Math 344 Lecture #35 5.6 and Banach Spaces 5.6.1 Pointwise and Uniform Convergence ∞ Definition 5.6.1. Let (fn)n=1 be a of functions with domain X and codomain a Y . ∞ Each x ∈ X gives a sequence in Y defined by (fn(x))n=1. ∞ ∞ We say (fn)n=1 converges pointwise on X if for every x the sequence (fn(x))n=1 converges in Y . ∞ A pointwise convergent sequence (fn)n=1 determines a f : X → Y defined by ∞ f(x) = limn→∞ fn(x), called the pointwise of (fn)n=1. ∞ ∞ Definition 5.6.3. Let (fn)n=1 be a sequence of functions in L (X,Y ) with domain X and codomain a normed linear space (Y, k · kY ). Recall that the L∞-norm on L∞(X,Y ) is

kfk∞ = sup kf(x)kY . x∈X

∞ ∞ We say that (fn)n=1 converges uniformly to a function f ∈ L (X,Y ) if kfn − fk∞ → 0 as n → ∞. Proposition 5.6.5. Uniform convergence implies pointwise convergence. The proof of this is HW (Exercise 5.35). 5.6.2 Banach Spaces Definition 5.6.6. A complete normed linear space is called a .

Theorem 5.6.8. The normed linear space (C([a, b], F), k · k∞) is a Banach space. ∞ Proof. Suppose (fk)k=1 is a Cauchy sequence in (C([a, b], F), k · k∞). ∞ The argument of showing that (fk)k=1 converges in C([a, b], F) proceeds in three steps. ∞ Step 1. We show that (fk)k=1 converges pointwise to a function f :[a, b] → F. ∞ By the assumption of (fk)k=1 being Cauchy, we have for  > 0 the existence of N ∈ N such that kfm − fnk∞ <  when m, n ≥ N. This implies for a fixed x ∈ [a, b] that

|fm(x) − fn(x)| ≤ sup |fm(y) − fn(y)| = kfm − fnk∞ <  y∈[a,b] whenever m, n ≥ N. ∞ Thus the sequence (fk(x))k=1 is Cauchy in F. Since F is complete, this Cauchy sequence converges to a value f(x) ∈ F. ∞ Since x is arbitrary, the Cauchy sequence (fk)k=1 converges pointwise to a function f : [a, b] → F. ∞ Step 2. We show that (fk)k=1 converges uniformly to f.

For  there exists N ∈ N such that kfm − fnk∞ < /2 whenever m, n ≥ N. This says that for each x ∈ [a, b] we have for m, n ≥ N that  |fn(x) − fm(x)| ≤ sup |fn(y) − fm(y)| = kfm − fnk∞ < . y∈[a,b] 2 By Corollary 5.3.22 we can pass the limit through pointwise, i.e., for each x ∈ [a, b] we have  |f(x) − fn(x)| = lim |fm(x) − fn(x)| ≤ m→∞ 2 when m, n ≥ N (which happens for m because m → ∞). This gives  kf − fnk∞ = sup |f(x) − fn(x)| ≤ < , x∈[a,b] 2 so that kf − fnk∞ → 0 as n → ∞. Step 3. We show that f :[a, b] → F is continuous. For  > 0 there exists N ∈ N such that kfm − fnk∞ < /3 whenever m, n ≥ N.

Fixing m = N and passing the limit as n → ∞ through this gives kfN − fk∞ ≤ /3. This says that for every x ∈ [a, b] there holds  |f (x) − f(x)| ≤ kf − fk ≤ . N N ∞ 3

The function fN is continuous on the compact [a, b], so by Theorem 5.5.9, the function fN is uniformly continuous on [a, b].

Thus there exists δ > 0 such that |fN (x) − fN (y)| < /3 whenever |x − y| < δ and x, y ∈ [a, b]. For a fixed x ∈ [a, b] and y ∈ [a, b] satisfying |x − y| < δ we then have

|f(x) − f(y)| ≤ |f(x) − fN (x)| + |fN (x) − fN (y)| + |fN (y) − f(y)|    < + + 3 3 3 = .

This says that f is continuous at x, and since x ∈ [a, b] is arbitrary, that f is continuous on [a, b], hence that f ∈ C([a, b], F).  ∞ Corollary 5.6.9. If a sequence (fn)n=1 in C([a, b], F) converges uniformly to a function f :[a, b] → F, then f ∈ C([a, b], F). Nota Bene. This Corollary holds in a more generally setting: for a compact subset ∞ K of a metric space (X, d), and a Banach space (Y, k · kY ), if a sequence (fn)n=1 of functions in C(K,Y ) converges uniformly to f : K → Y , the limit function f ∈ C(K,Y ). The same argument in the proof of Theorem 5.6.8 proves that the normed linear space (C(K,Y ), k · k∞) is a Banach space. 5.6.3 Sums in Banach Spaces Throughout this subsection we assume that (X, k · k) is a Banach space. ∞ Definition 5.6.10. For a sequence (xk)k=1 in X, we say the

∞ X xk k=1

∞ converges in X if the sequence of partial sums (sk)k=1 defined by

k X sk = xi, i=1 converges in X; otherwise we say the series diverges. ∞ P∞ Definition 5.6.11. For a sequence (xk)k=1 in X, we say the series k=1 xk converges absolutely if the series ∞ X kxkk k=1 converges in R. P∞ Remark 5.6.12. If a series k=1 xk converges absolutely, then for every  > 0 there P∞ exists N ∈ N such that k=n kxkk <  for all n ≥ N. ∞ P∞ Proposition 5.6.13. For a sequence (xk)k=1 in X, if the series k=1 xk converges P∞ absolutely, then the series k=1 converges in X. P∞ Proof. Suppose k=1 kxkk converges in R. P∞ Then by Remark 5.6.12, there is for every  > 0 an N ∈ N such that k=n kxkk <  whenever n ≥ N. Thus for n > m ≥ N we have

n n ∞ X X X ksm − snk = xk ≤ kxkk ≤ kxkk < . k=m+1 k=m+1 k=m+1

∞ This show that the sequence of partial sums (sk)k=1 is Cauchy in X. By the completeness of X, the Cauchy sequence of partial sums converges.  n−1 Example. For each n ∈ N let fn ∈ C([0, 1/2], R) be defined by fn(x) = x . P∞ ∞ The series n=1 fn(x) converges absolutely in the L -norm because

∞ ∞ ∞ ∞ X X X X 1 kf (x)k = sup |xn−1| = (1/2)n−1 = (1/2)k = = 2, n ∞ 1 − 1/2 n=1 n=1 x∈[0,1/2] n=1 k=0 where we recognize a convergent geometric series. P∞ Theorem 5.6.15. If k=1 xk converges absolutely to x ∈ X, then for any rearrange- ment of the terms, the rearranged series converges to x as well; that is, if f : N → N is a Pn bijection, then k=1 xf(k) converges to x. P∞ Proof. For  > 0 choose N ∈ N so that n=N kxnk < /2. Since N is finite, there is M ≥ N such that {1, 2,...,N} is a subset of {f(1), f(2), . . . , f(M)}.

For any n ≥ M let En = {f(1), f(2), . . . , f(M)}\{1, 2,...N}.

The finite set En is a subset of {N + 1,N + 2,... }. Thus we have

n N N n X X X X x − xf(k) = x − xk + xk − xf(k) k=1 k=1 k=1 k=1

N N n X X X ≤ x − xk + xk − xf(k) k=1 k=1 k=1

∞ X X = xk + xk . k=N+1 k∈En To pass the norm to the inside of the sum in the first term, we make use of the continuity of the norm. For L ≥ N + 1 we have by the triangle inequality that

L L X X xk ≤ kxkk. k=N+1 k=N+1 The sum on the inside of the norm on the left side converges as L → ∞, and the right side converges as L → ∞ as well. Since limits preserve inequalities we obtain

L L X X lim xk ≤ lim kxkk. L→∞ L→∞ k=N+1 k=N+1 Since the sum inside the norm on the left-hand side converges, passing the limit inside it gives ∞ L L ∞ X X X X xk = lim xk ≤ lim kxkk = kxkk. L→∞ L→∞ k=N+1 k=N+1 k=N+1 k=N+1

Thus since En is a subset of {N + 1,N + 2,... } we have whenever n ≥ M that

n ∞ X X X   x − x ≤ kx k + kx k < + = . f(k) k k 2 2 k=1 k=N+1 k∈En This shows that the rearranged series converges to x. That the rearranged series converges absolutely follows by applying a simpler argument than that above: for all n ≥ M we have

∞ n ∞ N N n X X X X X X kxkk − kxf(k)k = kxkk − kxkk + kxkk − kxf(k)k k=1 k=1 k=1 k=1 k=1 k=1 ∞ X X   = kx k + kx k < + =  k k 2 2 k=N+1 k∈En

This gives the result.  Remark 5.6.16. The converse of Theorem 5.6.15 is false. In any infinite dimensional Banach space there are series that converge however they are rearranged, but are not absolutely convergent.