Normed Spaces & Banach Spaces

Home , Banach space, Bounded set, Norm (mathematics), Sequence space

Contents

1 Normed Spaces. Banach Spaces. 2 1.1 Vector Space...... 2 1.2 NormedSpace.BanachSpace...... 10 1.3 FurtherPropertiesofNormedSpaces...... 16 1.4 FiniteDimensionalNormedSpacesandSubspaces...... 26 1.5 Linear Operators...... 32 1.6 Bounded and Continuous Linear Operators...... 38 1.7 LinearFunctionals...... 46 1.8 Linear Operators and Functionals on Finite Dimensional Spaces...... 55 1 Normed Spaces. Banach Spaces.

1.1 Vector Space. Definition 1.1. 1. An arbitrary subset M of a vector space X is said to be linearly independent if every non-empty finite subset of M is linearly independent. 2. A vector space X is said to be finite dimensional if there is a positive integer n such that X contains a linearly independent set of n vectors whereas any set of n +1 or more vectors of X is linearly dependent. n is called the dimension of X, written n = dim X. 3. If X is any vector space, not necessarily finite dimensional, and B is a linearly independent subset of X which spans X, then B is called a basis (or Hamel basis) of X. Hence if B is a basis for X, then every nonzero x X has a unique repre- • sentation as a linear combination of (finitely many!2 ) elements of B with nonzero scalars as coecients.

Theorem 1.2. Let X be an n-dimensional vector space. Then any proper subspace Y of X has dimension less than n.

1. Show that the set of all real numbers, with the usual addition and multiplication, constitutes a one-dimensional real vector space, and the set of all complex numbers constitutes a one-dimensional complex vector space.

Solution: The usual addition on R and C are commutative and associative, while scalar multiplication on R and C are also associative and distributive. For R,the zero vector is 0R =0 R,theidentityscalaris1 =1 R,andtheadditive 2 R 2 inverse is x for any x R.ForC,thezerovectoris0 =0+0i C,the 2 C 2 identity scalar is 1 =1+0i C and the additive inverse is z for all z C. C 2 2

2. Prove that 0x = 0, ↵0 = 0 and ( 1)x = x.

Solution: 0x =(0+0)x =0x +0x = 0 =0x +( (0x)) ) =0x +0x +( (0x)) =0x + 0 =0x. ↵0 = ↵(0 + 0)=↵0 + ↵0 = 0 = ↵0 +( (↵0)) ) = ↵0 + ↵0 +( (↵0)) = ↵0 + 0 = ↵0. ( 1)x =( 1(1))x = 1(1x)= x.

Page 2 3. Describe the span of M = (1, 1, 1), (0, 0, 2) in R3. { }

Solution: The span of M is

1 0 span M = ↵ 1 + 0 : ↵, R 8 2 3 2 3 2 9 < 1 2 = 4 ↵5 4 5 : ; = ↵ : ↵, R . 82 3 2 9 < ↵ +2 = 4 5 We see that span M corresponds: to the plane x = y on; R3.

4. Which of the following subsets of R3 constitute a subspace of R3? Here, x = (⇠1,⇠2,⇠3).

(a) All x with ⇠1 = ⇠2 and ⇠3 =0.

Solution: For any x, y W and any ↵, R, 2 2

⇠1 ⌘1 ↵⇠1 + ⌘1 ↵x + y = ↵ ⇠ + ⌘ = ↵⇠ + ⌘ W. 2 23 2 23 2 2 23 0 0 0 2 4 5 4 5 4 5

(b) All x with ⇠1 = ⇠2 +1.

2 3 Solution: Choose x = 1 W , x = 2 W ,then 1 2 3 2 2 3 0 2 0 2 4 5 4 5 2 3 5 x + x = 1 + 2 = 3 / W 1 2 2 3 2 3 2 3 0 0 0 2 4 5 4 5 4 5 since 5 =3+1. 6

Solution: Choose ↵ = 1, then for any x W , ↵x / W . 2 2 (d) All x with ⇠ ⇠ + ⇠ = k. 1 2 3 Solution: For any x, y W , 2

⇠1 + ⌘1 x + y = ⇠ + ⌘ . 2 2 23 ⇠3 + ⌘3 4 5

Page 3 Since

⇠ + ⌘ (⇠ + ⌘ )+(⇠ + ⌘ )=(⇠ ⇠ + ⇠ )+(⌘ ⌘ + ⌘ )=2k. 1 1 2 2 3 3 1 2 3 1 2 3 we see that W is a subspace of R3 if and only if k =0.

j 5. Show that x1,...,xn ,wherexj(t)=t ,isalinearlyindependentsetinthespace C[a, b]. { }

Solution: This is a simple consequence of Fundamental Theorem of Alge- bra. Fix a ﬁnite n>1. Suppose that for all t [a, b], we have 2 n n j jxj(t)= jt =0. j=1 j=1 X X

Suppose n = 0. Fundamental Theorem of Algebra states that any polynomials with degree 6n can have at most n real roots. Since the equation above is true for n all t [a, b], and the set of points in the interval [a, b]isuncountable, tj has 2 j j=1 to be the zero polynomial. Since n 1wasarbitrary(butﬁnite),thisshowsthatX any non-empty ﬁnite subset of xj j ⇤,⇤acountable/uncountableindexingset, 2 is linearly independent. { }

6. Show that in an n-dimensional vector space X, the representation of any x as a linear combination of a given basis vectors e1,...,en is unique.

Solution: Let X is an n-dimensional vector space, with a basis e1,...,en . Suppose any x X has a representation as a linear combination{ of the basis} vectors, we claim2 that the representation is unique. Indeed, if x X has two representations 2

x = ↵1e1 + ...+ ↵nen = 1e1 + ...+ nen.

subtracting them gives

n (↵ )e + ...+(↵ )e = (↵ )e = 0. 1 1 1 n n n j j j j=1 X

Since e1,...,en is a basis of X,bydeﬁnitionitislinearlyindependent,which implies{ that ↵ } =0forallj =1,...,n,i.e.therepresentationisunique. j j

7. Let e1,...,en be a basis for a complex vector space X. Find a basis for X regarded as a{ real vector} space. What is the dimension of X in either case?

Page 4 Solution: A basis for X regarded as a real vector space is e1,...,en, ie1, . . . , ien . The dimension of X is n as a complex vector space and 2n{ as a real vector space.}

8. If M is a linearly dependent set in a complex vector space X,isM linearly dependent in X,regardedasarealvectorspace?

1 i Solution: No. Let X = 2,withK = ,andconsiderx = and y = . C C i 1   x, y is a linearly dependent set in X since ix = y. Now suppose K = R,and { } ↵ + i 0+0i ↵x + y = = 0 = . ↵i 0+0i   Since ↵, can only be real numbers, we see that (↵,)=(0, 0) is the only solution to the equation. Hence x, y is a linearly independent set in X = C2 over R. { }

9. On a ﬁxed interval [a, b] R,considerthesetX consisting of all polynomials with real coecients and of degree⇢ not exceeding a given n,andthepolynomialx =0(for which a degree is not deﬁned in the usual discussion of degree). (a) Show that X, with the usual addition and the usual multiplication by real numbers, is a real vector space of dimension n + 1. Find a basis for X.

Solution: Let X be the set given in the problem. It is clear that X is a real vector space. Indeed, for any P, Q X,withdeg(P ), deg(Q) n, 2  deg(P + Q) n and deg(↵P) n for any ↵ R.Asimilarargumentfrom Problem 5 shows that 1,t,t2,...,t n is a linearly2 independent set in X,and since 1,t,t2,...,tn spans{ X,itisabasisof} X and X has dimension n +1. { } (b) Show that we can obtain a complex vector space X˜ in a similar fashion if we let those coecients be complex. Is X asubspaceofX˜?

Solution: No. Consider P (t)=t X,choose↵ = i,then↵P(t)=it / X. 2 2

10. If Y and Z are subspaces of a vector space X,showthatY Z is a subspace of X, but Y Z need not be one. Give examples. \ [

Solution: Let Y and Z be subspaces of a vector space X.Takeanyx, y Y Z, note that x, y are both elements of Y and Z.Forany↵, K, ↵x +2y \ Y (since Y is a subspace of X)and↵x+y Z (since Z is a subspace2 of X). Hence2 ↵x + y Y Z. 2 2 \

Page 5 ↵ 0 For the second part, consider Y = : ↵ R and Z = : ↵ R .It 0 2 ↵ 2 ⇢ ⇢ can be (easily) deduced that Y and Z are subspaces of R2,butY Z is not a 1 0 [ 1 subspace of R2.Toseethis,choosex = and y = ,thenx + y = / 0 1 1 2 Y Z.    [

11. If M = ? is any subset of a vector space X,showthatspanM is a subspace of X. 6

Solution: This is immediate since a (scalar) field K is closed under addition and sums of two finite sums remain finite.

12. (a) Show that the set of all real two-rowed square matrices forms a vector space X. What is the zero vector in X?

Solution: This follows from Problem 1 and the deﬁnition of matrix addition and matrix scalar multiplication: we prove that R is a vector space over R 00 or .ThezerovectorinX is 0 = . C 00 

(b) Determine dim X. Find a basis for X.

Solution: We claim that dim X = 4. To prove this, consider the following four vectors in X 10 01 00 00 e = e = e = e = . 1 00 2 00 3 10 4 01     ↵ ↵ Suppose ↵ e + ...+ ↵ e = 0 = 1 2 ,wehave↵ = ↵ = ↵ = ↵ =0, 1 1 4 4 ↵ ↵ 1 2 3 4  3 4 i.e. e1,e2,e3,e4 is a linearly independent set in X. However, any set of 5 { } ab or more vectors of X is linearly dependent, since any x = X can be cd2  written as a linear combination of e1,e2,e3,e4 ,i.e. x = ae1 +be2 +ce3 +de4. Hence, a basis for X is e ,e ,e ,e{ . } { 1 2 3 4} (c) Give examples of subspaces of X.

↵ 0 Solution: An example is W = : ↵ R . 00 2 ⇢

(d) Do the symmetric matrices x X form a subspace of X? 2

Page 6 a b a b Solution: Yes. Consider any symmetric matrices x = 1 1 ,y = 2 2 . b d b d  1 1  2 2 For any ↵, R (or C), 2 ↵a + a ↵b + b ↵x + y = 1 2 1 2 ↵b + b ↵d + d  1 2 1 2 which is a symmetric matrix.

(e) Do the singular matrices x X form a subspace of X? 2 11 32 Solution: No. To see this, consider x = and y = ;both 11 64 x and y are singular matrices since they have zero determinant. However, 43 x + y = is not a singular matrix since det(x + y)=20 21 = 1 =0. 75 6 

13. (Product) Show that the Cartesian product X = X1 X2 of two vector spaces over the same ﬁeld becomes a vector space if we deﬁne the⇥ two algebraic operations by

(x1,x2)+(y1,y2)=(x1 + y1,x2 + y2),

↵(x1,x2)=(↵x1,↵x2).

Solution: This is a simple exercise. We ﬁrst verify vector addition:

(x1,x2)+(y1,y2)=(x1 + y1,x2 + y2)

=(y1 + x1,y2 + x2)

=(y1,y2)+(x1,x2).

(x1,x2)+[(y1,y2)+(z1,z2)] = x1 +(y1 + z1),x2 +(y2 + z2) ⇣ ⌘ = (x1 + y1)+z1, (x2 + y2)+z2

=(⇣x1 + y1,x2 + y2)+(z1,z2) ⌘

= (x1,x2)+(y1,y2) +(z1,z2).

(x1,x2)=(hx1 + 0,x2 + 0) i

=(x1,x2)+(0, 0). (0, 0)=(x +( x ),y +( y )) 1 1 1 1 =(x ,y )+( x , y ). 1 1 1 1 Next, we verify scalar vector multiplication:

(x1,x2)=(1K x1, 1K x2)

=1K (x1,x2).

Page 7 ↵ (x1,x2) = ↵(x1,x2) h i = ↵(x1),↵(x2) ⇣ ⌘ = (↵)x1, (↵)x2

=(⇣↵)(x1,x2). ⌘

↵ (x1,x2)+(y1,y2) = ↵(x1 + y1,x2 + y2) h i = ↵(x1 + y1),↵(x2 + y2)

=(⇣↵x1 + ↵y1,↵x2 + ↵y2)⌘

=(↵x1,↵y1)+(↵x2,↵y2)

= ↵(x1,y1)+↵(x2,y2).

(↵ + )(x1,x2)= (↵ + )x1, (↵ + )x2

=(⇣↵x1 + x1,↵x2 + x2⌘)

=(↵x1,↵x2)+(x1,x2)

= ↵(x1,x2)+(x1,x2).

14. (Quotient space, codimension) Let Y be a subspace of a vector space X.The coset of an element x X with respect to Y is denoted by x + Y and is deﬁned to be the set 2 x + Y = x: v = x + y, y Y . { 2 } (a) Show that the distinct cosets form a partition of X.

Solution:

(b) Show that under algebraic operations deﬁned by

(w + Y )+(x + Y )=(w + x)+Y ↵(x + Y )=↵x + Y

these cosets constitute the elements of a vector space. This space is called the quotient space (or sometimes factor space)ofX by Y (or modulo Y )andis denoted by X/Y . Its dimension is called the codimension of Y and is denoted by codim Y ,thatis, codim Y =dim(X/Y ).

Solution:

3 15. Let X = R and Y = (⇠1, 0, 0): ⇠1 R . Find X/Y , X/X, X/ 0 . { 2 } { }

Page 8 Solution: First, X/X = x + X : x X ;sincex + X X for any x X, we see that X/X = 0 . Next,{ X/ 02 = }x + 0: x X 2= x: x X =2 X. For X/Y ,weareabletodeduce(geometrically)thatelementsof{ } { } { 2 } { X/Y2 are} lines parallel to the ⇠1-axis. More precisely, by deﬁnition,X/Y = x + Y : x X ;for 0 0 0 { 2 } aﬁxedx0 =(⇠1 ,⇠2 ,⇠3 ),

0 0 0 x0 + Y = (⇠1 ,⇠2 ,⇠3 )+(0, 0,⇠3): ⇠3 R { 2 } 0 0 ˜ ˜ = (⇠1 ,⇠2 , ⇠3): ⇠3 R . { 2 } which corresponds to a line parallel to ⇠1-axis.

Page 9 1.2 Normed Space. Banach Space. Deﬁnition 1.3. A norm on a (real or complex) vector space X is a real-valued function on X whose value at an x X is denoted by x and which has the properties 2 k k (N1) x 0. k k (N2) x =0 x =0. k k () (N3) ↵x = ↵ x . k k | |k k (N4) x + y x + y .(Triangle inequality) k kk k k k Here, x and y are arbitrary vectors in X and ↵ is any scalar.

AnormonX deﬁnes a metric d on X which is given by • d(x, y)= x y ,x,y, X k k 2 and is called the metric induced by the norm.

The norm is continuous, that is, x x is a continuous mapping of (X, )intoR. • 7! k k k·k

Theorem 1.4. A metric d induced by a norm on a normed space X satisﬁes (a) d(x + a, y + a)=d(x, y).

(b) d(↵x, ↵y)= ↵ d(x, y). for all x, y, a X| and| every scalar ↵. 2 This theorem illustrates an important fact: Every metric on a vector space might • not necessarily be obtained from a norm.

Acounterexampleisthespaces consisting of all (bounded or unbounded) sequences • of complex numbers with a metric d deﬁned by

1 1 ⇠ ⌘ d(x, y)= | j j| . 2j 1+ ⇠ ⌘ j=1 j j X | |

Page 10 1. Show that the norm x of x is the distance from x to 0. k k

Solution: x = x y = d(x, 0), which is precisely the distance from x to k k k k y=0 0. 2. Verify that the usual length of a vector in the plane or in three dimensional space has the properties (N1) to (N4) of a norm.

3 2 2 2 Solution: For all x R ,deﬁne x = x1 + x2 + x3. (N1) to (N3) are obvious. (N4) is an easy consequence2 ofk thek Cauchy-Schwarz inequality for sums. p More precisely, for x =(x1,x2,x3)andy =(y1,y2,y3) we have:

2 2 2 2 x + y =(x1 + y1) +(x2 + y2) +(x3 + y3) k k 2 2 2 2 2 2 =(x1 + x2 + x3)+(y1 + y2 + y3)+2(x1y1 + x2y2 + x3y3) x 2 + y 2 +2 x y =( x + y )2. k k k k k kk k k k k k Taking square root on both sides yields (N4).

3. Prove (N4) implies y x y x . k kk k k k Solution: From triangle inequality of norm we have the following two inequalities:

x = x y + y x y + x k k k kk k k k = x y x y . )kkk kk k y = y x + x y x + x k k k kk k k k = y x x y . )kkk kk k Combining these two yields the desired inequality.

4. Show that we may replace (N2) by x =0 = x =0withoutalteringtheconcept of a norm. Show that nonnegativityk ofk a norm) also follows from (N3) and (N4).

Solution: For any x X, 2 x = x + x x x + x + x =2 x + x =3 x k k k kk k k k k k k k k k = 0 2 x = 0 x . )  k k ) k k

n 1/2 5. Show that x = ⇠ 2 = ⇠ 2 + ...+ ⇠ 2 deﬁnes a norm. k k | j| | 1| | n| j=1 ! X p

Page 11 Solution: (N1) to (N3) are obvious. (N4) follows from the Cauchy-Schwarz inequality for sums, the proof is similar to that in Problem 3.

6. Let X be the vector space of all ordered pairs x =(⇠1,⇠2), y =(⌘1,⌘2), of real numbers. Show that norms on X are deﬁned by ··· x = ⇠ + ⇠ k k1 | 1| | 2| x =(⇠2 + ⇠2)1/2 k k2 1 2 x =max ⇠1 , ⇠2 . k k1 {| | | |}

Solution: (N1) to (N3) are obvious for each of them. To verify (N4), for x = (⇠1,⇠2)andy =(⌘1,⌘2),

x + y = ⇠ + ⌘ + ⇠ + ⌘ k k1 | 1 1| | 2 2| ⇠ + ⌘ + ⇠ + ⌘ = x + y . | 1| | 1| | 2| | 2| k k1 k k1

2 2 2 x + y 2 =(⇠1 + ⌘1) +(⇠2 + ⌘2) k k 2 2 2 2 =(⇠1 + ⇠2 )+(⌘1 + ⌘2)+2(⇠1⌘1 + ⇠2⌘2) x + y +2 x y =( x + y )2 k k2 k k2 k k2k k2 k k2 k k2 = x + y x + y . )k k2 k k2 k k2

x + y =max ⇠1 + ⌘1 , ⇠2 + ⌘2 k k1 {| | | |} max ⇠ + ⌘ , ⇠ + ⌘  {| 1| | 1| | 2| | 2|} max ⇠1 , ⇠2 +max ⌘1 , ⌘2 = x + y .  {| | | |} {| | | |} k k1 k k1 where we use the inequality a max a , b for any a, b R. | | {| | | |} 2

1/p 1 7. Verify that x = ⇠ p satisﬁes (N1) to (N4). k k | j| j=1 ! X

Solution: (N1) to (N3) are obvious. (N4) follows from Minkowski inequality for sums.Moreprecisely,forx =(⇠j)andy =(⌘j),

1 1 1 p p p 1 1 1 x + y = ⇠ + ⌘ p ⇠ p + ⌘ p . k k | j j|  | k| | m| j=1 ! ! m=1 ! X Xk=1 X

8. There are several norms of practical importance on the vector space of ordered n- tuples of numbers, notably those deﬁned by x = ⇠ + ⇠ + ...+ ⇠ k k1 | 1| | 2| | n|

Page 12 1/p x = ⇠ p + ⇠ p + ...+ ⇠ p (1

Solution: This is a generalisation of Problem 6, with the proof being almost identicall. The only thing that di↵ers is we use Minkowski inequality for sums to prove (N4) for . k·kp

9. Verify that x =max x(t) deﬁnes a norm on the space C[a, b]. k k t [a,b] | | 2

Solution: (N1) and (N2) are clear, as we readily see. For (N3), for any scalars ↵ we have: ↵x =max ↵x(t) = ↵ max x(t) = ↵ x . k k t [a,b] | | | | t [a,b] | | | |k k 2 2 Finally, for (N4),

x + y =max x(t)+y(t) max x(t) +max y(t) = x + y . k k t [a,b] | |t [a,b] | | t [a,b] | | k k k k 2 2 2

10. (Unit Sphere) The sphere S (0) = x X : x =1 . 1 { 2 k k } in a normed space X is called the unit sphere.ShowthatforthenormsinProblem 4 4 1/4 6andforthenormdeﬁnedby x 4 =(⇠1 + ⇠2 ) ,theunitsphereslookasshownin ﬁgure. k k

Solution: Refer to Kreyszig, page 65.

11. (Convex set, segment) AsubsetA of a vector space X is said to be convex if x, y A implies 2 M = z X : z = ↵x +(1 ↵)y, 0 ↵ 1 A. { 2   }⇢ M is called a closed segment with boundary points x and y;anyotherz M is called an interior point of M.Showthattheclosed unit ball in a normed space2X is convex.

Solution: Choose any x, y B˜ (0), then for any 0 ↵ 1, 2 1   ↵x +(1 ↵)y ↵ x +(1 ↵) y ↵ +1 ↵ =1. k k k k k k This shows that the closed unit ball in X is convex.

Page 13 12. Using Problem 11, show that

2 (x)= ⇠ + ⇠ | 1| | 2| ⇣p p ⌘ does not deﬁne a norm on the vector space of all ordered pairs x =(⇠1,⇠2)ofreal numbers. Sketch the curve (x)=1.

Solution: Problem 11 shows that if is a norm, then the closed unit ball in a normed space X =(X, )isconvex.Choosex =(1, 0) and y =(0, 1), x, y are elements of the closed unit ball in (X, )since (x)= (y) = 1. However, if we choose ↵ =0.5,

2 (0.5x +0.5y)= 0.5 + 0.5 =(2p0.5)2 =2> 1. | | | | ⇣p p ⌘ This shows that for 0.5x +0.5y is not an element of the closed unit ball in (X, ), and contrapositive of result from Problem 11 shows that (x)doesnotdeﬁnea norm on X.

13. Show that the discrete metric on a vector space X = 0 cannot be obtained from a norm. 6 { }

Solution: Consider a discrete metric space X = 0 . Choose distinct x, y X, for ↵ =2,d(2x, 2y)=1but 2 d(x, y)=2.Thestatementthenfollowsfrom6 { } 2 theorem. | |

14. If d is a metric on a vector space X = 0 which is obtained from a norm, and d˜ is deﬁned by 6 { } d˜(x, x)=0, d˜(x, y)=d(x, y)+1 (x = y), 6 show that d˜ cannot be obtained from a norm.

Solution: Consider a metric space X = 0 .Chooseanyx X,for↵ =2, d˜(2x, 2x)=d(2x, 2x)+1=1but 2 d˜(x,6 x)=2({ } d(x, x)+1)=2.Thestatement2 then follows from theorem. | |

15. (Bounded set) Show that a subset M in a normed space X is bounded if and only if there is a positive number c such that x c for every x M. k k 2

Solution: Suppose a subset M in a normed space X is bounded. By deﬁnition, the diameter (M)ofM is ﬁnite, i.e.

(M)= sup d(x, y)= sup x y < . x,y M x,y M k k 1 2 2

Page 14 Fix an y M,thenforanyx M, 2 2 x = x y + y x y + y (M)+ y < . k k k kk k k k k k 1 Choosing c = (M)+ y yields the desired result. Conversely, suppose there exists an c>0suchthatk kx c for all x M.Thenforanyx, y M, k k 2 2 d(x, y)= x y x + y 2c. k kk k k k Taking supremum over x, y M on both sides, we obtain (M) 2c< .This shows that M (in a normed2 space X)isbounded.  1

Page 15 1.3 Further Properties of Normed Spaces. Deﬁnition 1.5. A subspace Y of a normed space X is a subspace of X considered as a vector space, with the norm obtained by restricting the norm on X to the subset Y . This norm on Y is said to be induced by the norm on X.

Theorem 1.6. A subspace Y of a Banach space X is complete if and only if the set Y is closed in X.

Deﬁnition 1.7.

1. If (xk) is a sequence in a normed space X, we can associate with (xk) the sequence (sn) of partial sums sn = x1 + x2 + ...+ xn where n =1, 2,....If(s ) is convergent, say, s s as n , then the n n ! ! 1 1 inﬁnite series xk is said to converge or to be convergent, s is called the k=1 sum of the inﬁniteX series and we write

1 s = xk = x1 + x2 + .... Xk=1

1 2. If x + x +...converges, the series x is said to be absolutely convergent. k 1k k 2k k Xk=1 3. If a normed space X contains a sequence (e ) with the property that for every x X, n 2 there is a unique sequence of scalars (↵n) such that

x (↵ e + ...+ ↵ e ) 0 as n . k 1 1 n n k! ! 1

1 then (en) is called a Schauder basis for X. The series ↵jej which has the j=1 X sum x is called the expansion of x with respect to (en), and we write

1 x = ↵jej. j=1 X If X has a Schauder basis, then it is separable. The converse is however not • generally true.

Theorem 1.8. Let X =(X, ) be a normed space. There exists a Banach space Xˆ and an isometry A from X ontok·k a subspace W of Xˆ which is dense in Xˆ. The space Xˆ is unique, except for isometries.

Page 16 1. Show that c l1 is a vector subspace of l1 and so is c0, the space of all sequences of scalars converging⇢ to zero.

Solution: The space c consists of all convergent sequences x =(⇠j)ofcomplex numbers. Choose any x =(⇠j),y =(⌘j) c l1,withlimit⇠,⌘ C respectively. For ﬁxed scalars ↵,,theresultistrivialiftheyarezero,sosupposenot.Given2 ⇢ 2 any ">0, there exists N1,N2 N such that 2 " ⇠ ⇠ < for all j>N. | j | 2 ↵ 1 |" | ⌘ ⌘ < for all j>N. | j | 2 2 | | Choose N =max N ,N ,thenforallj>Nwe have that { 1 2} ↵⇠ + ⌘ ↵⇠ ⌘ = ↵(⇠ ⇠)+(⌘ ⌘) | j j | | j j | ↵ ⇠j ⇠ + ⌘j ⌘ | || " | | "|k k < ↵ + = ". | |2 ↵ | |2 | | | | This shows that the sequence ↵x + y =(↵⇠ + ⌘ )isconvergent,hencex c. j j 2 Since ↵, were arbitrary scalar, this proves that c is a subspace of l1.By replacing ⇠ = ⌘ = 0 as limit, the same argument also shows that c0 is a subspace of l1.

2. Show that c0 in Problem 1 is a closed subspace of l1,sothatc0 is complete.

Solution: Consider any x =(⇠ ) c¯ ,theclosureofc.Thereexistsx =(⇠n) j 2 0 n j 2 c0 such that xn x in l1. Hence, given any ">0, there exists an N N such that for all n !N and all j we have 2 " ⇠n ⇠ x x < . | j j|k n k 2 in particular, for n = N and all j.Sincex c ,itsterms⇠N form a convergent N 2 0 j sequence with limit 0. Thus there exists an N1 N such that for all j N1 we have 2 " ⇠N < . | j | 2 The triangle inequality now yields for all j N the following inequality: 1 " " ⇠ ⇠ ⇠N + ⇠N < + = ". | j|| j j | | j | 2 2 This shows that the sequence x =(⇠ ) is convergent with limit 0. Hence, x c . j 2 0 Since x c¯ was arbitrary, this proves closedness of c in l1. 2 0 0

Page 17 3. In l1,letY be the subset of all sequences with only ﬁnitely many nonzero terms. Show that Y is a subspace of l1 but not a closed subspace.

Solution: Consider any x =(⇠j),y =(⌘j) Y l1,thereexistsNx,Ny N 2 ⇢ 2 such that ⇠j =0forallj>Nx and ⌘j =0forallj>Ny.Thusforanyscalars ↵,, ↵⇠ + ⌘ =0forallj>N=max N ,N ,and↵x + y Y .Thisshows j j { x y} 2 that Y is a subspace of l1. However, Y is not a closed subspace. Indeed, consider asequencex =(⇠n) Y deﬁned by n j 2 1 if j n, ⇠n = j  j 8 < 0 if j>n. 1 : Let x =(⇠ )= ,thenx x in l1 since j j n ! ✓ ◆ 1 xn x l1 =sup ⇠j = 0asn . k k j>n | | n +1 ! ! 1

but x/Y since x has inﬁnitely many nonzero terms. 2

4. (Continuity of vector space operations) Show that in a normed space X,vector addition and multiplication by scalars are continuous operation with respect to the norm; that is, the mappings deﬁned by (x, y) x+y and (↵, x) ↵x are continuous. 7! 7!

Solution: Consider any pair of points (x0,y0) X X.Givenany">0, choose " 2 ⇥ = = > 0. Then for all x satisfying x x < and all y satisfying 1 2 2 k 0k 1 y y <, k 0k 2 x + y (x + y ) x x + y y < + = ". k 0 0 kk 0k k 0k 1 2

Since (x0,y0) X X was arbitrary, the mapping deﬁned by (x, y) x + y is continuous with2 respect⇥ to the norm. 7!

Choose any scalar ↵ .Consideranynonzerox X.Givenany">0, choose " 0 0 2 " 1 = > 0and2 > 0suchthat(1 + ↵0 )2 = .Thenforall↵ satisfying 2 x0 | | 2 ↵ ↵k 0and2 = > 0. Then for all ↵ satisfying 2 1+ ↵0 ↵ ↵ < and all x satisfying x <, | | | 0| 1 k k 2 ↵x = ↵ x ↵ ↵ + ↵ x k k | |k k | 0| | 0| k k ⇣ ⌘ < + ↵ 1 | 0| 2 ⇣ ⌘ " = 1+ ↵0 ⇠⇠⇠ = ". | | ⇠1+⇠ ↵0 ⇣ ⌘ | | Since ↵0 and x0 were arbitrary scalars and vectors in K and X,themapping deﬁned by (↵, x) ↵x is continuous with respect to the norm. 7!

5. Show that xn x and yn y implies xn + yn x + y.Showthat↵n ↵ and x x implies! ↵ x !↵x. ! ! n ! n n !

Solution: If x x and y y,then n ! n ! x + y x y x x + y y 0asn . k n n kk n k k n k! ! 1 If ↵ ↵ and x x,then n ! n ! ↵ x ↵x = ↵ x ↵ x + ↵ x ↵x k n n k k n n n n k = ↵ (x x)+(↵ ↵)x k n n n k ↵ x x + ↵ ↵ x | n|k n k | n |k k C x x + ↵ ↵ x  k n k | n | k k 0 0 ! ! where we use the fact that convergent| sequences{z } | are{z bounded} for the last inequality.

6. Show that the closure Y¯ of a subspace Y of a normed space X is again a vector subspace.

Solution: If x, y Y¯ ,thereexistssequencesxn,yn Y such that xn x and y y.Thenforanyscalars2 ↵,, 2 ! n ! ↵x + y (↵x + y) ↵ x x + y y 0asn . k n n k| |k n k | |k n k! ! 1

This shows that the sequence (↵xn + yn) Y converges to ↵x + y,which implies that ↵x + y Y¯ . 2 2

Page 19 7. (Absolute convergence) Show that convergence of y + y + y + ... may k 1k k 2k k 3k not imply convergence of y1 + y2 + y3 + ....

Solution: Let Y be the set of all sequences in l1 with only ﬁnitely many nonzero n terms, which is a normed space. Consider the sequence (yn)=(⌘j ) Y deﬁned by 2 1 if j = n, ⌘n = j2 j 8 0ifj = n. < 6 1 1 1 : Then y = and < . However, y + y + y + ... y,where k nk n2 n2 1 1 2 3 ! n=1 1 X y = ,since n2 ✓ ◆ n 1 yj y = 0asn . (n +1)2 ! ! 1 j=1 X but y/Y . 2

8. If in a normed space X,absoluteconvergenceofanyseriesalwaysimpliesconvergence of that series, show that X is complete.

Solution: Choose (xn)beanyCauchysequenceinX.Givenanyk N,there 1 2 exists Nk N such that for all m, n Nk, xm xn < ;byconstruction,(Nk) 2 k k 2k is an increasing sequence. Consider the sequence (yk)deﬁnedbyyk = xNk+1 xNk . Then 1 1 1 1 y = x x < < . k kk k Nk+1 Nk k 2k 1 Xk=1 Xk=1 Xk=1 1 This shows that the series yk is absolute convergent, which is also convergent k=1 by assumption. Thus, X

n 1 yk =lim yk n k=1 !1 k=1 X Xn

=lim xN xN n k+1 k !1 Xk=1 =limxN xN < . n n+1 1 !1 1

Hence, (xNn+1 )isaconvergentsubsequenceof(xn), and since (xn)isaCauchy sequence, (xn)isconvergent.Since(xn)wasanarbitraryCauchysequence,X is complete.

Page 20 9. Show that in a Banach space, an absolutely convergent series is convergent.

1 Solution: Let x be any absolutely convergent series in a Banach space k kk k=1 X.SinceaBanachspaceisacompletenormedspace,itsuX cestoshowthatthe sequence (sn) of partial sums sn = x1 + x2 + ...+ xn is Cauchy. Given any ">0, there exists an N N such that 2 1 x <". k kk k=XN+1 For any m>n>N,

s s = x + x + ...+ x k m nk k n+1 n+2 mk xn+1 + xn+2 + ...+ xm km k k k k k = x k kk k=Xn+1 1 x  k kk k=Xn+1 1 x <".  k kk k=XN+1

This shows that (sn)isCauchyandthedesiredresultfollowsfromcompleteness of X.

10. (Schauder basis) Show that if a normed space has a Schauder basis, it is separable.

Solution: Suppose X has a Schauder basis (en). Given any x X, there exists auniquesequenceofscalars( ) K such that 2 n 2 x ( e + ...+ e ) 0asn . k 1 1 n n k! ! 1 e Consider the sequence (f ) X deﬁned by f = n . Note that f =1forall n ⇢ n e k nk k nk n 1and(fn)isaSchauderbasisforX;indeed,ifwechooseµj = j ej K, then k k2 n n

x µjfj = x jej 0asn . ! ! 1 j=1 j=1 X X In particular, for any x X,givenany ">0, there exists an N N such that 2 2 n " x µjfj < for all n>N. 2 j=1 X

Page 21 Deﬁne M to be the set

n ˜ M = ✓jfj : ✓j K,n N . 2 2 ( j=1 ) X

where K˜ is a countable dense subset of K.Sinceµj K,givenany">0, there " 2 exists an ✓ K˜ such that µ ✓ < for all j =1,...,n.Then j 2 | j j| 2n

n n n n

x ✓jfj x µjfj + µjfj ✓jfj  j=1 j=1 j=1 j=1 X Xn X X " < + µ ✓ f 2 | j j|k jk j=1 X " " n ◆ < + ◆1 2 2⇢n ◆ ◆j=1 " " X = + = ". 2 2 This shows that there exists an y M in any "-neighbourhood of x.Sincex X was arbitrary, M is a countable dense2 subset of X and X is separable. 2

11. Show that (e ), where e =( ), is a Schauder basis for lp,where1 p<+ . n n nj  1

p Solution: Let x =(⇠j)beanysequenceinl ,wehave

1 p 1 ⇠ p 0asn . | j| ! ! 1 j=n+1 ! X

Now choose a sequence of scalars (n) C deﬁned by j = ⇠j, 2 1 p 1 x ( e + ...+ e ) = ⇠ p 0asn . k 1 1 n n k | j| ! ! 1 j=n+1 ! X p This shows that (en)=(nj)isaSchauderbasisforl . Uniqueness?

12. (Seminorm) A seminorm on a vector space X is a mapping p: X R satisfying (N1), (N3), (N4). (Some authors call this a pseudonorm.) Show that!

p(0) = 0, p(y) p(x) p(y x). | | (Hence if p(x)=0 = x = 0,thenp is a norm.) )

Page 22 Solution: Using (N3),

p(0)=p(0x)=0p(x)=0.

Using (N4), for any x, y X, 2 p(y) p(y x)+p(x).  p(x) p(x y)+p(y)=p(y x)+p(y).  = p(y) p(x) p(y x). )| |

13. Show that in Problem 12, the elements x X such that p(x)=0formasubspace N of X and a norm on X/N is deﬁned by 2xˆ = p(x), where x xˆ andx ˆ X/N. k k0 2 2

Solution: Let N be the set consisting of all elements x X such that p(x)=0. For any x, y N and scalars ↵,, 2 2 0 p(↵x + y) p(↵x)+p(y)= ↵ p(x)+ p(y)=0.   | | | |

This shows that N is a subspace of X. Consider xˆ 0 = p(x), where x xˆ and xˆ X/N.Westartbyshowing is well-deﬁned.k k Indeed, for any u,2 v xˆ, 2 k·k0 2 there exists nu,nv N such that u = x + nu and v = x + nv.SinceN is a subspace of X, 2

0 p(u) p(v) p(u v) = p(n n ) =0. | || | | u v | xˆ = p(x) 0. •kk0

Supposex ˆ = 0ˆ = N,then xˆ 0 = p(x) = 0. Now suppose xˆ 0 =0,then • p(x)=0 = x N = kxˆ =k 0ˆ. Thus, (N2) is satisfied. k k ) 2 ) For any nonzero scalars ↵,anyy ↵xˆ can be written y = ↵x + n for some • n N.Thus, 2 2 n ↵xˆ = p(↵x + n)= ↵ p x + k k0 | | ↵ = ↵ xˆ⇣ . ⌘ | |k k0 If ↵ =0,then0ˆx = N and 0ˆx =0bydefinitionofN. k k0 Lastly, for anyx, ˆ yˆ X/N, • 2 xˆ +ˆy = p(x + y) p(x)+p(y) k k0  = xˆ + yˆ . k k0 k k0 Thus, (N4) is satisfied.

Page 23 14. (Quotient space) Let Y be a closed subspace of a normed space (X, ). Show that a norm on X/Y is deﬁned by k·k k·k0

xˆ 0 =inf x . x xˆ k k 2 k k wherex ˆ X/Y ,thatis,ˆx is any coset of Y . 2

Solution: Deﬁne xˆ 0 as above. Also, recall that X/Y = xˆ = x + Y : x X and its algebraic operationsk k are deﬁned by { 2 }

uˆ +ˆv =(u + Y )+(v + Y )=(u + v)+Y = u[+ v. ↵uˆ = ↵(u + Y )=↵u + Y = ↵u.

(N1) is obvious. • c

Ifx ˆ = 0ˆ = Y ,then xˆ 0=0 since 0 Y .Conversely,suppose xˆ 0 = • inf x =0.Propertiesofinﬁmumgivesthatthereexistsaminimisingk k 2 k k x xˆ 2 k k sequence (xn) xˆ such that xn 0 0, with limit x = 0.SinceY is closed, anyx ˆ 2X/Y is closed,k thisk implies! that 0 xˆ,andˆx = 0ˆ.Thus, (N2) is satisﬁed.2 2

For any nonzero scalars ↵, •

↵xˆ 0 =inf ↵x + y k k y Y k k 2 y = ↵ inf x + | | y Y ↵ 2 = ↵ inf x + y y Y | | 2 k k = ↵ xˆ . | |k k0 If ↵ =0,then 0ˆx = 0x = 0ˆ =0=0 xˆ . Thus, (N3) is satisﬁe.d k k0 k k0 k k0 k k0 For anyu, ˆ vˆ X/Y , • 2 c

uˆ +ˆv 0 =inf u + y1 + v + y2 y1,y2 Y k k 2 k k inf u + y1 + v + y2 y1,y2 Y  2 k k k k =inf u + y1 + inf v + y2 y1 Y y2 Y 2 k k 2 k k = uˆ + vˆ . k k0 k k0

Page 24 15. (Product of normed spaces) If (X1, 1)and(X2, 2)arenormedspaces,show that the product vector space X = X k·Xk becomes ak· normedk space if we deﬁne 1 ⇥ 2 x =max x , x , where x =(x ,x ). k k k 1k1 k 2k2 1 2 n o

Solution: (N1) to (N3) are obvious. To verify (N4), for x =(x1,x2),y = (y ,y ) X X , 1 2 2 1 ⇥ 2 x + y =max x + y , x + y k k k 1 1k1 k 2 2k2 n o max x + y , x + y  k 1k1 k 1k1 k 2k2 k 2k2 n o max x , x +max y , y  k 1k1 k 2k2 k 1k1 k 2k2 = x +n y . o n o k k k k where we use the inequality a max a , b for any a, b R. | | {| | | |} 2

Page 25 1.4 Finite Dimensional Normed Spaces and Subspaces.

Lemma 1.9. Let x1,...,xn be a linearly independent set of vectors in a normed space X (of any dimension).{ There} exists a number c>0 such that for every choice of scalars ↵1,...,↵n we have

↵ x + ...+ ↵ x c ↵ + ...+ ↵ . k 1 1 n nk | 1| | n| ⇣ ⌘ Roughly speaking, it states that in the case of linear independence of vectors, we • cannot ﬁnd a linear combination that involves large scalars but represents a small vector.

Theorem 1.10. Every ﬁnite dimensional subspace Y of a normed space X is complete. In particular, every ﬁnite dimensional normed space is complete.

Theorem 1.11. Every ﬁnite dimensional subspace Y of a normed space X is closed in X.

Deﬁnition 1.12. A norm on a vector space X is said to be equivalent to a norm on X if there are positivek·k constants a and b such that for all x X we have k·k0 2 a x x b x . k k0 k k k k0 Equivalent norms on X deﬁne the same topology for X. •

Theorem 1.13. On a ﬁnite dimensional vector space X, any norm is equivalent to any other norm . k·k k·k0

2 1. Given examples of subspaces of l1 and l which are not closed.

Solution:

2. What is the largest possible c in (1) if 2 (a) X = R and x1 =(1, 0),x2 =(0, 1),

Solution:

3 (b) X = R and x1 =(1, 0, 0),x2 =(0, 1, 0),x3 =(0, 0, 1).

Solution:

3. Show that in the deﬁnition of equivalance of norms, the axioms of an equivalence relation hold.

Page 26 Solution: We say that on X is equivalent to 0 on X,denotedby 0 if there exists positive constantsk·k a, b > 0suchthatforallk·k x X we havek·k ⇠ k·k 2 a x x b x . k k0 k k k k0 Reﬂexivity is immediate. • Suppose .Thereexistsa, b > 0suchthatforallx X we have • k·k⇠k·k0 2 1 1 a x x b x = x x x . k k0 k k k k0 ) b k kk k0  ak k This shows that ,andsymmetryisshown. k·k0 ⇠k·k

Suppose 0 and 0 1.Thereexistsa, b, c, d > 0suchthat • for all x k·Xk⇠kwe have·k k·k ⇠k·k 2 a x x b x . k k0 k kk k0 c x x d x . k k1 k k0  k k1 On one hand, x b x bd x . k k k k0  k k1 On the other hand, x a x ac x . k k k k0 k k1 Combining them yields for all x X 2 ac x x bd x . k k1 k k0  k k1 This shows tht ,andtransitivityisshown. k·k⇠k·k1

4. Show that equivalent norms on a vector space X induce the same topology for X.

Solution: Suppose and 0 are equivalent norms on a vector space X. There exists positivek constants·k ka,·k b > 0 such that for all x X we have 2 a x x b x . k k0 k k k k0 To show that they induce the same topology for X,wewanttoshowthatthe open sets in (X, )and(X, ) are the same. Consider the identity map k·k k·k0 I :(X, ) (X, ). k·k ! k·k0

Let x0 be any point in X.Givenany">0, choose = a"> 0. For all x satisfying x x <, we have k 0k 1 ⇢a" x x0 0 x x0 < = ". k k  ak k ⇢a

Page 27 Since x X is arbitrary, this shows that I is continuous. Hence, if M X 0 2 ⇢ is open in (X, 0), its preimage M again is also open in (X, ). Similarly, consider the identityk·k map k·k

I˜:(X, ) (X, ). k·k0 ! k·k " Let x be any point in X.Givenany">0, choose = > 0. For all x satisfying 0 b x x <,wehave k 0k0 b" x x0 b x x0 0 < = ". k k k k b

Since x0 X is arbitrary, this shows that I˜ is continuous. Hence, if M X is open in (X,2 ), its preimage M is also open in (X, ). ⇢ k·k k·k0

Remark: The converse is also true, i.e. if two norms and 0 on X give the same topology, then they are equivalent norms on Xk.·k k·k

5. If and 0 are equivalent norms on X,showthattheCauchysequencesin (X,k·k )and(k·X,k )arethesame. k·k k·k0

Solution: Suppose on X is equivalent to 0 on X.Thereexistspositive constants a, b > 0 suchk·k that for all x X we havek·k 2 a x x b x . k k0 k k k k0 Let (x ) be any Cauchy sequence in (X, ). Given any ">0, there exists n k·k N1 N such that 2 x x < a" for all m, n > N . k m nk 1 which implies

1 ⇢a" xm xn 0 xm xn < = ". for all m, n > N1. k k  ak k ⇢a This shows that (x )isalsoaCauchysequencein(X, ). Conversely, let (x ) n k·k0 n be any Cauchy sequence in (X, 0). Given any ">0, there exists N2 N such that k·k 2 " x x < for all m, n > N . k m nk0 b 1 which implies

b" xm xn b xm xn 0 < = " for all m, n > N2. k k k k b This shows that (x )isalsoaCauchysequencein(X, ). n k·k

Page 28 6. Theorem 2.4.5 implies that 2 and are equivalent. Give a direct proof of this 1 fact. k·k k·k

n Solution: Let X = R ,andx =(⇠j)beanyelementofX.Ononehand,

2 2 2 2 2 x = max ⇠j ⇠1 + ...+ ⇠n = x 2. k k1 j=1,...,n | | | | | | k k ⇣ ⌘ Taking square roots of both sides yields x x 2.Ontheotherhand, k k1 k k 2 2 2 2 2 x 2 = ⇠1 + ...+ ⇠n n max ⇠j = n x . k k | | | |  j=1,...,n | | k k1 ⇣ ⌘ Taking square roots of both sides yields x 2 pn x . Hence, combining these 1 inequalities gives for all x X k k  k k 2

x x 2 pn x . k k1 k k  k k1

7. Let 2 be as in Problem 8, section 2.2, and let be any norm on that vector space,k· callk it X. Show directly that there is a b>0k such·k that x b x for all x. k k k k2

n Solution: Let X = R ,and e1,...,en be the standard basis of X deﬁned by n { } ⇠j = jn. Any x =(⇠j)inX has a unique representation x = ⇠1e1 + ...+ ⇠2e2. Thus,

x = ⇠1e1 + ...+ ⇠nen ⇠1 e1 + ...+ ⇠n en k k k k|n |k k | |k k = ⇠ e | j|k jk j=1 X 1 1 n 2 n 2 ⇠ 2 e 2  | j| k jk j=1 ! j=1 ! X X = b x . k k2 where we use Cauchy-Schwarz inequality for sums in the last inequality. Since x X was arbitrary, the result follows. 2

8. Show that the norms and satisfy k·k1 k·k2 1 x x x . pnk k1 k k2 k k1

Page 29 n Solution: Let X = R ,andx =(⇠j)beanyelementofX. Note that if x = 0, the inequality is trivial since x = x =0bydeﬁnitionofanorm.So,pick k k1 k k2 any nonzero x Rn. Using Cauchy-Schwarz inequality for sums, 2 1 1 n n 2 n 2 x = ⇠ 12 ⇠ 2 = pn x . k k1 | j| | j| k k2 j=1 j=1 ! j=1 ! X X X ⇠ On the other hand, since x =0,deﬁney =(⌘ ), where ⌘ = j .Then k k2 6 j j x k k2 1 1 n 2 1 n 2 y = ⌘ 2 = ⇠ 2 k k2 | j| x 2 | j| j=1 ! 2 j=1 ! X k k X 1 = x =1. x k k2 k k2 n 1 n y = ⌘ = ⇠ k k1 | j| x | j| j=1 ! 2 j=1 X k k X x = k k1 . x k k2 and

n n 2 2 y 2 = ⌘j max ⌘i ⌘j y 1 k k | |  i=1,...,n | | | |k k j=1 j=1 X  X x = 1 y = k k1 ) k k1 x k k2 = x x . )kk2 k k1 To justfiy the second inequality on the first line, note that it suces to prove that ⌘i 1foralli =1,...,n,orequivalently, ⇠i x 2 for all i =1,...,n.From the| | definition of , | |k k k·k2 n x 2 = ⇠ 2 ⇠ 2 for all i =1,...,n. k k2 | j| | i| j=1 X Taking square roots of both sides yields x ⇠ for all i =1,...,n. k k2 | i|

Remark: Alternatively,

n 2 n n x 2 = ⇠ = ⇠ 2 + ⇠ ⇠ k k1 | j| | j| | i|| j| j=1 ! j=1 ! i=j ! X X X6 n ⇠ 2 = x 2. | j| k k2 j=1 ! X

Page 30 9. If two norms and on a vector space X are equivalent, show that k·k k·k0 x x 0 x x 0. k n k! () k n k0 !

Solution: Suppose two norms and 0 on a vector space X are equivalent, there exists positive constant a,k b· >k 0suchthatforallk·k x X we have 2 a x x b x . k k0 k k k k0 If x x 0, then k n k! 1 x x x x 0asn 0. k n k0  ak n k! ! Conversely, if x x 0, then k n k0 ! x x b x x 0asn 0. k n k k n k0 ! !

10. Show that all complex m n matrices A =(↵jk)withﬁxedm and n constitute an mn-dimensional vector space⇥ Z. Show that all norms on Z are equivalent. What would be the analogues of 1, 2 and for the present space Z? k·k k·k k·k1

Solution:

Page 31 1.5 Linear Operators. Deﬁnition 1.14. A linear operator T is an operator such that (a) the domain (T ) of T is a vector space and the range (T ) lies in a vector space over the sameD ﬁeld, R

(b) for all x, y (T ) and scalars ↵,, 2D T (↵x + y)=↵T (x)+T(y).

Note that the above formula expresses the fact that a linear operator T is a ho- • momorphism of a vector space (its domain) into another vector space, that is, T preserves the two operations of a vector space.

On the LHS, we ﬁrst apply a vector space operation (addition or scalar multiplica- • tion) and then map the resulting vector into Y , whereas on the RHS we ﬁrst map x and y into Y and then perform the vector space operations in Y ,theoutcome being the same.

Theorem 1.15. Let T be a linear operator. Then: (a) The range (T ) is a vector space. R (b) If dim (T )=n< , then dim (T ) n. D 1 R  (c) The null space (T ) is a vector space. N An immediate consequence of part (b) is worth noting: Linear operators preserve • linear dependence.

Theorem 1.16. Let X, Y be a vector spaces, both real or both complex. Let T : (T ) Y be a linear operator with domain (T ) X and range (T ) Y . Then: D ! 1 D ⇢ R ⇢ (a) The inverse T : (T ) (T ) exists if and only if Tx = 0 = x = 0. R ! D ) 1 (b) If T exists, it is a linear operator.

1 (c) If dim (T )=n< and T exists, then dim (T ) = dim (T ). D 1 R D

Theorem 1.17. Let T : X Y and S : Y Z be bijective linear operators, where ! ! 1 X, Y, Z are vector spaces. Then the inverse (ST) : Z X of the composition ST exists, and ! 1 1 1 (ST) = T S .

Page 32 1. Show that the identity operator, the zero operator and the di↵erentiation operator (on polynomials) are linear.

Solution: For any scalars ↵, and x, y X, 2

IX (↵x + y)=↵x + y = ↵IX x + IX y. 0(↵x + y)=0 = ↵0x + 0y.

T (↵x(t)+y(t)) = (↵x(t)+y(t))0

= ↵x0(t)+y0(t)=↵T x(t)+Ty(t).

2 2 2. Show that the operators T1,...,T4 from R into R deﬁned by T :(⇠ ,⇠ ) (⇠ , 0) 1 1 2 7! 1 T :(⇠ ,⇠ ) (0,⇠ ) 2 1 2 7! 2 T :(⇠ ,⇠ ) (⇠ ,⇠ ) 3 1 2 7! 2 1 T :(⇠ ,⇠ ) (⇠ ,⇠ ) 4 1 2 7! 1 2 respectively, are linear, and interpret these operators geometrically.

Solution: Denote x =(⇠1,⇠2)andy =(⌘1,⌘2). For any scalars ↵,,

T1(↵x + y)=(↵⇠1 + ⌘1, 0)

= ↵(⇠1, 0) + (⌘1, 0) = ↵T1(x)+T1(y).

T2(↵x + y)=(0,↵⇠2 + ⌘2)

= ↵(0,⇠2)+(0,⌘2)=↵T2(x)+T2(y).

T3(↵x + y)=(↵⇠2 + ⌘2,↵⇠1 + ⌘1)

=(↵⇠2,↵⇠1)+(⌘2,⌘1)

= ↵(⇠2,⇠1)+(⌘2,⌘1)=↵T3(x)+T3(y).

T4(↵x + y)= (↵⇠1 + ⌘1),(↵⇠2 + ⌘2)

=(⇣↵⇠1,↵⇠2)+(⌘1,⌘2) ⌘

= ↵(⇠1,⇠2)+(⌘1,⌘2)=↵T4(x)+T4(y).

T1 and T2 are both projection to x-axis and y-axis respectively, while T4 is a scaling transformation. T3 ﬁrst rotates the vector 90 anti-clockwise about the origin, then reﬂects across the y-axis.

3. What are the domain, range and null space of T1,T2,T3 in Problem 2?

2 Solution: The domain of T1,T2,T3 is R ,andtherangeisthex-axis for T1,the 2 y-axis for T2 and R for T3.Thenullspaceistheline⇠1 =0forT1,theline⇠2 =0 for T2 and the origin (0, 0) for T3.

Page 33 4. What is the null space of T4 in Problem 2? Of T1 and T2 in 2.6-7? Of T in 2.6-4?

2 Solution: The null space of T4 is R if =0andtheorigin(0, 0) if = 0. Fix 3 6 avectora =(a1,a2,a3) R . Consider the linear operators T1 and T2 deﬁned by 2 ⇠ a ⇠ a ⇠ a 1 1 2 3 3 2 T1x = x a = ⇠2 a2 = ⇠3a1 ⇠1a3 . ⇥ 2⇠ 3 ⇥ 2a 3 2⇠ a ⇠ a 3 3 3 1 2 2 1 T x = x a = ⇠4 a 5+ ⇠4a +5 ⇠ a4. 5 2 · 1 1 2 2 3 3

i.e. T1 and T2 are the cross product and the dot product with the ﬁxed vector a respectively. The null space of T1 is any scalar multiple of the vector a,whilethe 3 null space of T2 is the plane ⇠1a1 + ⇠2a2 + ⇠3a3 =0inR .Forthedi↵erentiation operator, the null space is any constant functions x(t)fort [a, b]. 2

5. Let T : X Y be a linear operator. ! (a) Show that the image of a subspace V of X is a vector space.

Solution: Denote the image of a subspace V of X under T by Im(V ), it suces to show that Im(V )isasubspaceofX.Chooseanyy ,y Im(V ), 1 2 2 there exists x1,x2 V such that Tx1 = y1 and Tx2 = y2.Foranyscalars ↵,, 2 ↵y1 + y2 = ↵T x1 + Tx2 = T (↵x1 + x2).

This shows that ↵y1 + y2 Im(V )since↵x1 + x2 V due to V being a subspace of X. 2 2

(b) Show that the inverse image of a subspace W of Y is a vector space.

Solution: Denote the inverse image of a subspace W of Y under T by PIm(W ), it suces to show that PIm(W )isasubspaceofY .Chooseany x1,x2 PIm(W ), there exists y1,y2 W such that Tx1 = y1 and Tx2 = y2. For any2 scalars ↵,, 2

T (↵x1 + x2)=↵T x1 + Tx2 = ↵y1 + y2.

This shows that ↵x1 + x2 PIm(W )since↵y1 + y2 W due to W being asubspaceofY . 2 2

6. If the composition of two linear operators exists, show that it is linear.

Solution: Consider any two linear operators T : X Y , S : Y Z.Forany x, y X and scalars ↵,, ! ! 2 ST(↵x + y)=S(↵T x + Ty) by linearity of T. h i

Page 34 = ↵(ST)x + (ST)y by linearity of S. h i

7. (Commutativity) Let X be any vector space and S : X X and T : X X any operators. S and T are said to commute if ST = TS,thatis,(! ST)x =(!TS)x for all x X.DoT and T in Problem 2 commute? 2 1 3

Solution: No. Choose x =(1, 2), then

(T1T3)(1, 2) = T1(2, 1) = (2, 0).

(T3T1)(1, 2) = T3(1, 0) = (0, 1).

8. Write the operators in Problem 2 using 2 2matrices. ⇥

Solution: 10 00 01 0 T = T = T = T = . 1 00 2 01 3 10 4 0    

9. In 2.6-8, write y = Ax in terms of components, show that T is linear and give examples.

Solution: For any j =1,...,r,wehavethat⌘j = ajk⇠k = aj1⇠1 +...+ajn⇠n. k=1 To see that T is linear, for any j =1,...,r, X

A(↵x + y) = ajk(↵⇠k + ⌘k) j k=1 ⇣ ⌘ Xn n

= ↵ ajk⇠k + ajk⌘k = ↵(Ax)j + (Ay)j. Xk=1 Xk=1

10. Formulate the condition in 2.6-10(a) in terms of the null space of T .

Solution: Let X, Y be vector spaces, both real or both complex. Let T : (T ) Y be a linear operator with domain (T ) X and (T ) Y .TheinverseD ! 1 D ⇢ R ⇢ T : (T ) (T )existsifandonlyifthenullspaceofT , (T )= 0 . R ! D N { }

Page 35 11. Let X be the vector space of all complex 2 2matricesanddeﬁneT : X X by Tx = bx,whereb X is ﬁxed and bx denotes⇥ the usual product of matrices.! Show 2 1 that T is linear. Under what condition does T exists?

Solution: For any x, y X and scalars ↵,, 2 b b ↵⇠ + ⌘ ↵⇠ + ⌘ T (↵x + y)= 1 2 1 1 2 2 b b ↵⇠ + ⌘ ↵⇠ + ⌘  3 4 3 3 4 4 b b ⇠ ⇠ ⌘ ⌘ = 1 2 ↵ 1 2 + 1 2 b b ⇠ ⇠ ⌘ ⌘  3 4⇢  3 4  3 4 = ↵bx + by = ↵T x + Ty.

1 This shows that T is linear. T exists if and only if b is a non-singular 2 2 complex matrix. ⇥

12. Does the inverse of T in 2.6-4 exist?

Solution: The inverse of the di↵erentiation operator T does not exist because (T ) = 0 ,thezerofunction. N 6 { }

13. Let T : (T ) Y be a linear operator whose inverse exists. If x ,...,x is D ! { 1 n} alinearlyindependentsetin (T ), show that the set Tx1,...,Txn is linearly independent. D { }

Solution: Let T : (T ) Y be a linear operator whose inverse exists. Suppose D !

↵1Tx1 + ...+ ↵nTxn = 0Y .

By linearity of T ,theequationaboveisequivalentto

T (↵1x1 + ...+ ↵nxn)=0Y .

1 Since T exists, we must have “ Tx = 0 = x = 0 ”. Thus, Y ) X

↵1x1 + ...+ ↵nxn = 0X .

but x ,...,x is a linearly independent set in (T ), so this gives ↵ = ... = { 1 n} D 1 ↵n =0.

14. Let T : X Y be a linear operator and dim X =dimY = n< .Showthat ! 1 1 (T )=Y if and only if T exists. R

Page 36 Solution: Suppose (T )=Y ,bydeﬁnition,forally Y ,thereexistsx X such that Tx = y,i.e.R T is surjective. We now show that2 T is injective. Since2 dim (X)=dim(Y )=n< ,theRank-Nullity Theorem gives us 1 dim( (T )) + dim( (T )) = dim(X). R N but by assumption, (T )=Y ,so R dim(Y )+dim( (T )) = dim(X). N = dim( (T )) = dim(X) dim(Y )=0. ) N = (T )= 0 . )N { X }

This shows that T is injective. Indeed, suppose for any x1,x2,wehaveTx1 = Tx2. By linearity of T , Tx1 Tx2 = T (x1 x2)=0Y = x1 x2 = 0X = x1 = x2. ) ) 1 Since T is both injective and surjective, we conclude that the inverse of T , T , exists.

1 Conversely, suppose T exists. From Problem 10, this means that (T )= 0 = dim( (T )) = 0. Invoking the Rank-Nullity Theorem givesN { X } ) N dim( (T )) + dim( (T )) = dim(X). R N = dim( (T )) = dim(X)=n. ) R This implies that (T )=Y since any proper subspace W of Y has dimension less than n. R

15. Consider the vector space X of all real-valued functions which are deﬁned on R and have derivatives of all orders everywhere on R.DeﬁneT : X X by y(t)=Tx(t)= 1 ! x0(t). Show that (T )isallofX but T does not exist. Compare with Problem 14 and comment. R

Solution: For any y(t) (T ), deﬁne x(t)= t y(s) ds X; Fundamental 2R 2 Theorem of Calculus gives that x (t)=Tx(t)=1y(t). On the other hand, T 1 0 R does not exist since the null space of T consists of every constant functions on R. However, it doesn’t contradict Problem 14 since X is an inﬁnite-dimensional vector space.

Page 37 1.6 Bounded and Continuous Linear Operators. Deﬁnition 1.18.

1. Let X and Y be normed spaces and T : (T ) Y a linear operator, where (T ) X. The operator T is said to be boundedD if! there is a nonnegative numberDC such⇢ that for all x (T ), Tx C x . 2D k k k k This also shows that a bounded linear operator maps bounded sets in (T ) • onto bounded sets in Y . D

2. The norm of a bounded linear operator T is deﬁned as

Tx T =supk k. k k x (T ) x 2xD=0 k k 6 This is the smallest possible C for all nonzero x (T ). • 2D With C = T , we have the inequality Tx T x . • k k k kk kk k If (T )= 0 , we deﬁne T =0. • D { } k k

Lemma 1.19. Let T be a bounded linear operator. Then: (a) An alternative formula for the norm of T is

T =sup Tx . k k x (T ) k k 2xD=1 k k

(b) T is a norm. k k

Theorem 1.20. If a normed space X is ﬁnite dimensional, then every linear operator on X is bounded.

Theorem 1.21. Let T : (T ) Y be a linear operator, where (T ) X and X, Y are normed spaces. D ! D ⇢ (a) T is continuous if and only if T is bounded.

(b) If T is continuous at a single point, it is continuous.

Corollary 1.22. Let T be a bounded linear operator. Then: (a) x x [where x ,x (T )] implies Tx Tx. n ! n 2D n ! (b) The null space (T ) is closed. N It is worth noting that the range of a bounded linear operator may not be closed. •

Page 38 Deﬁnition 1.23.

1. Two operators T1 and T2 are deﬁned to be equal, written T1 = T2 if they have the same domain (T )= (T ) and if T x = T x for all x (T )= (T ). D 1 D 2 1 2 2D 1 D 2 2. The restriction of an operator T : (T ) Y to a subset B (T ) is denoted by T and is the operator deﬁned byDT :!B Y , satisfying ⇢D |B |B ! T x = Tx for all x B. |B 2 3. The extension of an operator T : (T ) Y to a superset M (T ) is an D ! D operator T˜ : M Y such that T˜ (T ) = T , that is, Tx˜ = Tx for all x (T ). ! |D 2D [Hence T is the restriction of T˜ to (T ).] D If (T ) is a proper subset of M, then a given T has many extensions; of prac- • ticalD interest are those extensions which preserve linearity or boundedness.

Theorem 1.24 (Bounded linear extension). Let T : (T ) Y be a bounded linear operator, where (T ) lies in a normed space X and DY is a! Banach space. Then T has an extension T˜D: (T ) Y , where T˜ is a bounded linear operator with norm T˜ = T . D ! k k k k The theorem concerns an extension of a bounded linear operator T to the closure • (T )ofthedomainsuchthattheextendedoperatorisagainboundedandlinear, andD even has the same norm.

This includes the case of an extension from a dense set in a normed space X to all • of X.

It also includes the case of an extension from a normed space X to its completion. •

n n 1. Prove T1T2 T1 T2 and T T (n N)forboundedlinearoperators T : X k Ykk, T : Y kk kZ andkT : XkkXk ,where2 X, Y, Z are normed spaces. 2 ! 1 ! !

Solution: Using boundedness of T1 and T2,

(T T )x = T (T x) T T x T T x . k 1 2 k k 1 2 kk 1kk 2 kk 1kk 2kk k The ﬁrst inequality follows by taking supremum over all x of norm 1. A similar argument also shows the second inequality.

2. Let X and Y be normed spaces. Show that a linear operator T : X Y is bounded if and only if T maps bounded sets in X into bounded sets in Y . !

Page 39 Solution: Suppose T : X Y is bounded, there exists an C>0suchthat Tx C x for all x !X.TakeanyboundedsubsetA of X,thereexists Mk k> 0suchthatk k x 2M for all x A.Foranyx A, A k k A 2 2 Tx C x CM . k k k k A This shows that T maps bounded sets in X into bounded sets in Y .

Conversely, suppose a linear operator T : X Y maps bounded sets in X into bounded sets in Y . This means that for any! ﬁxed R>0, there exists a constant MR > 0suchthat x R = Tx MR.Wenowtakeanynonzeroy X and set k k )k k 2 y x = R = x = R. y )kk k k Thus,

R R Ty = T y = Tz M . y k k y k k R k k ✓k k ◆ M = Ty R y . )k k R k k where we crucially used the linearity of T .Rearrangingandtakingsupremum over all y of norm 1 shows that T is bounded.

3. If T =0isaboundedlinearoperator,showthatforanyx (T )suchthat x < 1 we have6 the strict inequality Tx < T . 2D k k k k k k

Solution: We have Tx T x < T . k kk kk k k k

4. Let T : (T ) Y be a linear operator, where (T ) X and X, Y are normed spaces.D Show! that if T is continuous at a single point,D it⇢ is continuous on (T ). D

Solution: Suppose T is continuous at an arbitrary x (T ). This means that 0 2D given any ">0, there exists a >0suchthat Tx Tx0 " for all x (T ) satisfying x x . Fix an y (T ), andk set k 2D k 0k 0 2D y y x x = 0 = x x = . 0 y y )k 0k k 0k Since T is linear, for any y (T )satisfying y y , 2D k 0k T (y y ) = T (y y ) = T (x x ) " y y k 0 k y y 0 k 0 k 0 ✓ 0 ◆ k k k k y y0 " = T (y y0) "k k = ". )k k 

This shows that T is continuous at y0.Sincey0 (T )isarbitrary,thestatement follows. 2D

Page 40 5. Show that the operator T : l1 l1 deﬁned by y =(⌘j)=Tx, ⌘j = ⇠j/j, x =(⇠j), is linear and bounded. !

Solution: For any x, z l1 and scalars ↵,, 2 ⇠  ⇠  T (↵x + z)= ↵ j + j = ↵ j + j = ↵T x + Tz. j j j j ✓ ◆ ✓ ◆ ✓ ◆

For any x =(⇠ ) l1, j 2

⇠j ⇠j sup ⇠j = x . j | | j N | | k k 2 Taking supremum over j N on both sides yields Tx x .Weconcludethat T is a bounded linear operator.2 k kk k

6. (Range) Show that the range (T ) of a bounded linear operator T : X Y need not be closed in Y . R !

n Solution: Deﬁne (xn)tobeasequenceinthespacel1,wherexn =(⇠j )and

n pj if j n, ⇠j =  ( 0ifj>n.

n Consider the operator T : l1 l1 in Problem 5. Then Tx = y =(⌘ ), where ! n n j 1 ⇠n if j n, ⌘n = j = pj  j j 8 < 0ifj>n.

We now have our sequence (yn) (T:) l1.Weclaimthatitconvergestoy in 2R ⇢ 1 l1,wherey is a sequence in l1 deﬁned as y =(⌘ ), ⌘ = .Indeed, j j pj

n 1 yn y l1 =sup ⌘j ⌘j = 0asn . k k j N | | pn +1 ! ! 1 2

However, y/ (T ). Indeed, if there exists an x l1 such that Tx = y,then 2R 2 x must be the sequence (⇠j), with ⇠j = pj,whichisclearlynotinthespacel1. Hence, (T )isnotclosedinl1. R

7. (Inverse operator) Let T be a bounded linear operator from a normed space X onto a normed space Y .Ifthereisapositiveb such that

Tx b x for all x X, k k k k 2 1 show that then T : Y X exists and is bounded. !

Page 41 1 Solution: We ﬁrst show that T is injective, and therefore the inverse T exists since T is bijective (T is surjective by assumption). Indeed, choose any x1,x2 X and suppose x = x ,then x x > 0. Since T is linear, 2 1 6 2 k 1 2k Tx Tx = T (x x ) b x x > 0= Tx = Tx . k 1 2k k 1 2 k k 1 2k ) 1 6 2 1 We are left to show there exists an C>0suchthat T y C y for all y Y .SinceT is surjective, for any y Y ,thereexistsxk X suchk thatk ky = Tx; 2 1 1 2 2 existence of T then implies x = T y.Thus,

1 1 1 1 T (T (y)) b T y = T y y . k k k k )k k b k k 1 where C = > 0. b

1 8. Show that the inverse T : (T ) X of a bounded linear operator T : X Y need not be bounded. R ! !

Solution: Consider the operator T : l1 l1 deﬁned by y = Tx =(⌘ ), ⌘ = ! j j ⇠j/j, x =(⇠j). We shown in Problem 6 that T is a bounded linear operator. We ﬁrst show that T is injective. For any x ,x l1,supposeTx = Tx .Forany 1 2 2 1 2 j N, 2 ⇠1 ⇠2 1 (Tx ) =(Tx ) = j = j = ⇠1 = ⇠2 since =0. 1 j 2 j ) j j ) j j j 6

This shows that x1 = x2 and T is injective. Thus, there exists an inverse 1 1 T : (T ) l1 deﬁned by x = T y =(⇠j), ⇠j = j⌘j, y =(⌘j). Let’s verifyR that this! is indeed the inverse operator.

1 1 1 ⇠j ⇠j T (Tx)=T y = T = j =(⇠ )=x. j j j ✓ ◆ ✓ ◆ 1 j⌘j T (T y)=Tx = T ((j⌘ )) = =(⌘ )=y. j j j ✓ ◆ 1 We claim that T is not bounded. Indeed, let yn =(jn)j1=1,wherejn is the Kronecker delta function. Then y =1and k nk 1 1 T yn T y = (j ) = n = k k = n. k nk k jn k ) y k nk Since n N is arbitrary, this shows that there is no ﬁxed number C>0such 21 T yn 1 that k k C,i.e.T is not bounded. y  k nk

Page 42 9. Let T : C[0, 1] C[0, 1] be deﬁned by ! t y(t)= x(s) ds. Z0 1 1 Find (T )andT : (T ) C[0, 1]. Is T linear and bounded? R R !

Solution: First, Fundamental Theorem of Calculus yields

(T )= y(t) C[0, 1]: y(t) C1[0, 1],y(0) = 0 C[0, 1]. R { 2 2 }⇢ 1 Next, we show that T is injective and thus the inverse T : (T ) C[0, 1] exists. Indeed, suppose for any x ,x C[0, 1], Tx = Tx .ThenR ! 1 2 2 1 2 t t Tx1 = Tx2 = x1(s) ds = x2(s) ds ) 0 0 Z t Z = x (s) x (s) ds =0 ) 1 2 Z0 = x(s)=h y(s)forallis [0,t]. ) 2 where the last implication follows from x y being a continuous function in 1 1 1 [0,t] [0, 1]. The inverse operator T is deﬁned by T y(t)=y0(t), i.e. T is the⇢ di↵erentiation operator. Since di↵erentiation is a linear operation, so is 1 1 n T . However, T is not bounded. Indeed, let yn(t)=t ,wheren N.Then 2 yn =1and k k 1 1 n 1 T yn T y = nt = n = k k = n. k nk k k ) y k nk Since n N is arbitrary, this shows that there is no ﬁxed number C>0such 21 T yn 1 that k k C,i.e.T is not bounded. y  k nk

10. On C[0, 1] deﬁne S and T

1 Sx(t)=y(t)=t x(s) ds T x(t)=y(t)=tx(t). Z0 respectively. Do S and T commute? Find S , T , ST and TS . k k k k k k k k

Solution: They do not commute. Take x(t)=t C[0, 1]. Then 2 1 t (ST)x(t)=S(t2)=t s2 ds = . 3 Z0 t t2 (TS)x(t)=T = . 2 2 ✓ ◆

Page 43 11. Let X be the normed space of all bounded real-valued functions on R with norm deﬁned by x =sup x(t) , k k t R | | 2 and let T : X X deﬁned by ! y(t)=Tx(t)=x(t ) where > 0isaconstant.(Thisiamodelofadelay line,whichisanelectricdevice whose output y is a delayed version of the input x,thetimedelaybeing.)IsT linear? Bounded?

Solution: For any x, z X and scalars ↵,, 2 T (↵x + z)=↵x(t ) + z(t ) = ↵T x + Tz. This shows that T is linear. T is bounded since

Tx =sup x(t ) = x . k k t R | | k k 2

12. (Matrices) We know that an r n matrix A =(↵jk)definesalinearoperatorfrom the vector space X of all ordered⇥n-tuples of numbers into the vector space Y of all ordered r-tuples of numbers. Suppose that any norm is given on X and any k·k1 norm 2 is given on Y .RememberfromProblem10,Section2.4,thatthereare variousk· normsk on the space Z of all those matrices (r and n fixed). A norm on Z is said to be compatible with and if k·k k·k1 k·k2 Ax A x . k k2 k kk k1 Show that the norm defined by Ax A =supk k2 k k x X x 1 x2=0 k k 6 is compatible with and .Thisnormisoftencalledthenatural norm defined k·k1 k·k2 by 1 and cdot 2.IFwechoose x 1 =maxj ⇠j and y 2 =maxj ⌘j ,showthat thek natural·k normk isk k k | | k k | | n

A =max ↵jk . k k j | | Xk=1

Solution:

13. Show that in 2.7-7 with r = n, a compatible norm is deﬁned by

1 n n 2 A = ↵2 , k k jk j=1 ! X Xk=1

Page 44 but for n>1thisisnot the natural norm deﬁned by the Euclidean norm on Rn.

Solution:

14. If in Problem 12, we choose

n r x = ⇠ y = ⌘ , k k1 | k|kk2 | j| j=1 Xk=1 X show that a compatible norm is deﬁned by

A =max ↵jk . k k k | | j=1 X

Solution:

15. Show that for r = n,thenorminProblem14isthenaturalnormcorrespondingto and as deﬁned in that problem. k·k1 k·k2

Solution:

Page 45 1.7 Linear Functionals. Definition 1.25. 1. A linear functional f is a linear operator with domain in a vector space X and range in the scalar field K of X; thus, f : (f) K, where K = R if X is real D ! and K = C if X is complex. 2. A bounded linear functional f is a bounded linear operator with range in the scalar field of the normed space X in which the domain (f) lies. Thus there exists a nonnegative number C such that for all x (f), fD(x) C x . Furthermore, the norm of f is 2D | | k k f(x) f =sup| | =sup f(x) . k k x (f) x x (f) | | 2xD=0 k k 2xD=1 6 k k As before, we have that f(x) f x . • | |k kk k

Theorem 1.26. A linear functional f with domain (f) in a normed space X is continuous if and only if f is bounded. D

Definition 1.27. 1. The set of all linear functionals defined on a vector space X can itself be made into a vector space. This space is denoted by X⇤ and is called the algebraic dual space of X. Its algebraic operations of vector space are defined in a natural way as follows.

(a) The sum f1 + f2 of two functionals f1 and f2 is the functional s whose value at every x X is 2

s(x)=(f1 + f2)(x)=f1(x)+f2(x).

(b) The product ↵f of a scalar ↵ and a functional f is the functional p whose value at every x X is 2 p(x)=(↵f)(x)=↵f(x).

2. We may also consider the algebraic dual (X⇤)⇤ of X⇤, whose elements are the linear functionals deﬁned on X⇤. We denote (X⇤)⇤ by X⇤⇤ and call it the second algebraic dual space of X.

We can obtain an interesting and important relation between X and X⇤⇤, as • follows. We can obtain a g X⇤⇤, which is a linear functional deﬁned on X⇤, by choosing a ﬁxed x X and2 setting 2

g(f)=g (f)=f(x) where x X ﬁxed, f X⇤ variable. x 2 2 g is linear. Indeed, • x

gx(↵f1 + f2)=(↵f1 + f2)(x)=↵f1(x)+f2(x)=↵gx(f1)+gx(f2).

Hence, g is an element of X⇤ , by the deﬁnition of X⇤⇤. x ⇤

Page 46 To each x X there corresponds a g X⇤⇤. This deﬁnes a mapping • 2 x 2 C : X X⇤⇤, C : x g ; C is called the canonical mapping of X into ! 7! x X⇤⇤. C is linear since its domain is a vector space and we have

(C(↵x + y))(f)=g↵x+y(f) = f(↵x + y) = ↵f(x)+f(y)

= ↵gx(f)+gy(f) = ↵(Cx)(f)+(Cy)(f).

C is also called the canonical embedding of X into X⇤⇤. 3. An metric space isomorphism T of a metric space X =(X, d) onto a metric space X˜ =(X,˜ d˜) is a bijective mapping which preserves distance, that is, for all x, y X, d˜(Tx,Ty)=d(x, y). X˜ is then called isomorphic with X. 2 4. An vector space isomorphism T of a vector space X onto a vector space X˜ over the same ﬁeld is a bijective mapping which preserves the two algebraic operations of vector space; thus, for all x, y X and scalars ↵, 2 T (x + y)=Tx+ Ty and T (↵x)=↵T x, that is, T : X X˜ is a bijective linear operator. X˜ is then called isomorphic with X, and X!and X˜ are called isomorphic vector spaces. 5. If X is isomorphic with a subspace of a vector space Y , we say that X is embeddable in Y . It can be shown that the canonical mapping C is injective. Since C is linear, • it is a vector space isomorphism of X onto the range (C) X⇤⇤. R ⇢ Since (C) is a subspace of X⇤⇤, X is embeddable in X⇤⇤, and C is also called • R the canonical embedding of X into X⇤⇤.

If C is surjective (hence bijective), so that (C)=X⇤⇤, then X is said to be • algebraically reﬂexive. R

1. Show that the functionals in 2.8-7 and 2.8-8 are linear.

Solution: Choose a ﬁxed t0 J =[a, b]andsetf1(x)=x(t0), where x C[a, b]. For any x, y C[a, b]andscalars2 ↵,, 2 2

f1(↵x + y)=(↵x + y)(t0)=↵x(t0)+y(t0)=↵f1(x)+f1(y).

1 Choose a ﬁxed a =(a ) l2 and set f(x)= ⇠ ↵ ,wherex =(⇠ ) l2.For j 2 j j j 2 j=1 X any x =(⇠ ),y =(⌘ ) l2 and scalars ↵,, j j 2

1 1 1 f(↵x + y)= (↵⇠j + ⌘j)↵j = ↵ .⇠j↵j + ⌘j↵j = ↵f(x)+f(y). j=1 j=1 j=1 X X X

Page 47 Note we can split the inﬁnite sum because the two inﬁnite sums are convergent by Cauchy-Schwarz inequality.

2. Show that the functionals deﬁned on C[a, b]by

b f (x)= x(t)y (t) dt (y C[a, b]ﬁxed.) 1 0 0 2 Za f2(x)=↵x(a)+x(b)(↵, ﬁxed.)

are linear and bounded.

Solution: For any x, y C[a, b]andscalars,, 2 b f1(x + y)= x(t)+y(t) y0(t) dt a Z hb i b = x(t)y0(t) dt + y(t)y0(t) dt. Za Za = f1(x)+f1(y).

f2(x + y)=↵(x + y)(a)+(x + y)(b) = ↵(x(a)+y(a)) + (x(b)+y(b)) = (↵x(a)+x(b)) + (↵y(a)+y(b))

= f2(x)+f2(y).

To show that f and f are bounded, for any x C[a, b], 1 2 2 b b f1(x) = x(t)y0(t) dt max x(t) y0(t) dt | |  t [a,b] | | Za 2 Za b = y (t) dt x . 0 k k ✓Za ◆ f2(x) = ↵x(a)+x(b) ↵ max x(t) + max x(t) | | | | t [a,b] | | t [a,b] | | 2 2 =(↵ + ) x . k k

3. Find the norm of the linear functional f deﬁned on C[ 1, 1] by 0 1 f(x)= x(t) dt x(t) dt. 1 0 Z Z

Page 48 Solution:

0 1 f(x) = x(t) dt x(t) dt | | 1 0 Z Z 0 1 x(t) dt + x(t) dt  1 0 Z Z 0 1 x dt + x dt k k 1 k k 0 Z Z =2 x k k Taking the supremum over all x of norm 1, we obtain f 2. To get f 2, we choose the particular x(t) C[ 1, 1] deﬁned by k k k k 2 1 2t 2if 1 t ,  2 8 1 1 x(t)=> 2t if t , > 2   2 <> 1 2t +2 if t 1. 2   > > Note that x =1and :> k k f(x) f | | = f(x) =2. k k x | | k k

4. Show that for J =[a, b],

f1(x)=maxx(t) f2(x)=minx(t) t J t J 2 2 deﬁne functionals on C[a, b]. Are they linear? Bounded?

Solution: Both f1 and f2 deﬁne functionals on C[a, b]sincecontinuousfunc- tions attains its maximum and minimum on closed interval. They are not linear. t a (t a) Choose x(t)= and y(t)= .Then b a b a

f1(x + y)=0 but f1(x)+f1(y)=1+0=1. f (x + y)=0 but f (x)+f (y)=0 1= 1. 2 2 2 They are, however, bounded, since for any x C[a, b], 2

f1(x) =maxx(t) max x(t) = x . t J t J | | 2  2 | | k k f2(x) =minx(t) max x(t) = x . t J t J | | 2  2 | | k k

Page 49 5. Show that on any sequence space X we can deﬁne a linear functional f by setting f(x)=⇠n (n ﬁxed), where x =(⇠j). Is f bounded if X = l1?

Solution: For any x =(⇠ ),y =(⌘ ) X and scalars ↵,, j j 2

f(↵x + y)=↵⇠n + ⌘n = ↵f(x)+f(y).

f is bounded if X = l1.Indeed,foranyx l1, 2

f(x) = ⇠n sup ⇠j = x . | | | | j N | | k k 2

Remark: In fact, we can show that f =1. Taking supremum over all x of norm 1 on previous equation yields kf k 1. To get f 1, we choose the particular x =(⇠ )=( ), note that k xk=1and k k j jn k k f(x) f | | = f(x) =1. k k x | | k k

6. (Space C1[a, b]) The space C1[a, b]isthenormedspaceofallcontinuouslydi↵erentiable functions on J =[a, b]withnormdeﬁnedby

x =max x(t) +max x0(t) . t J t J k k 2 | | 2 | | (a) Show that the axioms of a norm are satisﬁed.

Solution: (N1) and (N3) are obvious. For (N2), if x(t) 0, then x =0. ⌘ k k On the other hand, if x =0,sincebothmaxx(t) and max x0(t) are t J t J k k 2 | | 2 | | nonnegative, we must have x(t) =0and x0(t) =0forallt [a, b]which implies x(t) 0. Finally, (N4)| follows| from| | 2 ⌘

x + y =max x(t)+y(t) +max x0(t)+y0(t) t J t J k k 2 | | 2 | | max x(t) +max y(t) +max x0(t) +max y0(t) t J t J t J t J  2 | | 2 | | 2 | | 2 | | = x + y . k k k k

a + b (b) Show that f(x)=x (c), c = ,deﬁnesaboundedlinearfunctionalon 0 2 C1[a, b].

Solution: For any x, y C1[a, b]andscalars↵,, 2 f(↵x + y)=↵x0(c)+y0(c)=↵f(x)+f(y). To see that f is bounded,

f(x) = x0(c) max x0(t) x . t J | | | | 2 | |k k

Page 50 (c) Show that f is not bounded, considered as functional on the subspace of C[a, b] which consists of all continuously di↵erentiable functions.

Solution:

7. If f is a bounded linear functional on a complex normed space, is f¯ bounded? Linear? (The bar denotes the complex conjugate.)

Solution: f¯ is bounded since f(x) = f¯(x) ,butitisnotlinearsinceforany x X and complex numbers ↵,| f(↵x| )=| ↵f(x| )=¯↵f(x) = ↵f(x). 2 6

8. (Null space) The null space (M ⇤)ofasetM ⇤ X⇤ is deﬁned to be the set of all N ⇢ x X such that f(x)=0forallf M ⇤.Showthat (M ⇤)isavectorspace. 2 2 N

Solution: Since X is a vector space, it suces to show that (M ⇤)isasubspace N of X. Note that all element of M ⇤ are linear functionals. For any x, y (M ⇤), 2N we have f(x)=f(y)=0forallf M ⇤.Thenforanyf M ⇤ and scalars ↵,, 2 2

f(↵x + y)=↵f(x)+f(y)=0. by linearity of f M ⇤ 2 h i This shows that ↵x + y (M ⇤)andthestatementfollows. 2N

9. Let f =0beanylinearfunctionalonavectorspaceX and x0 any ﬁxed element of X 6 (f), where (f)isthenullspaceoff.Showthatanyx X has a unique representation\N x =N↵x + y,wherey (f). 2 0 2N

Solution: Let f =0beanylinearfunctionalonX and x0 any ﬁxed element of X (f). We6 claim that for any x X,thereexistsascalar↵ such that x = ↵\Nx + y,wherey (f). First, applying2 f on both sides yields 0 2N f(x)=f(↵x + y)=↵f(x )+f(y)= f(y)=f(x) ↵f(x ). 0 0 ) 0 f(x) By choosing ↵ = (which is well-deﬁned since f(x0) =0),weseethat f(x0) 6

f(x) f(y)=f(x) f(x0)=0 = y (f). f(x0) ) 2N

To show uniqueness, suppose x has two representations x = ↵1x0 +y1 = ↵2x0 +y2, where ↵1,↵2 are scalars and y1,y2 (f). Subtracting both representations yields 2N (↵ ↵ )x = y y . 1 2 0 2 1 Applying f on both sides gives

f((↵ ↵ )x )=f(y y ) 1 2 0 2 1

Page 51 (↵ ↵ )f(x )=f(y ) f(y )=0 1 2 0 2 1

by linearity of f and y1,y2 (f). Since f(x0) = 0, we must have ↵1 ↵2 =0. This also implies y y =0.2N 6 1 2

10. Show that in Problem 9, two elements x1,x2 X belong to the same element of the quotient space X/ (f)ifandonlyiff(x )=2f(x ). Show that codim (f)=1. N 1 2 N

Solution: Suppose two elements x ,x X belong to the same element of the 1 2 2 quotient space X/ (f). This means that there exists an x X and y1,y2 (f) such that N 2 2N x1 = x + y1 and x2 = x + y2. Substracting these equations and applying f yields

x x = y y = f(x x )=f(y y ) 1 2 1 2 ) 1 2 1 2 = f(x ) f(x )=f(y ) f(y )=0. ) 1 2 1 2 where we use the linearity of f and y ,y (f). Conversely, suppose f(x ) 1 2 2N 1 f(x2)=0;linearityoff gives

f(x x )=0 = x x (f). 1 2 ) 1 2 2N

This means that there exists y (f)suchthatx1 x2 = 0 + y,whichimplies that x ,x X must belong to2 theN same coset of X/ (f). 1 2 2 N

Codimension of (f)isdeﬁnedtobethedimensionofthequotientspaceX/ (f). Choose anyx ˆ NX/ (f), there exists an x X such thatx ˆ = x + (f).N Since f = 0, there exists2 anN x X (f)suchthat2 f(x ) =0.LookingatProblem9,N 6 0 2 \N 0 6 we deduce thatx ˆ has a unique representationx ˆ = ↵x0 + (f)=↵(x0 + (f)). This shows that x + (f)isabasisforX/ (f)andcodimN (f)=1. N 0 N N N

11. Show that two linear functionals f1 =0andf2 =0whicharedeﬁnedonthesame vector space and have the same null6 space are proportional.6

Solution: Let x, x0 X and consider z = xf1(x0) x0f1(x). Clearly, f1(z)= 0= z (f )=2 (f ). Thus, ) 2N 1 N 2 0=f (z)=f (x)f (x0) f (x0)f (x). 2 2 1 2 1

Since f =0,thereexistssomex0 X (f )suchthatf (x0) =0;wealsohave 1 6 2 \N 1 1 6 f (x0) =0since (f )= (f ). Hence, for such an x0, we have 2 6 N 1 N 2 f2(x0) f2(x)= f1(x). f1(x0) Since x X is arbitrary, the result follows. 2

Page 52 12. (Hyperplane) If Y is a subspace of a vector space X and codimY =1,thenevery element of X/Y is called a hyperplane parallel to Y .Showthatforanylinearfunc- tional f =0onX,thesetH1 = x X : f(x)=1 is a hyperplane parallel to the null space6 (f)off. { 2 } N

Solution: Since f =0onX, H1 is not empty. Fix an x0 H1,andconsider the coset x + (f).6 Note that this is well-deﬁned irrespective2 of elements in H . 0 N 1 Indeed, for any y H1,y = x0, y x0 (f)sincef(y x0)=f(y) f(x0)= 1 1=0;thisshowsthat2 6 x + (f)=2y N+ (f)foranyx, y H . N N 2 1 For any x x + (f), there exists an y (f)suchthatx = x + y. • 2 0 N 2N 0 Since f is linear, f(x)=f(x0 + y)=f(x0)+f(y)=1 = x H1. This shows that x + (f) H . ) 2 0 N ⇢ 1

For any x H1, x = x +0=x + x0 x0 = x0 +(x x0) x0 + (f)since • f(x x )=2 f(x) f(x )=1 1=0.Thisshowsthat H2 x N+ (f). 0 0 1 ⇢ 0 N

Finally, combining the two set inequality gives H1 = x0 + (f)andthestatement follows. N

13. If Y is a subspace of a vector space X and f is a linear functional on X such that f(Y )isnotthewholescalarﬁeldofX,showthatf(y)=0forally Y . 2

Solution: The statement is trivial if f is the zero functional, so suppose f =0. Suppose, by contradiction, that f(y) =0forally Y ,thenthereexistsan6 6 2 y0 Y such that f(y0)=↵ for some nonzero ↵ K.SinceY is a subspace of a vector2 space X, y Y for all K.Bylinearityof2 f, 0 2 2 f(y )=f(y )=↵ f(Y ). 0 0 2 Since K is arbitrary, this implies that f(Y )=K;thisisacontradictionto the assumption2 that f(Y ) = K. Hence, by proof of contradiction, f(y)=0for all y Y . 6 2

14. Show that the norm f of a bounded linear functional f =0onanormedspaceX can be interpreted geometricallyk k as the reciprocal of the distance6 d˜=inf x : f(x)= 1 of the hyperplane H = x X : f(x)=1 from the origin. {k k k 1 { 2 }

Solution:

15. (Half space) Let f = 0 be a bounded linear functional on a real normed space 6 X.Thenforanyscalarc we have a hyperplane Hc = x X : f(x)=c ,andHc determines the two half spaces { 2 }

X = x X : f(x) c and X = x X : f(x) c . c1 { 2  } c2 { 2 }

Page 53 Show that the closed unit ball lies in Xc1,wherec = f ,butforno">0, the half space X with c = f " contains that ball. k k c1 k k

Solution:

Page 54 1.8 Linear Operators and Functionals on Finite Dimensional Spaces. AlinearoperatorT : X Y determines a unique matrix representing T with • respect to a given basis for!X and a given basis for Y ,wherethevectorsofeachof the bases are assumed to be arranged in a ﬁxed order. Conversely, any matrix with r rows and n columns determines a linear operator which it represents with respect to given bases for X and Y .

Let us now turn to linear functionals on X,wheredimX = n and e ,...,e • { 1 n} is a basis for X.Foreveryf X⇤ and every x = ⇠ e X,wehave 2 j j 2 n n P n f(x)=f ⇠jej = ⇠jf(ej)= ⇠j↵j. j=1 ! j=1 j=1 X X X

where ↵j = f(ej)forj =1,...,n.Weseethatf is uniquely determined by its val- ues ↵j ath the n basis vectors of X.Conversely,everyn-tuple of scalars ↵1,...,↵n determines a linear functional on X.

Theorem 1.28. Let X be an n-dimensional vector space and E = e ,...,e a basis { 1 n} for X. Then F = f ,...,f given by f (e )= is a basis for the algebraic dual X⇤ { 1 n} k j jk of X, and dim X⇤ = dim X = n.

f ,...,f is called the dual basis of the basis e ,...,e for X. •{1 n} { 1 n}

Lemma 1.29. Let X be a ﬁnite dimensional vector space. If x X has the property 0 2 that f(x )=0for all f X⇤, then x =0. 0 2 0

Theorem 1.30. A ﬁnite dimensional vector space is algebraically reﬂexive.

1. Determine the null space of the operator T : R3 R2 represented by ! 132 . 210 

Solution: Performing Gaussian elimination on the matrix yields

1320 1320 . 2100 ! 0740  

This yields a solution of the form (⇠1,⇠2,⇠3)=t( 2, 4, 7) where t R is a free variable. Hence, the null space of T is the span of ( 2, 4, 7). 2

Page 55 3 3 2. Let T : R R be deﬁned by (⇠1,⇠2,⇠3) (⇠1,⇠2, ⇠1 ⇠2). Find (T ), (T ) and a matrix! which represents T . 7! R N

Solution: Consider the standard basis for X,givenbye1 =(1, 0, 0),e2 =(0, 1, 0), e =(0, 0, 1). The matrix representing T with respect to e ,e ,e is 3 { 1 2 3} 100 A = 010. 2 3 1 10 4 5 The range of T , (T )istheplane⇠1 + ⇠2 + ⇠3 =0.ThenullspaceofT , (T )is span of (0, 0, 1). R N

3. Find the dual basis of the basis (1, 0, 0), (0, 1, 0), (0, 0, 1) for R3. { }

3 j Solution: Consider a basis e1,e2,e3 of R deﬁned by ej =(⇠n)=jn for { } 3 3 k j =1, 2, 3. Given any x =(⌘j)inR ,letfk(x)= ↵j ⌘j be the dual basis j=1 of e ,e ,e . From the deﬁnition of a dual basis, weX require that f (e )= . { 1 2 3} k j jk More precisely, for f1,werequirethat

1 f1(e1)=↵1 =1. 1 f1(e2)=↵2 =0. 1 f1(e3)=↵3 =0.

which implies that f1(x)=⌘1.Repeatingthesamecomputationforf2 and f3, we ﬁnd that f2(x)=⌘2 and f3(x)=⌘3. Hence,

f1 =(1, 0, 0) f2 =(0, 1, 0) f3 =(0, 0, 1).

3 4. Let f1,f2,f3 be the dual basis of e1,e2,e3 for R ,wheree1 =(1, 1, 1),e2 = (1, 1,{ 1),e =(1} , 1, 1). Find f (x){,f (x),f}(x), where x =(1, 0, 0). 3 1 2 3

Solution: Note that we can write x as 1 1 1 1 1 1 1 1 x = 0 = 1 +0 1 + 1 = e + e . 2 3 2 2 3 2 3 2 2 3 2 1 2 3 0 1 1 1 4 5 4 5 4 5 4 5 Thus, using the deﬁntiion of a dual basis fk(ej)=jk, we have 1 1 1 f (x)= f (e )+ f (e )= . 1 2 1 1 2 1 3 2

Page 56 1 1 f (x)= f (e )+ f (e )=0. 2 2 2 1 2 2 3 1 1 1 f (x)= f (e )+ f (e )= . 3 2 3 1 2 3 3 2

where we use linearity of f1,f2,f3.

5. If f is a linear functional on an n-dimensional vector space X,whatdimensioncan the null space (f)have? N

Solution: The Rank-Nullity theorem states that

dim ( (f)) = dim (X) dim ( (f)) = n dim ( (f)). N R R If f is the zero functional, then (f)=X and (f)hasdimensionn;iff is not the zero functional, then (Nf)=K which hasN dimension 1, so (f)has dimension n 1. Hence, (f)hasdimensionR n or n 1. N N

3 6. Find a basis for the null space of the functional f deﬁned on R by f(x)=⇠1 +⇠2 ⇠3, where x =(⇠1,⇠2,⇠3).

Solution: Let x =(⇠1,⇠2,⇠3)beanypointinthenullspaceoff,theymust satisfy the relation ⇠ + ⇠ ⇠ =0.Thus, 1 2 3

⇠1 ⇠1 1 0 x = ⇠ = ⇠ = ⇠ 0 + ⇠ 1 . 2 23 2 2 3 1 2 3 2 2 3 ⇠3 ⇠1 + ⇠2 1 1 4 5 4 5 4 5 4 5 Hence, a basis for (f)isgivenby (1, 0, 1), (0, 1, 1) . N { }

7. Same task as in Problem 6, if f(x)=↵ ⇠ + ↵ ⇠ + ↵ ⇠ ,where↵ =0. 1 1 2 2 3 3 1 6

Solution: Let x =(⇠ ,⇠ ,⇠ )beanypointin (f), they must satisfy the relation 1 2 3 N ↵2 ↵3 ↵1⇠1 + ↵2⇠2 + ↵3⇠3 =0 ⇠1 = ⇠2 ⇠3. () ↵1 ↵1 Rewriting x using this relation yields

↵2 ↵3 ⇠1 ⇠2 ⇠3 ↵2 ↵3 ↵1 ↵1 ⇠2 ⇠3 x = ⇠2 = 2 ⇠ 3 = ↵1 + 0 . 2 3 2 ↵1 2 3 ↵1 2 3 ⇠3 ⇠ 0 ↵1 6 3 7 4 5 4 5 4 5 4 5 Hence, a basis for (f)isgivenby ( ↵ ,↵ , 0), ( ↵ , 0,↵ ) . N { 2 1 3 1 }

Page 57 8. If Z is an (n 1)-dimensional subspace of an n-dimensional vector space X, show that Z is the null space of a suitable linear functional f on X,whichisuniquely determined to within a scalar multiple.

Solution: Let X be an n-dimensional vector space, and Z an (n 1)-dimensional subspace of X.ChooseabasisA = z1,...,zn 1 of Z, here we can obtain a basis { } B = z1,...,zn 1,zn of X,whereB is obtained by extending A using sifting { } method. Any z Z can be written uniquely as z = ↵1z1 + ...+ ↵n 1zn 1.Ifwe 2 want to ﬁnd f X⇤ such that (f)=Z,usinglinearityoff this translates to 2 N f(z)=f(↵1z1 + ...+ ↵n 1zn 1)=↵1f(z1)+...+ ↵n 1f(zn 1)=0. Since this must be true for all z Z,itenforcestheconditionf(z )=0forall 2 j j =1,...,n 1. A similar argument shows that f(zn) =0,otherwise (f)= X = Z.Thus,thefunctionalwearelookingformustsatisfythefollowingtwo 6 N conditions:6 (a) f(z )=0forallj =1,...,n 1. j (b) f(z ) =0. n 6

Consider a linear functional f : X K deﬁned by f(zj)=jn for all j = 1,...,n.Weclaimthat (f)=Z.Indeed,! N Choose any z Z,thereexistsauniquesequenceofscalars( )suchthat • 2 j z = 1z1 + ...+ n 1zn 1. Using linearity of f,wehave n 1 n 1 f(z)=f jzj = jf(zj)=0. j=1 ! j=1 X X This shows that Z (f). ⇢N Choose any x X Z,thereexistsauniquesequenceofscalars( )such • 2 \ j that x = 1z1 + ...+ nzn and n =0.(otherwisex Z.) Using linearity of f,wehave 6 2 n n f(x)=f z = f(z )= f(z ) =0. j j j j n n 6 j=1 ! j=1 X X This shows that (X Z) (f)orequivalently (f) Z. \ 6⇢ N N ⇢ Hence, Z (f)and (f) Z implies Z = (f). ⇢N N ⇢ N

We are left to show that f is uniquely determined up to scalar multiple. Let be any nonzero scalars and consider the linear functional f. Any x X can be 2 written uniquely as x = ↵1z1 + ...+ ↵nzn. Using linearity of f, we have n n

(f)(x)=f(x)=f ↵jzj = ↵jf(zj) = ↵nf(zn). j=1 ! j=1 ! X X If ↵n =0,thenx Z and (f)(x)=0;if↵n =0,thenx X Z and (f)(x)=↵ =0. 2 6 2 \ n 6

Page 58 9. Let X be the vector space of all real polynomials of a real variable and of degree less than a given n,togetherwiththepolynomialx =0(whosedegreeisleftundefined in the usual discussion of degree). Let f(x)=x(k)(a), the value of the kth derivative (k fixed) of x X at a fixed a R.Showthatf is a linear functional on X. 2 2

Solution: This follows from the algebra rules of di↵erentiation. More precisely, for any x, y X and scalars ↵,, 2 f(↵x + y)=(↵x + y)(k)(a)=↵x(k)(a)+y(k)(a)=↵f(x)+f(y).

10. Let Z be a proper subspace of an n-dimensional vector space X,andletx X Z. 0 2 \ Show that there is a linear functional f on X such that f(x0)=1andf(x)=0for all x Z. 2

Solution: Fix an nonzero x X Z,notethatx = 0 then such a functional 0 2 \ 0 doesn’t exist since f(x0)=0.LetX be an n-dimensional vector space, and Z be an m-dimensional subspace of X,withm

x = ↵1z1 + ...+ ↵mzm + ↵m+1x0 + ↵m+2zm+2 + ...+ ↵nzn.

Consider the linear functional f : X K deﬁned by f(x)=↵ + ...+ ↵ , ! m+1 n where ↵j is the j-th scalar of x with respect to the basis B for all j = m+1,...,n. We claim that f(x0)=1andf(Z)=0.Indeed,

If x = x ,then↵ =1and↵ =0forallj = m +1. • 0 m+1 j 6 If x Z,then↵ =0forallj = m +1,...,n. • 2 j If x X Z,atleastoneof ↵ ,...,↵ is non-zero. • 2 \ { m+1 n}

Remark: The functional f we constructed here is more tight, in the sense that (f)=Z. If we only require that Z (f), then f(x)=↵ will do the job. N ⇢N m+1

11. If x and y are di↵erent vectors in a ﬁnite dimensional vector space X,showthat there is a linear functional f on X such that f(x) = f(y). 6

Solution: Let X be an n-dimensional vector space and e1,...,en abasisof X. Any x, y X can be written uniquely as { } 2 n n

x = ↵jej and y = jej. j=1 j=1 X X

Page 59 Since x = y,thereexistsatleastonej0 1,...,n such that ↵j0 = j0 .Consider the linear6 functional f : X K deﬁned2{ by f(e )=} . Using linearity6 of f, ! j jj0 n n

f(x)=f ↵jej = ↵jf(ej)=↵j0 . j=1 ! j=1 X X n n

f(y)=f jej = jf(ej)=j0 . j=1 ! j=1 X X Clearly, f(x) = f(y). 6

12. If f1,...,fp are linear functionals on an n-dimensional vector space X, where p

Solution:

13. (Linear extension) Let Z be a proper subspace of an n-dimensional vector space X,andletf be a linear functional on Z.Showthatf can be extended linearly to X,thatis,thereisaalinearfunctionalf˜ on X such that f˜ = f. |Z

Solution:

2 14. Let the functional f on R be deﬁned by f(x)=4⇠1 3⇠2,wherex =(⇠1,⇠2). Regard 2 3 ˜ R as the subspace of R given by ⇠3 =0.Determinealllinearextensionsf of f from R2 to R3.

Solution:

3 15. Let Z R be the subspace represented by ⇠2 =0andletf on Z be deﬁned by ⇢ ⇠1 ⇠3 ˜ 3 ˜ f(x)= . Find a linear extension f of f to R such that f(x0)=k (a given 2 constant), where x0 =(1, 1, 1). Is f˜ unique?

Solution:

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