1.
Prove that, for any finite \(a, M > 0\text{,}\) the set \(\{ \mathbf x:=\{\mathbf x(k)\}_{k=1}^{\infty}\in l^{p}: \sum_{k=1}^{\infty} k^{a} |\mathbf x(k)|^{p}\le M\}\) is compact in \(l^{p}\text{.}\)
Exercises 2.8 Exercises
2.
Suppose that \(\{f_n(\bx)\}\) is a sequence of real valued functions defined on a compact set \(K\) such that
-
Each \(f_n(x)\) is continuous on \(K\text{,}\)
-
\(\{f_n(x)\}\) converges pointwise to some continuous \(f(x)\) on \(K\text{,}\)
-
\(f_n(x)\ge f_{n+1}(x)\) for all \(x\in K\) and \(n=1, 2, \cdots\text{.}\)
Show that \(\{f_n(x)\}\to f(x) \) uniformly on \(K\text{.}\)
Question: Can the compactness of \(K\) be dropped? Where is the continuity of the limit function \(f\) in condition (b) used? Can it be dropped?
Solution. A proof using open covering argument
From the assumptions, we know that for any given \(\epsilon>0\text{,}\) and any \(x\in K\text{,}\) there exist \(N=N(x)\) and \(\delta_{1}=\delta(x)>0\) such that
\begin{equation*}
|f_{n}(x)-f(x)| < \epsilon \text{ for all } n\ge N\text{,}
\end{equation*}
\begin{equation*}
|f(t)-f(x)| < \epsilon \text{ for all } t \in B(x, \delta_{1})\text{,}
\end{equation*}
where \(B(x, \delta_{1})\) stands for the \(\delta\) neighborhood of \(x\text{.}\)
Take \(n=N(x)\text{,}\) then, using the continuity of \(f_{n}\) at \(x\text{,}\) there exists some \(\delta_{2}>0\text{,}\) such that
\begin{equation*}
|f_{n}(t)-f_{n}(x)| < \epsilon \text{ for all } t \in B(x, \delta_{2})\text{.}
\end{equation*}
Set \(\delta=\delta(x) :=\min\left\{\delta_{1},\delta_{2}\right\}\text{,}\) then for all \(t \in B(x, \delta)\text{,}\) we have
\begin{align*}
|f_{n}(t)-f(t)| \amp \le |f_{n}(t)-f_{n}(x)|+|f_{n}(x)-f(x)| +|f(x)-f(t)| \\
\amp \le \epsilon + \epsilon + \epsilon = 3\epsilon \text{.}
\end{align*}
Furthermore, using the assumption that \(f_{n+1}(t)\le f_{n}(t)\) for any \(t\text{,}\) we conclude that \(0\le f_{n+1}(t)-f(t)\le f_{n}(t)-f(t)\text{,}\) and
\begin{equation*}
|f_{m}(t)-f(t)|\le |f_{n}(t)-f(t)| \le 3\epsilon \text{ for all }
m\ge n, t\in B(x, \delta)\text{.}
\end{equation*}
Now \(\{ B(x, \delta): x \in K\}\) forms an open cover of \(K\text{.}\) Using the compactness of \(K\text{,}\) we find a finite subcover \(\cup_{i=1}^{k} B(x_{k}, \delta(x_{k}))\) of \(K\) for some \(\left\{ x_{i}\right\}_{i=1}^{k}\text{.}\) Set \(N =\max_{1\le i \le k} N(x_{i})\text{.}\) Then for any \(t\in K\text{,}\) we have \(t\in B(x_{i},\delta (x_{i}))\) for some \(i\text{,}\) therefore, for all \(m\ge N\text{,}\) we have
\begin{equation*}
|f_{m}(t)-f(t)|\le 3\epsilon\text{,}
\end{equation*}
which concludes our proof.
The above proof can be written in a more compact way. For any \(\epsilon>0\text{,}\) using the continuity of \(f_{n}, f\text{,}\) we know that the set \(O_{n,\epsilon}:=\left\{x\in K: f_{n}(t)-f(t) < \epsilon\right\}\) is open. Since for any \(x\in K\text{,}\) \(f_{n}(x)-f(x)\to 0\) as \(n\to \infty\text{,}\) we conclude that \(x\in O_{n,\epsilon}\) for some \(n\text{.}\) Thus \(\cup_{n}O_{n,\epsilon}\) is an open over of \(K\text{.}\) By the compactness of \(K\text{,}\) there exists some finite subcover. Since \(O_{n,\epsilon} \subset O_{m,\epsilon}\) for \(m\ge n\text{,}\) we have in fact \(K\subset O_{N,\epsilon}\) for some \(N\text{,}\) which implies that for all \(x\in K\) and all \(n\ge N\)
\begin{equation*}
0\le f_{n}(x)-f(x)\le f_{N}(x) -f(x) < \epsilon\text{.}
\end{equation*}
3.
Hint.
Solution.
First we have
\begin{equation*}
\frac{f(x)-f(0)}{x} = \sum_{k=1}^{\infty} \frac{\cos (kx) -1}{x k^2}= -2 \sum_{k=1}^{\infty} \frac{ \sin^2 (\frac{kx}{2})}{xk^2}= -2x \sum_{k=1}^{\infty} \left( \frac{ \sin (\frac{kx}{2})}{xk}\right)^2.
\end{equation*}
Note that for \(x>0\text{,}\) \(x \sum_{k=1}^{\infty} \left( \frac{ \sin (\frac{kx}{2})}{xk}\right)^2\) is a Riemann sum for the improper integral \(\int_0^{\infty} \left( \frac{ \sin (y/2) }{y} \right)^2 dy\text{,}\) so we expect
\begin{equation*}
\lim_{x\to 0+} x \sum_{k=1}^{\infty} \left( \frac{ \sin (\frac{kx}{2})}{xk}\right)^2 = \int_0^{\infty} \left( \frac{ \sin (y/2) }{y} \right)^2 dy >0.
\end{equation*}
Theorem 2.2.7 does not directly apply here. One way to justify the assertion above is to use the divide-and-conquer strategy. For any \(\epsilon >0\) given, first we find \(L>\epsilon^{-1}\) such that
\begin{equation*}
\int_{L} ^{\infty} \left( \frac{ \sin (y/2) }{y} \right)^2 dy < \epsilon.
\end{equation*}
Then, we break the summation into \(k \le x^{-1}L\) and \(k > x^{-1}L\text{.}\) We estimate
\begin{equation*}
x \sum_{k> x^{-1}L} \left( \frac{ \sin (\frac{kx}{2})}{xk}\right)^2
\le x^{-1} \sum_{k> x^{-1}L} \frac{1}{k^{2}} \le x^{-1} \int_{x^{-1}L}^{\infty} \frac{dy}{y^{2}}
\le L^{-1} \le \epsilon.
\end{equation*}
Finally, the summation \(x \sum_{k\le x^{-1}L } \left( \frac{ \sin (\frac{kx}{2})}{xk}\right)^2\) is a Riemann sum for the proper Riemann integral \(\int_0^{L} \left( \frac{ \sin (y/2) }{y} \right)^2 dy\text{,}\) so as the step size \(x\to 0\text{,}\) we have
\begin{equation*}
\left| x \sum_{k\le x^{-1}L } \left( \frac{ \sin (\frac{kx}{2})}{xk}\right)^2-
\int_0^{L} \left( \frac{ \sin (y/2) }{y} \right)^2 dy \right| < \epsilon.
\end{equation*}
This justifies the limiting process above and concludes that
\begin{equation*}
\lim_{x\to 0+} \frac{f(x)-f(0)}{x} = - 2 \int_0^{\infty} \left( \frac{ \sin (y/2) }{y} \right)^2 dy < 0.
\end{equation*}
Similarly,
\begin{equation*}
\lim_{x\to 0-} \frac{f(x)-f(0)}{x} = 2 \int_0^{\infty} \left( \frac{ \sin (y/2) }{y} \right)^2 dy > 0.
\end{equation*}
Thus \(f'(0)\) does not exist. This proves indirectly that the series for \(f'(x)\) can’t converge uniformly in a neighborhood of \(x=0\text{.}\)
