Exercise 6.4.7

Question

Let
  $$
  f(x)=\sum_{k=1}^{\infty} \frac{\sin (k x)}{k^{3}}
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Show that $f(x)$ is differentiable and that the derivative $f^{\prime}(x)$ is contimuous.
  \item Can we determine if $f$ is twice-differentiable?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Let $f_n(x) = \frac{\sin(nx)}{n^3}$. We have
    $$\left|f'_n(x)\right| = \left|\frac{\cos (nx)}{n^2}\right| \leq \frac{1}{n^2}$$
    and so $\sum^\infty_n f'_n(x)$ converges uniformly by the Weierstrass M-Test. We also have $f(x)$ converging at $x=0$ (since every term is zero), so by the differentiable limit theorem we have $f(x)$ differentiable with $f'(x) = \sum^\infty_{n=1} f'_n(x)$. Since this converges uniformly, $f'(x)$ is continuous.

\item Probably not easily - trying the same trick leaves us with trying to bound $\left|\frac{\sin(kx)}{k}\right|$ with $M_n$ where $\sum M_n$ converges, but $M_n = 1/k$ doesn't work as $\sum^\infty_{k=1} 1/k$ diverges.
  \end{enumerate}

Exercise 6.4.7

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Exercise 6.4.7

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