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1 Banach Spaces and Hilbert Spaces

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1 Banach Spaces and Hilbert Spaces 1 Banach spaces and Hilbert spaces We will first discuss Banach spaces, since much of what we say applies to Hilbert spaces, without change. Let V be a normed linear space over either the real or complex numbers. 1 A sequence of vectors fvjgj=1 is a map from the natural numbers to V . We say that vj converges to v 2 V if lim kvj − vk = 0: j!1 A sequence fvjg is said to be Cauchy if for each > 0, there exists a natural number N such that kvj − vkk < for every j; k ≥ N. Every convergent sequence is Cauchy, but there are many examples of normed linear spaces V for which there exists non-convergent Cauchy sequences. One such example is thep set of rational numbers Q. The sequence (1:4; 1:41; 1:414;::: ) converges to 2 which is not a rational number. We say a normed linear space is complete if every Cauchy sequence is convergent in the space. The real numbers are an example of a complete normed linear space. We say that a normed linear space is a Banach space if it is complete. We call a complete inner product space a Hilbert space. Consider the following examples: 1. Every finite dimensional normed linear space is a Banach space. Like- wise, every finite dimensional inner product space is a Hilbert space. 2. Let x = (x1; x2; :::; xn; :::) be a sequence. The following spaces of se- quences are Banach spaces: 1 p X p ` = fx : jxjj = kxk`p < 1g; 1 ≤ p < 1 (1.1) j=1 1 ` = fx : sup jxjj < 1g (1.2) j c = fx : lim xj existsg; kxkc = kxk1 (1.3) j!1 c0 = fx : lim xj = 0g; kxkc = kxk1 (1.4) j!1 0 Except for `1. the spaces above are separable { i.e., each has a countably dense subset, 1 3. The following function spaces are Banach spaces: C[0; 1]; kfkC[0;1] = max jf(x)j (1.5) x2[0;1] k (k) X j C [a; b]; kfkC(k)[a;b] = sup jf (x)j (1.6) j=0 x2[0;1] 1 Z p Lp(I); kfk = jf(x)jp (1.7) I 1 L (I); kfkL1(I) = ess-supx2I jf(x)j (1.8) There are two Hilbert spaces among the spaces listed: the sequence space `2 and the function space L2. In the result below, we will show that `1 is complete. After that, we will show that C[0; 1] is complete, relative to the sup-norm, kfkC[0;1] = max jf(x)j. Of course, this means that both of them are Banach spaces. Proposition 1.1. The space `1 is a Banach space. Proof. The norm on `1 is given by kxk1 = sup jx(j)j: j 1 1 1 Let fxngn=1 ⊂ ` denote a Cauchy sequence of elements in ` . We show 1 that this sequence converges to x 2 ` . Since fxng is Cauchy, for each > 0, there exists an N such that for all n; m ≥ N, kxn − xmk1 < . (1.9) This implies that jxn(j) − xm(j)j < for all j. Therefore, the sequence fxn(j)g is a Cauchy sequence of real numbers, and hence converges to some value x(j). That is, limn!1 xn(j) = x(j) exists. From (1.9), if we choose = 1, then for all n; m ≥ N, we have kxn − xmk1 < 1: In particular, it follows that kxnk < 1 + kxmk; n; m ≥ N: 2 Fix m. Then, for all n ≥ N, jxn(j)j ≤ kxnk1 < 1 + kxmk1. Letting n ! 1, we see that jx(j)j ≤ 1 + kxmk1 holds uniformly in j. Therefore, x 2 `1. To complete the proof, we need to show that xn converges to x in norm. We have jxn(j) − xm(j)j < 8j 2 N and n; m ≥ N: Let n ! 1. Then, xn(j) ! x(j) so we have that jx(j) − xm(j)j < 8j 2 N; m ≥ N: Since this holds for all j, it follows that kx − xmk1 < for all m ≥ N. 1 Therefore, the sequence xm converges to x 2 ` . Proposition 1.2. Relative to the sup norm, C[0; 1] is complete and is thus a Banach space. 1 Proof. Let ffn(x)gn=1 be a Cauchy sequence in C[0; 1]. Then, for every > 0, there exists an N such that kfn − fmk < for all n; m ≥ N. For any fixed t 2 [0; 1], this implies that jfn(t) − fm(t)j < 8 m; n ≥ N: (1.10) 1 Thus, for t fixed, ffn(t)gn=1 is a Cauchy sequence of real numbers, and so it converges. Define f(t) by the pointwise limit of this sequence: f(t) = lim fn(t); t 2 [0; 1]: (1.11) n!1 By taking the limit as m ! 1 in (1.10), we see that jfn(t) − f(t)j ≤ 8 n ≥ N; which holds uniformly for all t 2 [0; 1]. Consequently, kfn − fk = sup jfn(t) − f(t)j ≤ , 8 n ≥ N; t2[0;1] and so limn!1 kfn − fk = 0. What remains is to show that f 2 C[0; 1]. To do that, fix t and let > 0. Because fn 2 C[0; 1], there exists a δ > 0 such that jf (t + h) − f (t)j < 8 jhj < δ; n n 3 3 assuming, of course, that t+h 2 [0; 1]. Then applying the triangle inequality gives jf(t + h) − f(t)j ≤ jf(t + h) − fn(t + h)j + jfn(t + h) − fn(t)j + jfn(t) − f(t)j: From (1.11), we may choose n so large that both jf(t + h) − fn(t + h)j < 3 and jf(t) − fn(t)j < 3 . Once we have found an n such that this holds, we note that for these n jf(t + h) − f(t)j < + jf (t + h) − f (t)j + : 3 n n 3 Since the fn are continuous functions, we can find a δ such that jfn(t + h) − f(t)j < 3 whenever jhj < δ. It follows that, for this choice of δ, we have jf(t + h) − f(t)j < + + = 3 3 3 whenever jhj < δ, which implies that f 2 C[0; 1]. Just because a space is complete relative to one norm does not mean that the same space will also be complete in another. The example below illustrates this for the space of continuous functions, C[0; 1]. Example 1.1. If we replace the sup norm on C[0; 1] with the L1 norm, then C[0; 1] is not complete. Proof. To simplify the discussion, we will work with [−1; 1] rather than [0; 1]. Consider the sequence of continuous functions fn 2 C[−1; 1] defined piece- wise by 8 1 >−1 x 2 [−1; − n ] < 1 1 fn(t) = nx x 2 [− n ; n ] :> 1 1 x 2 [ n ; 1]: Let n > m. We note that jfn(t) − fm(t)j will be symmetric about t = 0. Using this, we see that 8 1 >nt − mt t 2 [0; n ] < 1 1 jfn(t) − fm(t)j = 1 − mt t 2 [ n ; − m ] :> 1 0 t 2 [ m ; 1] 4 Computing the integrals over [0; 1] yields 1 1 Z 1 Z n Z m jfn(t) − fm(t)j dt = (n − m)t dx + 1 − mt dx 1 0 0 n 1 1 1 m 1 m 1 = (n − m) + − − + 2n2 m n 2 m2 2 n2 1 1 1 = − 2 m n Making use of the symmetry in conjunction with the result above then gives 1 1 2 us kfn − fmkL1[−1;1] = m − n . Let N > . Then, for m; n ≥ N, we find that 1 1 kf − f k 1 < + < + = , n m L [−1;1] n m 2 2 and so the sequence is Cauchy in the L1[−1; 1] norm. However, the limit is not continuous. A computation shows that, for the step function f(t) defined by 8 −1 t 2 [−1; 0) <> f(t) = 0 t = 0 :>1 t 2 (0; 1]; 1 we have kf − fnkL1[−;1] = n , and so lim kf − fnkL1 = 0: n!1 Therefore, the sequence of functions fn 2 C[−1; 1] converges to a discontin- uous function under the L1[−1; 1] norm, and, consequently, C[−1; 1] is not complete under the L1[−1; 1] norm. 2 Approximating Continuous Functions Modulus of continuity Every f 2 C[0; 1] is uniformly continuous: for every > 0, there is a δ > 0 such that jf(t2) − f(t1)j < (2.1) as long as t1; t2 2 [0; 1] satisfy jt2 − t1j < δ. Using this fact, we are able to make the following definition: 5 Definition 2.1. Let f 2 C[0; 1]. The modulus of continuity for f is defined to be !(f; δ) = sup jf(t1) − f(t2)j; 8 t1; t2 2 [0; 1]: (2.2) jt1−t2|≤δ p p Example 2.1. Let f(x) = x, 0 ≤ x ≤ 1. Show that !(f; δ) ≤ C δ. Proof. Let 0 < s < t ≤ 1. We note that p p t − s 0 < t − s = p p t + s p p t − s = t − sp p t + s p p1 − s = t − s t 1 + p s p p t ≤ t − s = δ p Hence, !(f; δ) ≤ δ. To get equality, take s = 0 and t = δ. (1) 0 Example 2.2. Suppose that f 2 C [0; 1]. Show that !(f; δ) ≤ jjf jj1δ.
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