Continuous but nowhere differentiable function
Theorem
If $0<\alpha <1$, then the function
\[\begin{align*} f_\alpha(x) = \sum_{n=0}^\infty 2^{-n\alpha}e^{i2^nx} \end{align*}\]is continuous but nowhere differentiable.
The continuity is clear because the series are absolutely convergent by Weistrass M-test. In other words, we can use Corollary 8.3.2 to prove continuity of $f_\alpha$.
Recall that we have a couple of ways of summing the Fourier series. With Dirichlet kernel $D_N$,
\[\begin{align*} g*D_N (x)&= \frac{1}{2\pi}\int_{-\pi}^\pi g(y) \sum_{n=-N}^N e^{in(x-y)}dy \\ &=\sum_{n=-N}^N e^{inx}\frac{1}{2\pi} \int_{\pi}^\pi g(y) e^{-iny}dy \\ &=\sum_{n=-N}^N \hat{g}(n) e^{inx} \\ &=S_N(g). \end{align*}\]Or with Fejer kernel $F_N$,
\[\begin{align*} g*F(x) &=g*(\frac{1}{N} \sum_{k=0}^{N-1} D_k)(x) \\ &=\frac{1}{N} \sum_{k=0}^{N-1} (f*D_k)(x) \\ &=\frac{1}{N} \sum_{k=0}^{N-1}S_k(f)(x) \\ &=\sigma_N(f). \end{align*}\]If we look at the partial sum $S_N$ on the Fourier coefficient side, we see that
\[\begin{align*} \widehat{S_N(g)}(n) &= \widehat{g*D_N}(n) \\ &= \hat{g}(n) \cdot \hat{D}_N(n) \end{align*}\]where
\[\begin{align*} \hat{D}_N(n) &= \frac{1}{2\pi} \int_{-\pi}^\pi \sum_{k=-N}^N e^{ikx}e^{-inx}dx \\ &=\frac{1}{2\pi} \sum_{k=-N}^N \int_{-\pi}^\pi e^{i(k-n)x}dx \\ &=\begin{cases} 1 \quad \text{ if } \lvert n \rvert \leq N \\ 0 \quad \text{ otherwise} \end{cases} \end{align*}\]Fourier coefficient of Cesaro sum,
\[\begin{align*} \widehat{\sigma_N(g)}(n) &= \widehat{g* F_N }(n) \\ &=\hat{g}(n) \hat{F}(n) \\ &= \hat{g}(n) \widehat{\frac{1}{N} \sum_{k=0}^{N-1}D_k}(n) \\ &= \hat{g}(n) \frac{1}{N}\left(\hat{D_0}(n) + \hat{D_1}(n) + \cdots + \hat{D}_{N-1}(n)\right) \\ &=\begin{cases} \hat{g}(n) \frac{1}{N} \left(N - \lvert n\rvert\right) \quad \text{ if } \lvert n \rvert \leq N \\ 0 \quad \text{ otherwise} \end{cases} \end{align*}\]Equivalently,
\[\begin{align*} \sigma_N(g)(x)&=\frac{1}{N}\left(S_0(g)(x) + S_1(g)(x) + \cdots S_{N-1}(g)(x) \right) \\ &=\frac{1}{N} \sum_{l=0}^{N-1}\sum_{k=-l}^l \hat{g}(k) e^{ikx} \\ &= \frac{1}{N} \sum_{\lvert n \rvert \leq N}(N-\lvert n \rvert) \hat{g}(n) e^{inx}\\ &=\frac{1}{N} \sum_{\lvert n \rvert \leq N}(1- \lvert n \rvert /N)\hat{g}(n)e^{inx} \end{align*}\]Definition
We define the delayed means $\Delta_N(g)$ of the Fourier series of $g$ by
\[\begin{align*} \Delta_N(g) &:= 2\sigma_{2N}(g) - \sigma_N(g)\\ &=2g* F_{2N} - g* F_N \\ &=g*(2F_{2N}-F_N) \end{align*}\]On the Fourier coefficient side, it is easy to see that
\[\begin{align*} \widehat{\Delta_N(g)}(n) = \begin{cases} \hat{g}(n) \quad \text{ if } \lvert n \rvert \leq N \\ \hat{g}(n)2(1-\frac{\lvert n \rvert}{N}) \quad \text{ if } N < \lvert n\rvert \leq 2N \\ 0 \quad \text{ if } \lvert n \rvert > 2N \end{cases} \end{align*}\]For delayed means
\[\begin{align*} \Delta_N(g) &= 2\sigma_{2N}(g) - \sigma_N(g) \\ &= 2 \frac{1}{2N} \sum_{\lvert n \rvert \leq 2N} (2N - \lvert n \rvert) \hat{g}(n) e^{inx} - \frac{1}{N}\sum_{\lvert n \rvert \leq N}(N-\lvert n \rvert) \hat{g}(n) e^{inx} \\ &=\frac{1}{N} \sum_{\lvert n \rvert \leq N} N \hat{g}(n)e^{inx} + \sum_{N< \lvert n \rvert \leq 2N} (2- \lvert n \rvert /N)e^{inx}\\ &=S_N(g) + \sum_{N < \lvert n \rvert \leq 2N}(2-\lvert n \rvert / N)\hat{g}(n) e^{inx}. \end{align*}\]Recall that our function is
\[\begin{align*} f_\alpha(x) = \sum_{n=0}^\infty 2^{-n\alpha}e^{i2^nx} \end{align*}\]with $\alpha \in (0,1)$. The Fourier coefficient of $f_\alpha$ is
\[\begin{align*} \hat{f_\alpha}(n) = \begin{cases} 2^{-n\alpha} \quad \text{ if } n=2^k \\ 0 \quad \text{ otherwise}. \end{cases} \end{align*}\]Observation
For fixed $N$, if we choose the largest $k$ for which $2^k \leq N$, then
\[\begin{align*} \Delta_{2^k}(f_\alpha) &= \sum_{\lvert n \rvert \leq 2^k}\hat{f_\alpha}(n) e^{inx} + \sum_{2^k < \lvert n\rvert \leq 2^{k+1}} (2-\lvert n\rvert /2^k) \hat{f_\alpha}(n) e^{inx} \\ &= \sum_{\lvert n \rvert \leq 2^k}\hat{f_\alpha}(n) e^{inx} \\ &= S_N(f_\alpha) \end{align*}\]since there is no $\hat{f_\alpha}(n)$ for $2^k < \lvert n \rvert < 2^{k+1}$ by the definition of $f_\alpha$.
Dumb Observation
If $2N= 2^n$, then
\[\begin{align*} \Delta_{2N}(f) - \Delta_N(f) = 2^{-n\alpha} e^{i2^nx}. \end{align*}\]Our contradiction will be obtained as follows. By the above dumb observation, for any point $x_0$,
\[\begin{align*} \left\lvert \Delta_{2N}(f)^\prime(x_0) - \Delta_N(f)^\prime (x_0)\right\rvert &= \lvert i2^n 2^{-n\alpha}e^{i2^nx_0}\rvert \\ &=2^{n(1-\alpha)} = (2N)^{1-\alpha}. \end{align*}\]However, with the following lemma,
Lemma
Let $g$ be continuous. If $g$ is differentiable at $x_0$, then
\[\begin{align*} \sigma_N(g)^\prime(x_0) = O(\log N). \end{align*}\]Since $\Delta_N(g)^\prime (x_0)= 2\sigma_{2N}(g)^\prime (x_0) - \sigma(g)^\prime (x_0) = O(\log N)$, we would have a contradiction if we assume $f_\alpha$ is differentiable at $x_0$.
<Proof of lemma>
\[\begin{align*} \sigma_N(g)^\prime(x_0) &= (F_N * g)^\prime (x_0) \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi F^\prime (x_0-t) g(t) dt \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi F^\prime (t) g(x_0 - t) dt. \\ \end{align*}\]The second equality holds because of Leibniz rule.
Since $F_N$ is $2\pi$ periodic function,
\[\begin{align*} \int_{-\pi}^\pi F^\prime_N(t) dt = F_N(\pi) - F_N(-\pi) =0. \end{align*}\]Using this fact , we get
\[\begin{align*} \sigma_N(g)^\prime(x_0) = \frac{1}{2\pi} \int_{-\pi}^\pi F^\prime_N(t) [g(x_o-t) - g(x_0)]dt. \end{align*}\]Using the differentiability of $g$ at $x_0$,
\[\begin{align*} \lvert \sigma_N(g)^\prime(x_0)\rvert \leq C \int_{-\pi}^\pi F^\prime_N(t) \lvert t\rvert dt. \end{align*}\]FACTS
\(\begin{align*} \lvert F^\prime_N(t) \rvert &\leq AN^2 \\ \lvert F^\prime_N(t) \rvert &\leq \frac{A}{\lvert t \rvert^2} \end{align*}\)
Putting it all together, we see
\[\begin{align*} \lvert \sigma_N(g)^\prime(x_0)\rvert &\leq C \int_{-\pi}^\pi F^\prime_N(t) \lvert t\rvert dt \\ &\leq C \int_{\lvert t \rvert \geq \frac{1}{N}} \lvert F^\prime_N(t) \rvert \lvert t \rvert dt + C \int_{\lvert t \rvert \leq \frac{1}{N}} \lvert F^\prime_N(t) \rvert \lvert t \rvert dt \\ &=O(\log N) \end{align*}\] \[\tag*{$\square$}\]Reference
- Elias M. Stein and Rami Shakarchi 『Fourier Analysis: An Introduction』
- Math 139 Fourier Analysis Notes