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Home , Normed vector space, Norm (mathematics), Space (mathematics)

Introduction to Normed Vector

Audrey Terras March 29, 2009

1 Why worry about in…nite dimensional normed vector spaces?

We want to understand the integral from Lang’s perspective rather than that of your calculus book. Secondly we want to understand convergence of of functions - something that proved problematic for Cauchy in the 1800s. These things are important for many applications in physics, engineering, . We will be able to study vibrating things such as violin strings, drums, buildings, bridges, , planets, stock values. Quantum physics, for example, involves Hilbert , which is a type of normed with a where all Cauchy of vectors converge. The theory of such normed vector spaces was created at the same time as - the 1920s and 1930s. So with this chapter of Lang you are moving ahead hundreds of years from Newton and Leibnitz, perhaps 70 years from Riemann. Fourier series involve orthogonal sets of vectors in an in…nite dimensional :

C[a; b] = f :[a; b] R f continuous : f ! j g The L2 of a continuous f in C[a; b] is b 1=2 f = f(x) 2 dx : k k2 0 j j 1 Za @ A This is an analog of the usual idea of length of a vector f = (f(1); :::; (f(n)) Rn : 2 1=2 n f = f(j) 2 : k k2 0 j j 1 j=1 X @ A There are other natural norms for f C[a; b] such as: 2 b f = f(x) dx: k k1 j j Za

f = max f(x) : k k1 a x b j j   On …nite dimensional vector spaces such as Rn it does not matter what norm you use when you are trying to …gure out whether a of vectors has a limit. However, in in…nite dimensional normed vector spaces convergence can disappear if a di¤erent norm is used. Not all norms are equivalent in in…nite . See Homework 9, problem 6. Note that C[a; b] is in…nite dimensional since the 1; x; x2; x3; :::; xn; ::: is an in…nite set of linearly independent f g n j vectors. Prove this as follows. Suppose that we have a linear dependence relation cjx = 0; for all x in [a; b]. This j=0 X implies all the constants cj = 0: Why? Extra Credit. Prove this. In…nite dimensional vector spaces are thus more interesting than …nite dimensional ones. Each (inequivalent) norm leads to a di¤erent notion of convergence of sequences of vectors.

1 2 What is a Normed Vector Space?

In what follows we de…ne normed vector space by 5 . We will not put arrows on our vectors. We will try to keep vectors and scalars apart by using Greek letters for scalars. Our scalars will be real. Maybe by the end of next quarter we may allow complex scalars. It simpli…es Fourier series.

De…nition 1 A vector space V is a set of vectors v V which is closed under addition and closed under multiplication 2 by scalars R. This means u + v V and v V and the following 5 axioms must hold for all u; v; w V and all 2 2 2 2 ; R: VS1.2 u + (v + w) = (u + v) + w VS2. 0 V s.t. 0 + v = v 9 2 VS3. v V v0 V s.t. v + v0 = 0: VS4. v8 + 2u = u9+ v2 VS5. 1v = v; ( v) = ( )v; ( + )v = v + v; (u + v) = u + v

You may say we cheated by putting 4 axioms into VS5.

De…nition 2 A vector space V is a normed vector space if there is a norm function mapping V to the non-negative real numbers, written v ; for v V; and satisfying the following 3 axioms: k k 2 N1: v 0 v V and v = 0 if and only if v = 0: k k  8 2 k k N2: v = v ; v V and R: Here = of : k k j j k k 8 2 8 2 j j N3: u + v u + v ; u; v V: : k k  k k k k 8 2

De…nition 3 The between 2 vectors u; v in a normed vector space V is de…ned by d(u; v) = u v : k k Example 1. 3-Space. x1 3 R = x2 x1; x2; x3 R : 80 1 2 9 x3 < = @ A De…ne addition and multiplication by scalars as usual: : ; x1 y1 x1 + y1 x2 + y2 = x2 + y2 : 0 1 0 1 0 1 x3 y3 x3 + y3

@ A @ x1 A @ x1 A x2 = x2 ; R: 0 1 0 1 8 2 x3 x3 @ A @ A The usual norm is x1 2 2 2 x = x + x + x ifx = x2 : k k2 1 2 3 0 1 q x3 Other norms are possible: @ A

x 2 = x1 + x2 + x3 or x = max x1 ; x3 ; x3 : k k j j j j j j k k1 fj j j j j jg

The proof that these de…nitions make R3 a normed vector space is tedious. No doubt we will make it a homework problem.

2 Example 2. The space of continuous functions on an .

C[a; b] = f :[a; b] R f continuous : f ! j g For f; g C[a; b]; de…ne (f + g)(x) = f(x) + g(x) for all x [a; b] and de…ne for R ( f)(x) = f(x) for all x [a; b]: We2 leave it as an Exercise (see Homework 9) to check the2 axioms for a vector space.2 The most non-trivial one is the2 one that says f + g and f are both continuous functions on [a; b]. Again there are many possible norms. We will look at 3:

b 1=2 f = f(x) 2 dx : k k2 0 j j 1 Za @ A b f = f(x) dx: k k1 j j Za

f = max f(x) : k k1 a x b j j  

Most of the axioms for norms are easy to check. Let’sdo it for the f 1 norm. N1. v 0 v V and v = 0 if and only if v = 0. k k k k  8 2 k k N2. v = v ; v V and R . N3. Trianglek k j Inequality.j k k 8 u2+ v u8 +2 v ; u; v V: Proof of N1. k k  k k k k 8 2 Since f(x) 0 for all x we know that the integral is 0; because the integral preserves inequalities (Lang, Thm. 2.1, j j  b  p. 104 or earlier in these notes). Suppose that f(x) dx = 0: Since f is continuous by Theorem 2.4 of Lang, p. 104 or j j Za these notes near Figure 1, this implies f(x) = 0 for all x [a; b]: 2

Proof of N2. b b b Also for any R and f C[a; b]; we have f = f(x) dx = f(x) dx = f(x) dx = f : This 2 2 k k1 j j j j j j j j j j j j k k1 Za Za Za proves N2 for norms. Here we used the multiplicative property of absolute value as well as the linearity of the integral (i.e., scalars come out of the integral lecture notes p. 80).

Proof of N3. Using the de…nition of the 1-norm, and the triangle inequality for real numbers as well as the fact that the integral preserves ; we see that  b b f + g = f(x) + g(x) dx ( f(x) + g(x) ) dx: k k1 j j  j j j j Za Za To …nish the proof, use the linearity of the integral to see that

b b b ( f(x) + g(x) ) dx = f(x) dx + g(x) dx = f + g : j j j j j j j j k k1 k k1 Za Za Za

Putting it all together gives f + g f + g : k k1  k k1 k k1

3 3 Scalar Products.

You have seen the dot (or scalar or inner) product in R3: It is

x1 y1 x2 y2 = x1y1 + x2y2 + x3y3: 0 1  0 1 x3 y3 @ A @ A It turns out there is a similar thing for C[a; b]: First let’sde…ne the scalar product on a vector space and see how to get a norm if in addition the scalar product is positive de…nite.

De…nition 4 A (positive de…nite) scalar product < v; w > for vectors v; w in a vector space V is a < v; w > such that the following axioms hold:

SP1: < v; w >=< w; v >; v; w V (symmetry) 8 2 SP2: < u; v + w >=< u; v > + < u; w >; u; v; w V 8 2 SP3: < v; w >= < v; w >; v; w V and R 8 2 8 SP4: < v; v > 0; v V and < v; v >= 0 v = 0 (positive de…nite)  8 2 () Axioms SP1,2,3 say < v; w > is linear in each variable holding the other variable …xed. SP4 says the scalar product is positive de…nite. We will always want to assume SP4 because we want to be able to get a norm out of the scalar product.

De…nition 5 If V is a vector space with a (positive de…nite) scalar product < v; w > for v; w V; de…ne the associated 2 norm by v = p< v; v >; for all v V k k 2

Before proving that this really gives a norm, let’slook at some examples. Example 1. In R3 the scalar product is

x1 y1 < x; y >= x2 y2 = x1y1 + x2y2 + x3y3: 0 1  0 1 x3 y3 @ A @ A It is easy to check the axioms. For example, the positive de…niteness follows from the fact that of real numbers are 0 and sums of non-negative numbers are non-negative:  < x; x >= x2 + x2 + x2 0: 1 2 3  2 And 0 =< x; x > x implies xi = 0 for all i and thus x = 0.  i Example 2. C[a; b] = f :[a; b] R f is continuous f ! j g For f; g in C[a; b], de…ne the scalar product by

b < f; g >= fg: Za

Once more, it is not hard to use the properties of the integral to check axioms SP1,SP2,SP3 (extra credit exercise). To see SP4, note that f(x)2 0 for all x [a; b] implies by the fact that integrals preserve that  2  b < f; f >= f(x)2dx 0:  Za

4 Now suppose that < f; f >= 0. Then by the positivity property of the integral, we know that f(x)2 = 0 for all x [a; b] which says that f is the 0 function (the identity for addition in our vector space C[a; b]). 2 b 1=2 Then the norm associated to this scalar product is f = f(x) 2 dx : k k2 0 j j 1 Za @ A The following theorem is so useful people from lots of countries got their names attached.

Theorem 6 Cauchy-Schwarz (Bunyakovsky) Inequality Suppose that V is a vector space with scalar product < v; w >. Then using our de…nition of the norm v = p< v; v > , we have for all v; w V : k k 2 < v; w > v w : j j  k k k k Proof. (Compare Problem 8 in Homework 8). Let t R and look at 2 f(t) =< v + tw; v + tw > : By properties of the scalar product, we have 0 f(t) =< v; v > +2t < v; w > +t2 < w; w > : As a function of f; we see that f(t) = At2 + Bt + C; where A =< w; w >; B = 2 < v; w > and C =< v; v > : So the graph of f(t) is that of a parabola above or touching the t-axis. For example, we have drawn a parabola touching the t-axis at one .

Recall the quadratic formula for the roots r of f(t) = At2 + Bt + C = 0;  B pB2 4AC r =  :  2A Since we have at most one real root it follows that

B2 4AC 0:  Now plug in A =< w; w >; B = 2 < v; w > and C =< v; v > : This gives the Cauchy-Schwarz inequality.

Corollary 7 Under the hypotheses of the preceding theorem, v = p< v; v > de…nes a norm on V . k k Proof. We must prove: N1. v 0 v V and v = 0 if and only if v = 0. k k  8 2 k k N2. v = v ; v V and R . N3. Trianglek k j Inequality.j k k 8 u2+ v u8 +2 v ; u; v V: k k 2 k k k k 8 2 2 2 We get N2 from SP3:For then v =< v; v >= 2 < v; v >= v ; v V and R . k k j j k k 8 2 8

5 We get N1 from SP4: This says v 2 =< v; v > 0; v V and v 2 =< v; v >= 0 v = 0. To prove the triangle inequalityk N3,k we need to use8 the2 Cauchy-Schwarzk k inequality. This() proof goes as follows. By the linearity and symmetry of the scalar product we see that

u + v 2 = < u + v; u + v >=< u; u > +2 < u; v > + < v; v > k k = u 2 + 2 < u; v > + v 2 k k k k u 2 + 2 < u; v > + v 2 as x x  k k j j k k  j j u 2 + 2 u v + v 2 by Cauchy Schwarz  k k k k k k k k = ( u + v )2 : k k k k Now use the fact that the root p preserves inequalities to …nish the proof of the triangle inequality.

What’sthe good of all this? Now we can happily de…nite limits of sequences of vectors vn in our normed vector space f g V . Can you guess the de…nition of lim vn = L V ? n 2 + !1 Answer: " > 0; N" Z s.t. n N" implies vn L < ": That is, just replace absolute value in the old de…nition of limit with8 the norm.9 2  k k Similarly we can de…ne a vn in the normed vector space V . f g Another use of the scalar product is to de…ne orthogonal vectors in a vector space V with a scalar product.

De…nition 8 Two vectors v; w V; a vector space V with scalar product <; >, are de…ned to be orthogonal if the scalar product < v; w >= 0. 2

In a vector space with scalar product, you can also de…ne the angle  between 2 vectors v; w V; by 2 < v; w >= v w cos : k k k k You can draw the same picture you would draw in the since 2 vectors determine a plane and then use the cosine law from high school trig to see this. What is the cosine law? Using the triangles in Figure 2, it says

v w 2 = v 2 2 v w cos  + w 2 : k k k k k k k k k k You also need to see, using the axioms for scalar product, that

v w 2 =< v w; v w >= v 2 2 < v; w > + w 2 : k k k k k k

6 Figure 1: Visualizing vectors in a normed vector space and the angle between them.

7 4 Comparison of Norms

De…nition 9 Suppose vn is a sequence in a normed vector space V with norm ; We say f g kk lim vn = L n !1 and "vn converges to L V " 2 () lim vn L = 0: n !1 k k Note that the last limit is a sequence of real numbers.

Question. We know that there are lots of norms on V . How can we guarantee that 2 di¤erent norms and produce the same convergent sequences in V ? kk kk The answer is that equivalent norms produce the same convergent sequences where we de…ne equivalent as follows.

De…nition 10 2 norms and on the vector space V are equivalent i¤ there are constants A; B > 0 such that for all v V we have kk kk 2 A v v B v : k k  k k  k k In the preceding de…nition we are assuming that the constants A and B are independent of v V: Why do equivalent norms lead to the same convergent sequences? 2 Answer. Suppose vn is a convergent sequence for the -norm; i.e., for some L V we have lim vn L = 0: f g kk 2 n k k And suppose is an equivalent norm. Then !1 kk A vn L vn L B vn L : k k  k k  k k Since the outside sequences go to 0 as n ; it follows by the squeeze lemma that the guy in the middle has to go to 0 as well. For limits preserve inequalities and! we 1 would have

0 lim vn L 0: n  !1 k k  which implies lim vn L = 0: n !1 k k Similarly lim vn L = 0 implies lim vn L = 0: n k k n k k Moral. It!1 does not matter which of 2 equivalent!1 norms you use to test a sequence for convergence.

Theorem 11 All norms on Rn are equivalent. Proof. See Lang p. 145.

Thus, for our purposes, it does not matter which norm you use on …nite dimensional vector spaces. You get the same de…nition of convergence of sequences. However, things are very di¤erent for in…nite dimensional vector spaces. Look at Exercise 6 in Homework 9 for example. There you see a sequence of functions fn in C[0; 1] such that

1

lim fn 0 = 0 using the norm f = f(x) dx: n 1 1 !1 k k k k j j Z0 but fn 0 = 1 for all n, using the norm f = max f(x) : k k1 k k1 a x b j j   It follows that the norms 1 and on C[0; 1] are not equivalent. kk kk1 1 1=2 2 Extra Credit Exercise. Using the same example, show that lim fn 0 = 0 using the norm f = f(x) dx : n 2 2 !1 k k k k 0 j j 1 Z0 It follows that the norms 2 and on C[0; 1] are not equivalent. @ A kk kk1

8 b b 1=2 Proposition 12 The norms f = f(x) dx and f = f(x)2dx are not equivalent. However, we do have k k1 j j k k2 0 1 Za Za the inequality @ A f pb a f : k k1  k k2 Proof. To prove the inequality, use the Cauchy-Schwarz inequality on the functions f and g(x) = 1 for all x [a; b]: This gives j j 2 < f; g > f g : j j  k k2 k k2 So we have < f ; 1 > f 1 : Now this really means j j j j  kj jk2 k k2

b b 1=2 b 1=2 f = f(x) dx f(x)2dx 1dx k k1 j j  0 1 0 1 Za Za Za b @ 1=2 A @ A = pb a f(x)2dx = pb a f : 0 1 k k2 Za @ A To see that 1 and 2 are not equivalent norms, we take a = 0 and b = 1. Then we look at the following example. De…ne as in Lang,kk p. 147:kk pn; for 0 x 1 ; g (x) = n n 1 ; for 1  x  1:  px n   Note that gn is continuous. Why? Extra Credit Exercise. Show 1 gn = 2 k k1 pn and gn = 1 + log n: k k2 It follows that there cannot be a constant C > 0 such thatp v 2 C v 1 at least on the interval [0; 1]. Can you extend this idea to arbitrary intervals [a; b]? k k  k k

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