Tutorial note 6 (consequence of uniform convergence)

Example 1. For any $x\neq 0$ the series converges pointwise to \[
f(x)=x^2\sum_{n=0}^\infty \frac{1}{(1+x^2)^n}=x^2\cdot \frac{1+x^2}{x^2}=1+x^2.
\] If the convergence were uniform on $(0,1)$, then it would be uniform also on $[0,1]$ (easily seen by uniform Cauchy criterion, see the complete statement in the following remark), therefore $f$ would be continuous on $[0,1]$. However, \[
f(x)=\begin{cases}
0,&x=0,\\
1+x^2,&x\in (0,1].
\end{cases}
\] Therefore \[
f\text{ not continuous}\implies \text{the convergence is not uniform.}
\] Remark. Suppose that $f_1,f_2,\dots:[a,b]\to \R$ is a sequence of continuous functions such that it converges uniformly on $(a,b)$, then it also converges uniformly on $[a,b]$.

Proof. If $f_n$ converges uniformly on $(a,b)$, then it is uniformly Cauchy on $(a,b)$, thus for every $\epsilon>0$, there is an $N$ such that \[
m,n>N\implies \|f_m-f_n\|_{(a,b)}<\epsilon.
\] However, if $\|f_m-f_n\|_{(a,b)}<\epsilon$, then $|f_m(x)-f_n(x)|<\epsilon$ for every $x\in (a,b)$, by taking limit $x\to a^+$ and $x\to b^-$ respectively we have $|f_m-f_n|\leq \epsilon$ on $[a,b]$, thus $\|f_m-f_n\|_{[a,b]}\leq \epsilon$. We conclude that \[
m,n>N\implies \|f_m-f_n\|_{[a,b]}\leq \epsilon,
\] so $\{f_n\}$ converges uniformly on $[a,b]$.    $\inner{\hat \cdot \omega\hat\cdot}$

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Example 2.
(a) Since

• $\dis \sum_{k=1}^\infty \frac{1}{k}\brac{\sin \frac{x}{k}}'=\sum_{k=1}^\infty \frac{1}{k^2}\cos \frac{x}{k}$ converges uniformly on $(-1,1)$ by $M$-test.
• The series $\sum_{k=1}^\infty \frac{1}{k}\sin \frac{0}{k}$ converges with $0\in (-1,1)$.

By the Differentiation Theorem the series converges uniformly on $(-1,1)$.

(b) Although \[
\sum_{k=1}^\infty \frac{1}{k}\brac{\cos \frac{x}{k}}'=-\sum_{k=1}^\infty \frac{1}{k^2}\sin \frac{x}{k}
\] converges uniformly on $(-1,1)$, we cannot apply the Differentiation Theorem yet because we still need to find a point in $(-1,1)$ at which the series converges.

Actually such point does not exist: for every $x\in \R$, \[
\sum_{k=1}^\infty \frac{1}{k}\cos \frac{x}{k}=\infty.
\]

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Example 3. Note that LHS is an improper integral, we must switch to proper integral in order to apply Integration Theorem. For this, we use the following result:

Theorem. If $f:(a,b]\to \R$ is locally integrable and bounded near $a$, then $f$ is improper integrable, moreover, \[
\int_a^b f\,dx=\int_a^b \tilde f\,dx,
\] for any $\tilde f:[a,b]\to \R$ such that $f=\tilde f$ on $(a,b]$ and $\tilde f(a)\in \R$, i.e., for any function $\tilde f$ that extends $f$ to $[a,b]$.

Now for any $x\neq 0$ we have \[
\frac{1-\cos x^2}{x^4}=\sum_{k=1}^\infty \frac{(-1)^{k+1}x^{4k-4}}{(2k)!},
\] RHS is a function on $\R$, thus we conclude that RHS is a function that extends LHS to $[0,1]$, let's check that LHS is bounded near $0$: \[
\lim_{x\to 0}\frac{1-\cos x^2}{x^4}=\lim_{x\to 0}\frac{1}{2}\frac{\sin x^2}{x^2}=\frac{1}{2}.
\]  Hence by the theorem we have just quoted, \[
\int_0^1 \frac{1-\cos x^2}{x^4}\,dx = \int_0^1 \sum_{k=1}^\infty \frac{(-1)^{k+1}x^{4k-4}}{(2k)!}\,dx.
\] Now RHS is a proper integral, the rest is routine: check that the convergence is uniform on $[0,1]$ and switch the order of  integral sign and summation sign. This is very straightforward by $M$-test.

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Example 4.
(a) Write the series as \[
\sum_{k=1}^\infty (-1)^k \frac{(e^{-x})^k}{k}.\]
Since $x\in [0,\infty)$, we have $e^{-x}\in (0,1]$, by substitution property it is enough to show the power series \[
\sum_{k=1}^\infty (-1)^k \frac{x^k}{k}
\] converges uniformly on $[0,1]$. For this,  we prove that the power series converges pointwise  on $[0,1]$, after that Abel's Limit Theorem will guarantee the uniformity.

The radius of convergence is \[
\frac{1}{\lims \sqrt[k]{|(-1)^k\frac{1}{k}|}}=1.
\] Therefore the power series converges on $(-1,1)$, in particular, on $[0,1)$. When $x=1$, the power series becomes \[
\sum_{k=1}^\infty (-1)^k \frac{1}{k},
\] which also converges by Alternating Series Test, thus we are done.

(b) (i) We try to use Abel's Limit Theorem. Firstly we write the series as the standard form of power series as follows \[
\sum_{k=1}^\infty \frac{(x-2)^k}{2^kk},
\] its radius of convergence is $2$, hence it converges on $(0,4)$, thus on $(0,2]$. At $x=0$, the power series becomes \[
\sum_{k=1}^\infty \frac{(-1)^k}{k},
\] which also converges by Alternating Series Test, so by Abel's Limit Theorem, done.

(ii) This time we cannot use Abel's Limit Theorem directly because the series is not a power series, as $k+\sqrt{5}\not\in \N$ for every $k$. We observe that \[
 \sum_{k=1}^\infty (-1)^k \frac{(2-x)^{k+\sqrt{5}}}{2^kk} =  \underbrace{(2-x)^{\sqrt{5}}}_{\text{bounded on $[0,2]$}}\underbrace{\sum_{k=1}^\infty (-1)^k \frac{(2-x)^{k}}{2^kk}}_\text{converges uniformly by part (a)},
\] by Bounded Multiplier Property, done.