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Week 9 Discussion Notes

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Uniform Convergence and Differentiability

In Week 7, we saw that even if fn→ff_n \to f uniformly, ff is not guaranteed to be differentiability at even a single point. This tells us that uniform convergence of fnf_n is independent from the differentiability of ff, so our conditions will need to look relatively different:

Theorem

Assume that fn,g ⁣:[a,b]→R\func{f_n,g}{\br{a,b}}{\R} are functions such that:

  1. fn′f_n' exists for all n∈Nn \in \N
  2. fn′→gf_n' \to g uniformly on [a,b]\br{a, b}
  3. there exists x0∈[a,b]x_0 \in \br{a, b} such that {fn(x0)}n\set{f_n\p{x_0}}_n converges

Then fnf_n converges uniformly to a function ff on [a,b]\br{a, b} and ff is differentiable with f′=gf' = g on (a,b)\p{a, b}.
Remark.

The theorem tells us that under the conditions above, we can swap limits in the following:

lim⁡n→∞lim⁡h→0fn(x+h)−fn(x)h=lim⁡h→0lim⁡n→∞fn(x+h)−fn(x)h\lim_{n\to\infty} \lim_{h\to0} \frac{f_n\p{x+h} - f_n\p{x}}{h} = \lim_{h\to0} \lim_{n\to\infty} \frac{f_n\p{x+h} - f_n\p{x}}{h}

The proof of the theorem is quite hard, which tells you that swapping limits is hard in general, so always take a second to think before you try swapping things!

This theorem is very useful when dealing with power series. For now, I'll illustrate its usefulness with a concrete example:

Example 1.

Let f(x)=∑n=0∞xnf\p{x} = \sum_{n=0}^\infty x^n. Show that ff is differentiable on (−1,1)\p{-1, 1} and that

f′(x)=∑n=1∞nxn−1.f'\p{x} = \sum_{n=1}^\infty nx^{n-1}.
Solution.

Notice that the formula for f′f' says

ddx∑n=0∞xn=∑n=0∞ddxxn.\deriv{}{x} \sum_{n=0}^\infty x^n = \sum_{n=0}^\infty \deriv{}{x} x^n.

To take a derivative, we need to take a limit, and to calculate an infinite sum, we need to take another limit. This tells us that even though it looks obvious, this fact is not trivial since we need to interchange two limits.

To show that ff is differentiable on (−1,1)\p{-1, 1}, we can just show that ff is differentiable on any strict subset (−r,r)\p{-r, r}, where 0≤r<10 \leq r < 1. We want to restrict ourselves to a smaller set for two reasons:

  1. Differentiability is a local property (i.e., f′(x)f'\p{x} only depends on values of ff really close to xx), meaning we don't have to use the whole set to prove things about it.
  2. Bad things happen close to the endpoints −1-1 and 11, so we want to avoid them.

Let fN(x)=∑n=0Nxnf_N\p{x} = \sum_{n=0}^N x^n. We'll try to apply the theorem above to this sequence.

Because fNf_N is a finite sum, we already know

fN′(x)=∑n=1Nnxn−1,f_N'\p{x} = \sum_{n=1}^N nx^{n-1},

so (i) holds. To show (ii), we need to show that fN′f_N' converges uniformly, i.e., we need to show that an infinite series of functions converges uniformly. This tells us that we'll want to use the Weierstrass M-test, so first calculate

sup⁡x∈[−r,r]∣nxn−1∣=nrn−1.\sup_{x \in \br{-r,r}}{\abs{nx^{n-1}}} = nr^{n-1}.

To show that ∑n=1∞nrn−1\sum_{n=1}^\infty nr^{n-1} converges, we'll use the root test. Recall that lim⁡n→∞n1n=1\lim_{n\to\infty} n^{\frac{1}{n}} = 1, so by our limit laws,

lim⁡n→∞∣nrn−1∣1n=(lim⁡n→∞n1n)(lim⁡n→∞rn−1n)=r∈(0,1).\lim_{n\to\infty} \abs{nr^{n-1}}^{\frac{1}{n}} = \p{\lim_{n\to\infty} n^{\frac{1}{n}}}\p{\lim_{n\to\infty} r^{\frac{n-1}{n}}} = r \in \p{0, 1}.

(We also used the fact that rxr^x is continuous here.) This step is also where working on the compact set [−r,r]\br{-r, r} pays off, since the root test would be inconclusive if we tried to use it on the whole interval. The root test tells us

∑n=1∞sup⁡x∈[−r,r]∣nxn−1∣=∑n=1∞nrn−1<∞,\sum_{n=1}^\infty \sup_{x \in \br{-r,r}}{\abs{nx^{n-1}}} = \sum_{n=1}^\infty nr^{n-1} < \infty,

and so by the Weierstrass M-test, fN′f_N' converges uniformly to g(x)=∑n=1∞nxn−1g\p{x} = \sum_{n=1}^\infty nx^{n-1}, so (ii) holds.

Finally, a power series always converges at its center, i.e., fN(0)=1f_N\p{0} = 1, which converges as N→∞N \to \infty, so (iii) holds. By the theorem, ff is differentiable on (−r,r)\p{-r, r} and

f′(x)=∑n=1∞nxn−1.f'\p{x} = \sum_{n=1}^\infty nx^{n-1}.

for any x∈(−r,r)x \in \p{-r, r}.

Given any x∈(−1,1)x \in \p{-1, 1}, we can find 0≤r<10 \leq r < 1 (e.g., r=∣x∣+12r = \frac{\abs{x}+1}{2}) such that x∈(−r,r)x \in \p{-r, r}, so we conclude that ff is differentiable on all of (−1,1)\p{-1, 1} with the same formula above.
Exercise 1.

With the same ff as above, show that F(x)=∑n=0∞xn+1n+1F\p{x} = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} is an antiderivative for ff.

Hint: Show that F′=fF' = f (as opposed to trying to show that ∫f(x) dx=F(x)+C\int f\p{x} \,\diff{x} = F\p{x} + C). The proof should look mostly the same as the example.

Stone-Weierstrass

Theorem (Stone-Weierstrass)

Let (X,d)\p{X, d} be a compact metric space and let A⊆C(X,R)A \subseteq C\p{X, \R} be a subset such that:

  1. 1∈A1 \in A
  2. (AA is an R\R-algebra) for any f,g∈Af, g \in A and λ∈A\lambda \in A, we have λf+g,fg∈A\lambda f + g, fg \in A
  3. (AA separates points) for any x≠yx \neq y in XX, there exists f∈Af \in A such that f(x)≠g(x)f\p{x} \neq g\p{x}

Then AA is dense in C(X,R)C\p{X, \R}.

This theorem is important because continuous functions are determined uniquely by their values on dense sets:

Proposition

Let f∈C(X,R)f \in C\p{X, \R} and assume that AA is a dense subset of XX (i.e., that A‾=X\cl{A} = X). If f(a)=0f\p{a} = 0 for all a∈Aa \in A, then f(x)=0f\p{x} = 0 for all x∈Xx \in X.
Proof.

Let x∈Xx \in X. By density of AA, there exists a sequence {an}n⊆A\set{a_n}_n \subseteq A such that an→xa_n \to x. Thus, by continuity of ff,

f(x)=f(lim⁡n→∞an)=lim⁡n→∞f(an)=lim⁡n→∞0=0.f\p{x} = f\p{\lim_{n\to\infty} a_n} = \lim_{n\to\infty} f\p{a_n} = \lim_{n\to\infty} 0 = 0.

□\square
Corollary

Let f,g∈C(X,R)f, g \in C\p{X, \R} and assume that AA is dense a dense subset of XX. If f(a)=g(a)f\p{a} = g\p{a} for all a∈Aa \in A, then f(x)=g(x)f\p{x} = g\p{x} for all x∈Xx \in X.
Proof.

h=f−g∈C(X,R)h = f - g \in C\p{X, \R} and is 00 on a dense set, so h=0h = 0 on all of XX, i.e., f(x)=g(x)f\p{x} = g\p{x} for all x∈Xx \in X. □\square
Example 2.

Show that there is exactly one linear function F∈C(X,R)F \in C\p{X, \R}, where X=C([0,1],R)X = C\p{\br{0,1}, \R}, such that F(x2n)=1F\p{x^{2n}} = 1 for all n≥0n \geq 0.
Solution.

To prove something like this, we need to show two things:

  1. Existence: there is at least one FF that works.
  2. Uniqueness: there is at most one FF that works.

To show existence, we can just give an example of a function. Define F ⁣:X→R\func{F}{X}{\R} via F(f)=f(1)F\p{f} = f\p{1} (so FF sends a continuous function ff to a number by evaluating it at x=1x = 1). We need to show that FF is linear and continuous: let f,g∈Xf, g \in X and let λ∈R\lambda \in \R. Then

F(λf+g)=(λf+g)(1)=λf(1)+g(1)=λF(f)+F(g),\begin{aligned} F\p{\lambda f + g} &= \p{\lambda f + g}\p{1} \\ &= \lambda f\p{1} + g\p{1} \\ &= \lambda F\p{f} + F\p{g}, \end{aligned}

so FF is linear. To show continuity,

∣F(f)−F(g)∣=∣f(1)−g(1)∣≤sup⁡x∈[0,1]∣f(x)−g(x)∣=d∞(f,g),\begin{aligned} \abs{F\p{f} - F\p{g}} &= \abs{f\p{1} - g\p{1}} \\ &\leq \sup_{x \in \br{0,1}}{\abs{f\p{x} - g\p{x}}} \\ &= d_\infty\p{f, g}, \end{aligned}

i.e., FF is Lipschitz, hence continuous (by a old homework problem). This proves existence.

To show uniqueness, let GG be another such function, i.e., GG is a continuous linear function such that G(x2n)=1G\p{x^{2n}} = 1 for all n≥0n \geq 0. We need to show that F=GF = G on all of XX. Notice that

F(∑n=0Nanx2n)=∑n=0NanF(x2n)(F is linear)=∑n=0NanG(x2n)(F(x2n)=1=G(x2n))=G(∑n=0Nanx2n),(G is linear)\begin{aligned} F\p{\sum_{n=0}^N a_nx^{2n}} &= \sum_{n=0}^N a_n F\p{x^{2n}} && \p{F\text{ is linear}} \\ &= \sum_{n=0}^N a_n G\p{x^{2n}} && \p{F\p{x^{2n}} = 1 = G\p{x^{2n}}} \\ &= G\p{\sum_{n=0}^N a_n x^{2n}}, && \p{G\text{ is linear}} \end{aligned}

so F=GF = G on the set

A={∑n=0Nanx2n | N∈N, an∈R}⊆X.A = \set{\sum_{n=0}^N a_nx^{2n} \st N \in \N,\ a_n \in \R} \subseteq X.

If we can show that AA is dense in XX, then by the corollary above, we get F=GF = G on all of XX. To show that, we can use Stone-Weierstrass:

(i) holds since if n=0n = 0, then 1=x2n∈A1 = x^{2n} \in A. To show (ii), let f(x)=∑n=0Nanx2nf\p{x} = \sum_{n=0}^N a_nx^{2n}, g(x)=∑m=0Mbmx2mg\p{x} = \sum_{m=0}^M b_mx^{2m}, and λ∈R\lambda \in \R. Without loss of generality, we may assume that N≤MN \leq M (otherwise, we can just swap the roles of ff and gg), so we get

λf(x)+g(x)=∑n=0N(λan+bn)x2n+∑n=N+1Mbnx2n∈Af(x)g(x)=∑n=0N∑m=0M(anbm)x2(n+m)∈A,\begin{gathered} \lambda f\p{x} + g\p{x} = \sum_{n=0}^N \p{\lambda a_n + b_n} x^{2n} + \sum_{n=N+1}^M b_n x^{2n} \in A \\ f\p{x}g\p{x} = \sum_{n=0}^N \sum_{m=0}^M \p{a_nb_m} x^{2\p{n+m}} \in A, \end{gathered}

so AA is an R\R-algebra. Finally, for (iii), notice that x2∈Ax^2 \in A is injective, so if x≠yx \neq y in [0,1]\br{0, 1}, then x2≠y2x^2 \neq y^2. Thus, AA separates points.

By Stone-Weierstrass, AA is dense in XX, so F=GF = G on all of XX, which proves uniqueness.

True or False

Example 3.

True or false: If fn→ff_n \to f uniformly on compact sets of R\R, then fn→ff_n \to f uniformly on all of R\R.
Solution.

This is false. A simple example is a moving step function, i.e.,

fn(x)={1if x∈[n,n+1],0otherwise.f_n\p{x} = \begin{cases} 1 & \text{if } x \in \br{n, n+1}, \\ 0 & \text{otherwise}. \end{cases}

Any compact set K⊆RK \subseteq \R is bounded, so there exists N∈NN \in \N such that sup⁡K<N\sup{K} < N. Thus, fn(x)=0f_n\p{x} = 0 on KK for all n≥Nn \geq N, so fn→ff_n \to f uniformly on KK, hence on any compact set. However,

sup⁡x∈R∣fn(x)−0∣=1,\sup_{x \in \R}{\abs{f_n\p{x} - 0}} = 1,

so fnf_n does not converge uniformly to 00 on all of R\R.