Exercise problem 6.4.8 from the text Understanding Analysis by Stephen Abbott asks us to investigate the continuity, differentiability and twice-differentiability of the function $f(x)$ defined below. Is my proof attempt technically correct and rigorous?
Consider the function
\begin{equation*}
f( x) =\sum _{k=1}^{\infty }\frac{\sin( x/k)}{k}
\end{equation*}
Proof.
Well-definedness.
Fix $\displaystyle x=x_{0}$ and consider the absolute value series:
\begin{equation*}
\sum _{k=1}^{\infty }\left| \frac{\sin( x_{0} /k)}{k}\right|
\end{equation*}
We have that:
\begin{equation*}
0\leq \left| \frac{\sin( x_{0} /k)}{k}\right| \leq \frac{|x_{0} /k|}{k} =\frac{|x_{0} |}{k^{2}}
\end{equation*}
Since $\displaystyle \sum \frac{|x_{0} |}{k^{2}}$ is a convergent series, by the comparision test, $\displaystyle \sum _{k=1}^{\infty }\left| \frac{\sin( x_{0} /k)}{k}\right| $ is convergent. By the Absolute convergence test, $\displaystyle \sum _{k=1}^{\infty }\frac{\sin( x_{0} /k)}{k}$ also converges.
Consequently, $\displaystyle f$ is defined on $\displaystyle \mathbf{R}$.
Uniform Convergence and continuity.
Let $\displaystyle x\in [ -1,1]$. Define $\displaystyle M_{k}$ as:
\begin{equation*}
0\leq \left| \frac{\sin( x /k)}{k}\right| \leq \frac{|x/k|}{k} \leq \frac{1}{k^{2}}
\end{equation*}
Since $\displaystyle \sum M_{k}$ is a convergent series, by the Weierstrass $\displaystyle M$-Test, $\displaystyle \sum f_{k}$ converges uniformly on $\displaystyle [ -1,1]$. Since each $\displaystyle f_{k}$ is continuous on $\displaystyle [ -1,1]$, by the term-by-term continuity theorem, $\displaystyle f$ is continuous on $\displaystyle [ -1,1]$.
Differentiability.
We have:
\begin{equation*}
f_{k} '( x) =\frac{\cos x/k}{k^{2}}
\end{equation*}
Define $\displaystyle C_{k}$ as :
\begin{equation*}
0\leq \left| \frac{\cos x/k}{k^{2}}\right| \leq \frac{1}{k^{2}}
\end{equation*}
Since $\displaystyle \sum C_{k}$ is convergent, by the Weierstrass $\displaystyle M$-Test, $\displaystyle \sum f_{k} '$ is uniformly convergent on $\displaystyle \mathbf{R}$. $\displaystyle \sum f_{k}$ is convergent pointwise on atleast one point in $\displaystyle [ -1,1]$. Each $\displaystyle f_{k}$ is differentiable on $\displaystyle \mathbf{R}$. Hence, by the term-by-term differentiability theorem, $\displaystyle f$ is differentiable on $\displaystyle \mathbf{R}$ and $\displaystyle f'( x) =\sum f_{k} '( x)$ for all $\displaystyle x\in \mathbf{R}$. In turn, this implies $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$.
Twice Differentiability.
We have:
\begin{equation*}
f_{k} ''( x) =-\frac{\sin x/k}{k^{3}}
\end{equation*}
Define $\displaystyle D_{k}$ as :
\begin{equation*}
0\leq \left| -\frac{\sin x/k}{k^{3}}\right| \leq \frac{1}{k^{3}}
\end{equation*}
Since $\displaystyle \sum D_{k}$ is convergent, by the Weierstrass $\displaystyle M$-Test, $\displaystyle \sum f_{k} ''$ is uniformly convergent on $\displaystyle \mathbf{R}$. $\displaystyle \sum f_{k} '$ is convergent pointwise on atleast one point in $\displaystyle \mathbf{R}$. Each $\displaystyle f_{k} '$ is differentiable on $\displaystyle \mathbf{R}$. Hence, by the term-by-term differentiability theorem, $\displaystyle f$ is twice-differentiable on $\displaystyle \mathbf{R}$.
In fact, $\displaystyle f$ is infinitely differentiable.
Best Answer
Yes, this is correct. Another approach is to use power series:
$$f(x)=\sum_{k=1}^\infty\frac1k\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\left(\frac xk\right)^{2n+1}=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}\sum_{k=1}^\infty\frac1{k^{2n+2}},$$
with the change of the order of summation justified by the absolute convergence. Thus $f(x)$ has a power series (in $x$) that converges everywhere; this easily implies that $f$ is infinitely differentiable.