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Week 8 Discussion Notes

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Infinite Series

Definition

Let (X,dX)\p{X, d_X} be a metric space and (Y,∥ ⋅ ∥Y)\p{Y, \norm{\:\cdot\:}_Y} be a normed vector space. Let fn∈B(X,Y)f_n \in B\p{X, Y} be a sequence of functions. We say

  1. ∑n=1∞fn\sum_{n=1}^\infty f_n converges pointwise if for all x∈Xx \in X, ∑n=1Nfn(x)\sum_{n=1}^N f_n\p{x} converges as N→∞N \to \infty.
  2. ∑n=1∞fn\sum_{n=1}^\infty f_n converges uniformly if ∑n=1Nfn\sum_{n=1}^N f_n converges uniformly as N→∞N \to \infty.
  3. ∑n=1∞fn\sum_{n=1}^\infty f_n converges pointwise absolutely if for all x∈Xx \in X, ∑n=1N∥fn(x)∥Y\sum_{n=1}^N \norm{f_n\p{x}}_Y converges as N→∞N \to \infty.
  4. ∑n=1∞fn\sum_{n=1}^\infty f_n converges uniformly absolutely if ∑n=1Nsup⁡x∈X∥fn(x)∥Y\sum_{n=1}^N \sup_{x \in X}\norm{f_n\p{x}}_Y converges as N→∞N \to \infty.
Remark.

The main reason we want (Y,∥ ⋅ ∥Y)\p{Y, \norm{\:\cdot\:}_Y} to be a normed vector space instead of just a metric space is because we need to be able to add functions together, and in general metric spaces, we don't have a notion of addition.

Also, in most cases, (Y,∥ ⋅ ∥Y)=(R,∣ ⋅ ∣)\p{Y, \norm{\:\cdot\:}_Y} = \p{\R, \abs{\:\cdot\:}}.
Theorem (Weierstrass M-test)

Let (X,dX)\p{X, d_X}, (Y,∥ ⋅ ∥Y)\p{Y, \norm{\:\cdot\:}_Y}, and fn∈B(X,Y)f_n \in B\p{X, Y} be as above. If YY is complete (with respect to the metric induced by the norm), then (iv)   ⟹  \implies (ii).

Since R\R is complete, the Weierstrass M-test will be our main way of proving uniform convergence of series for most problems.

Example 1.

Let fn(x)=xnf_n\p{x} = x^n, where X=(−1,1)X = \p{-1, 1} and Y=RY = \R. Show that ∑n=1∞xn\sum_{n=1}^\infty x^n converges uniformly on any compact subset of (−1,1)\p{-1, 1}, but not uniformly on all of (−1,1)\p{-1, 1}.
Solution.

Without loss of generality, we may assume that our compact set is [−r,r]\br{-r, r}. Indeed, given any compact set K⊆(−1,1)K \subseteq \p{-1, 1}, we have

K⊆⋃r∈(0,1)(−r,r)K \subseteq \bigcup_{r \in \p{0, 1}} \p{-r, r}

so by compactness, there exist 0<r1<r2<⋯<rn<10 < r_1 < r_2 < \cdots < r_n < 1 such that

K⊆⋃i=1n(−ri,ri)=(−rn,rn).K \subseteq \bigcup_{i=1}^n \p{-r_i, r_i} = \p{-r_n, r_n}.

Since KK is closed, K=K‾⊆(−rn,rn)‾=[−rn,rn]K = \cl{K} \subseteq \cl{\p{-r_n, r_n}} = \br{-r_n, r_n}. Thus, if the series converges uniformly on [−rn,rn]\br{-r_n, r_n}, then it converges uniformly on KK also.

Now, let 0≤r<10 \leq r < 1. Then

sup⁡x∈[−r,r]∣xn∣=rn  ⟹  ∑n=1∞sup⁡x∈[−r,r]∣xn∣=∑n=1∞rn<∞,\sup_{x \in \br{-r,r}} \abs{x^n} = r^n \implies \sum_{n=1}^\infty \sup_{x \in \br{-r,r}} \abs{x^n} = \sum_{n=1}^\infty r^n < \infty,

since it is a geometric series with ratio ∣r∣<1\abs{r} < 1. By the Weierstrass M-test, we get ∑n=1∞xn\sum_{n=1}^\infty x^n converges uniformly on [−r,r]\br{-r, r}, hence on any compact subset of (−1,1)\p{-1, 1}.

To show that the convergence is not uniform on all of (−1,1)\p{-1, 1}, we proceed by contradiction. Assume that ∑n=1∞xn\sum_{n=1}^\infty x^n converges uniformly on (−1,1)\p{-1, 1}. Since the pointwise and uniform limits (when they exist) must agree, this means

∑n=1Nxn→N→∞11−x\sum_{n=1}^N x^n \xrightarrow{N\to\infty} \frac{1}{1 - x}

uniformly on (−1,1)\p{-1, 1}. But ∑n=1Nxn∈B(X,R)\sum_{n=1}^N x^n \in B\p{X, \R} and B(X,R)B\p{X, \R} is complete, i.e., uniform limits of bounded functions are bounded, which implies 11−x\frac{1}{1 - x} is bounded on (−1,1)\p{-1, 1}, a contradiction. Thus, the series does not converge uniformly on all of (−1,1)\p{-1, 1}.
Example 2.

Prove that ∑n=1∞nx1+n5x2\sum_{n=1}^\infty \frac{nx}{1 + n^5x^2} converges uniformly on R\R.
Solution.

Our main tool is the Weierstrass M-test, so let's try to calculate

sup⁡x∈R∣fn(x)∣wherefn(x)=nx1+n5x2.\sup_{x \in \R}{\abs{f_n\p{x}}} \quad\text{where}\quad f_n\p{x} = \frac{nx}{1 + n^5x^2}.

We'll try to maximize it using the first derivative test. Since fnf_n is an odd function, to maximize ∣fn∣\abs{f_n} on all of R\R, we can just maximize fnf_n for x≥0x \geq 0:

fn′(x)=n−n6x2(1+n5x2)2=0  ⟹  x=n−52f_n'\p{x} = \frac{n - n^6x^2}{\p{1 + n^5x^2}^2} = 0 \implies x = n^{-\frac{5}{2}}

Also,

fn(n−52)=12n−32>0.f_n\p{n^{-\frac{5}{2}}} = \frac{1}{2n^{-\frac{3}{2}}} > 0.

Now we need to prove that this actually gives us the global maximum of fnf_n. Notice that

lim⁡x→∞fn(x)=lim⁡x→∞nx1x2+n5=0n5=0.\lim_{x\to\infty} f_n\p{x} = \lim_{x\to\infty} \frac{\frac{n}{x}}{\frac{1}{x^2} + n^5} = \frac{0}{n^5} = 0.

(We're allowed to use limit laws since the limit of the denominator is non-zero.)

Thus, because fn(n−52)>0f_n\p{n^{-\frac{5}{2}}} > 0, there exists M>0M > 0 such that if x≥Mx \geq M, then fn(x)<fn(n−52)f_n\p{x} < f_n\p{n^{-\frac{5}{2}}}. The next part of the argument is a little tricky, so let me break it down into steps:

  1. If x≥Mx \geq M, then fn(x)f_n\p{x} is strictly smaller than fn(n−32)f_n\p{n^{-\frac{3}{2}}}, so if fnf_n has a maximum, it must occur on the interval [0,M]\br{0, M}.
  2. [0,M]\br{0, M} is compact and fnf_n is continuous, so fnf_n must have a maximum on [0,M]\br{0, M}. In other words, fnf_n has a global maximum.
  3. Since fnf_n is differentiable on [0,M]\br{0, M}, the maximizer xmaxx_{\mathrm{max}} of fnf_n must be a critical point, i.e., fn′(xmax)=0f_n'\p{x_{\mathrm{max}}} = 0.
  4. The only critical point of fnf_n is n−52n^{-\frac{5}{2}}, so xmax=n−52x_{\mathrm{max}} = n^{-\frac{5}{2}}.

In summary,

sup⁡x∈R∣fn(x)∣=fn(n−52)=12n32,\sup_{x \in \R}{\abs{f_n\p{x}}} = f_n\p{n^{-\frac{5}{2}}} = \frac{1}{2n^{\frac{3}{2}}},

so by the pp-series test with p=32>1p = \frac{3}{2} > 1,

∑n=1∞sup⁡x∈R∣fn(x)∣=∑n=1∞12n32<∞.\sum_{n=1}^\infty \sup_{x \in \R}{\abs{f_n\p{x}}} = \sum_{n=1}^\infty \frac{1}{2n^{\frac{3}{2}}} < \infty.

So by the Weierstrass M-test, the series converges uniformly on R\R.
Example 3.

Show that ∑n=1∞n2+x4n4+x2\sum_{n=1}^\infty \frac{n^2 + x^4}{n^4 + x^2} converges to a continuous function on R\R.
Solution.

Warning: It might be tempting to try to prove that the series converges uniformly on R\R (since that would imply that the series converges to a continuous function), but this series does not converge uniformly on R\R. Indeed, let f(x)=∑n=1∞n2+x4n4+x2f\p{x} = \sum_{n=1}^\infty \frac{n^2 + x^4}{n^4 + x^2}, and because all terms are positive,

∣f(x)−∑n=1∞n2+x4n4+x2∣=∑n=N+1∞n2+x4n4+x2≥(N+1)2+x4(N+1)4+x2.\abs{f\p{x} - \sum_{n=1}^\infty \frac{n^2 + x^4}{n^4 + x^2}} = \sum_{n=N+1}^\infty \frac{n^2 + x^4}{n^4 + x^2} \geq \frac{\p{N+1}^2 + x^4}{\p{N+1}^4 + x^2}.

(ff is larger than its partial sums, so we can drop the absolute value, and a sum of positive numbers is at least one of its terms, e.g., the first one.)

But

lim⁡x→∞(N+1)2+x4(N+1)4+x2=∞  ⟹  sup⁡x∈R∣f(x)−∑n=1∞n2+x4n4+x2∣=∞,\lim_{x\to\infty} \frac{\p{N+1}^2 + x^4}{\p{N+1}^4 + x^2} = \infty \implies \sup_{x \in \R}{\abs{f\p{x} - \sum_{n=1}^\infty \frac{n^2 + x^4}{n^4 + x^2}}} = \infty,

so the series does not converge uniformly.

Instead, our strategy is to prove that the series converges uniformly on compact intervals. If we manage to do this, then given any x∈Rx \in \R, there exists R>0R > 0 such that x∈(−R,R)x \in \p{-R, R}, so

∑n=1∞n2+x4n4+x2 converges uniformly on [−R,R]  ⟹  f is continuous on [−R,R]  ⟹  f is continuous at x.\begin{aligned} \sum_{n=1}^\infty \frac{n^2 + x^4}{n^4 + x^2} \text{ converges uniformly on } \br{-R, R} &\implies f \text{ is continuous on } \br{-R, R} \\ &\implies f \text{ is continuous at } x. \end{aligned}

This works for any x∈Rx \in \R, so we would show that ff is continuous on R\R. So, we need to show that the sum converges uniformly on [−R,R]\br{-R, R} for any R>0R > 0. As before, our main tool is the Weierstrass M-test, so if x∈[−R,R]x \in \br{-R, R}, then

n2+x4n4+x2≤n2+x4n4≤1n2+R4n4  ⟹  sup⁡x∈[−R,R]n2+x4n4+x2≤1n2+R4n4.\begin{gathered} \frac{n^2 + x^4}{n^4 + x^2} \leq \frac{n^2 + x^4}{n^4} \leq \frac{1}{n^2} + \frac{R^4}{n^4} \\ \implies \sup_{x \in \br{-R, R}} \frac{n^2 + x^4}{n^4 + x^2} \leq \frac{1}{n^2} + \frac{R^4}{n^4}. \end{gathered}

By the pp-series test again with p=2>1p = 2 > 1 and p=4>1p = 4 > 1, we get

∑n=1∞sup⁡x∈[−R,R]n2+x4n4+x2≤∑n=1∞(1n2+R4n4)<∞.\sum_{n=1}^\infty \sup_{x \in \br{-R, R}} \frac{n^2 + x^4}{n^4 + x^2} \leq \sum_{n=1}^\infty \p{\frac{1}{n^2} + \frac{R^4}{n^4}} < \infty.

Thus, by the Weierstrass M-test, the series converges uniformly on [−R,R]\br{-R, R}.