Exercise 6.4.3

Question

\begin{enumerate}[label=(\alph*)] 
  \item Show that
    $$
    g(x)=\sum_{n=0}^{\infty} \frac{\cos \left(2^{n} x\right)}{2^{n}}
    $$
    is continuous on all of $\mathbf{R}$.
  \item The function $g$ was cited in Section 5.4 as an example of a continuous nowhere differentiable function. What happens if we try to use Theorem $6.4 .3$ to explore whether $g$ is differentiable?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Define $g_n(x) = \frac{\cos(2^nx)}{2^n}$ and $M_n = 2^{-n} > \left|g_n(x)\right|$. By the Weierstrass M-test, $g(x)$ converges uniformly on $\mathbf{R}$. Since each $g_n(x)$ is continuous and $g(x)$ converges uniformly, $g(x)$ must also be continuous.
    \item $g'_n(x) = -\sin(2^nx)$ and thus $\sum^\infty_{n=1} g'_n(x)$ does not converge uniformly by Exercise 6.4.2 part a. (It might not converge pointwise either, but that seems more difficult to prove.) Therefore we cannot use Theorem 6.4.3.
   \end{enumerate}

Exercise 6.4.3

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Exercise 6.4.3

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