a) We have a result which says if then there exists a such that ( ) ( )

we get such that ( ) ( ) Let G =1/g

( ) √ ( )

√ ∑

( ∑ )

4 5

∑

then as we have |a,|≤1 |a2|≤ 2 Proof (b) Let us consider ( ) we define ( ) ( ) ( ) as ( )

114 | P a g e

( ), and hence is one to one so by define if and ( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) [ ]

( ) Also, ( ) ( ), ( ) - ( ) ( ) ( ), -

( ), -

( )

So that ( ) . / It follows that

( ) and ( ) Thus, Finally, we see that

| | | | | | | |

| | | |

| | , | | -

| | ( )

( ) ( ) ( )

This completes the proof.

5.8 Summary:  Canonical product is unique

 Order of

 Order of √

 For a finite sequence exponent of convergence is zero  for an entire function

115 | P a g e

 The order of a canonical product is equal to the convergence exponent of its zero.  If f is an entire function of finite order then f has finite genus  If f (z) is an entire function then order of ( ) and f (z) is same  Let f be an entire function that omits two values. Then f is constant.  Univalent function is one-one

5.9 Self assessment problems:

(1) Find the order of sin z, (2) Use Hadamard’s factorization theorem to show that

∑ 4 5

(3) For each and , 0 < < and 0 < < there is a constant ⊂ ( ) such that if f is an analytic function defined in some simply connected region containing ̅( ) that omits the values o and 1 and such that |f (0)| then | ( )| ( ) for | | (Schottky’s theorem)

(4) check Z+1/Z is univalent or not.

5.10 Points for Discussion ______

116 | P a g e

______5.11 Further Readings (1) Complex Analysis by Dr. H.K. Pathak (2) Complex Analysis by L.V. Ahlfors. (3) Complex and Real Analysis by walter Rudin.

117 | P a g e