Continuous but nowhere differentiable function
Theorem
If $0<\alpha<1$, then the function
\[f_\alpha(x)=\sum_{n=0}^{\infty}2^{-n\alpha}e^{i2^n x}\]is continuous but nowhere differentiable.
The series converges absolutely and uniformly by the Weierstrass M-test, since
\[\left\vert 2^{-n\alpha}e^{i2^n x}\right\vert =2^{-n\alpha}, \qquad \sum_{n=0}^{\infty}2^{-n\alpha}<\infty.\]Thus, $f_\alpha$ is a continuous $2\pi$-periodic function. We will prove that it is nowhere differentiable using delayed means of its Fourier series.
Recall two ways of summing the Fourier series of a continuous $2\pi$-periodic function $g$. We use the normalized convolution and Fourier coefficients
\[\begin{aligned} (g*h)(x)&:=\frac{1}{2\pi}\int_{-\pi}^{\pi}g(y)h(x-y)\,dy,\\ \widehat g(n)&:=\frac{1}{2\pi}\int_{-\pi}^{\pi}g(y)e^{-iny}\,dy. \end{aligned}\]With the Dirichlet kernel $D_N(x)=\sum_{n=-N}^{N}e^{inx}$, we obtain
\[\begin{aligned} (g*D_N)(x) &=\frac{1}{2\pi}\int_{-\pi}^{\pi}g(y)\sum_{n=-N}^{N}e^{in(x-y)}\,dy\\ &=\sum_{n=-N}^{N}e^{inx}\frac{1}{2\pi}\int_{-\pi}^{\pi}g(y)e^{-iny}\,dy\\ &=\sum_{n=-N}^{N}\widehat g(n)e^{inx}\\ &=S_N(g)(x). \end{aligned}\]With the Fejér kernel
\[F_N:=\frac{1}{N}\sum_{k=0}^{N-1}D_k, \qquad N\geq 1,\]we obtain the Cesàro mean
\[\begin{aligned} (g*F_N)(x) &=\frac{1}{N}\sum_{k=0}^{N-1}(g*D_k)(x)\\ &=\frac{1}{N}\sum_{k=0}^{N-1}S_k(g)(x)\\ &=\sigma_N(g)(x). \end{aligned}\]On the Fourier coefficient side,
\[\widehat{S_N(g)}(n) =\widehat g(n)\widehat D_N(n),\]where
\[\begin{aligned} \widehat D_N(n) &=\frac{1}{2\pi}\sum_{k=-N}^{N}\int_{-\pi}^{\pi}e^{i(k-n)x}\,dx\\ &=\begin{cases} 1,& \vert n\vert \leq N,\\ 0,& \vert n\vert >N. \end{cases} \end{aligned}\]Similarly,
\[\begin{aligned} \widehat{\sigma_N(g)}(n) &=\widehat g(n)\widehat F_N(n)\\ &=\widehat g(n)\frac{1}{N}\sum_{k=0}^{N-1}\widehat D_k(n)\\ &=\begin{cases} \left(1-\dfrac{\vert n\vert }{N}\right)\widehat g(n),& \vert n\vert \leq N,\\ 0,& \vert n\vert >N. \end{cases} \end{aligned}\]Equivalently,
\[\begin{aligned} \sigma_N(g)(x) &=\frac{1}{N}\sum_{l=0}^{N-1}\sum_{n=-l}^{l}\widehat g(n)e^{inx}\\ &=\frac{1}{N}\sum_{\vert n\vert \leq N}(N-\vert n\vert )\widehat g(n)e^{inx}\\ &=\sum_{\vert n\vert \leq N}\left(1-\frac{\vert n\vert }{N}\right)\widehat g(n)e^{inx}. \end{aligned}\]The terms with $\vert n\vert =N$ have weight zero, so including them in the summation does not change its value.
Definition
We define the delayed mean $\Delta_N(g)$ by
\[\begin{aligned} \Delta_N(g) &:=2\sigma_{2N}(g)-\sigma_N(g)\\ &=2g*F_{2N}-g*F_N\\ &=g*(2F_{2N}-F_N). \end{aligned}\]Its Fourier coefficients are
\[\widehat{\Delta_N(g)}(n) =\begin{cases} \widehat g(n),& \vert n\vert \leq N,\\ 2\left(1-\dfrac{\vert n\vert }{2N}\right)\widehat g(n),& N<\vert n\vert \leq 2N,\\ 0,& \vert n\vert >2N. \end{cases}\]Indeed,
\[\begin{aligned} \Delta_N(g)(x) &=2\sum_{\vert n\vert \leq 2N}\left(1-\frac{\vert n\vert }{2N}\right)\widehat g(n)e^{inx}\\ &\quad-\sum_{\vert n\vert \leq N}\left(1-\frac{\vert n\vert }{N}\right)\widehat g(n)e^{inx}\\ &=\sum_{\vert n\vert \leq N}\widehat g(n)e^{inx}\\ &\quad+\sum_{N<\vert n\vert \leq 2N}\left(2-\frac{\vert n\vert }{N}\right)\widehat g(n)e^{inx}\\ &=S_N(g)(x)\\ &\quad+\sum_{N<\vert n\vert \leq 2N}\left(2-\frac{\vert n\vert }{N}\right)\widehat g(n)e^{inx}. \end{aligned}\]Recall that
\[f_\alpha(x)=\sum_{k=0}^{\infty}2^{-k\alpha}e^{i2^k x}.\]By uniform convergence and orthogonality of the exponentials, its Fourier coefficients are
\[\widehat f_\alpha(n) =\begin{cases} 2^{-k\alpha},& n=2^k\text{ for some integer }k\geq 0,\\ 0,& \text{otherwise}. \end{cases}\]Observation
Fix an integer $N\geq 1$, and choose the largest integer $k\geq 0$ for which $2^k\leq N$. Then $2^k\leq N<2^{k+1}$, and
\[\begin{aligned} \Delta_{2^k}(f_\alpha)(x) &=\sum_{\vert n\vert \leq 2^k}\widehat f_\alpha(n)e^{inx}\\ &\quad+\sum_{2^k<\vert n\vert \leq 2^{k+1}} \left(2-\frac{\vert n\vert }{2^k}\right)\widehat f_\alpha(n)e^{inx}\\ &=\sum_{\vert n\vert \leq 2^k}\widehat f_\alpha(n)e^{inx}\\ &=S_N(f_\alpha)(x). \end{aligned}\]The second sum is zero because the Fourier coefficients vanish for $2^k<\vert n\vert <2^{k+1}$, while the multiplier $2-\vert n\vert /2^k$ vanishes at $\vert n\vert =2^{k+1}$. In particular, although $\widehat f_\alpha(2^{k+1})\neq 0$, its contribution to this sum is zero. The final equality follows because there are no nonzero Fourier coefficients with $2^k<\vert n\vert \leq N$.
A second observation
If $2N=2^n$ for an integer $n\geq 1$, then
\[\begin{aligned} \Delta_{2N}(f_\alpha)(x)-\Delta_N(f_\alpha)(x) &=S_{2N}(f_\alpha)(x)-S_N(f_\alpha)(x)\\ &=2^{-n\alpha}e^{i2^n x}. \end{aligned}\]Differentiating these trigonometric polynomials gives, for every $x_0\in\mathbb{R}$,
\[\begin{aligned} \left\vert \Delta_{2N}(f_\alpha)'(x_0)-\Delta_N(f_\alpha)'(x_0)\right\vert &=\left\vert i2^n2^{-n\alpha}e^{i2^n x_0}\right\vert \\ &=2^{n(1-\alpha)}\\ &=(2N)^{1-\alpha}. \end{aligned}\]We will obtain a contradiction from the following lemma.
Lemma
Let $g$ be continuous and $2\pi$-periodic. If $g$ is differentiable at $x_0$, then
\[\sigma_N(g)'(x_0)=O(\log N) \qquad\text{as }N\to\infty.\]Proof. Since $F_N$ is a trigonometric polynomial, we may differentiate the kernel under the integral. A change of variables, using periodicity, gives
\[\begin{aligned} \sigma_N(g)'(x_0) &=\frac{1}{2\pi}\int_{-\pi}^{\pi}F_N'(x_0-t)g(t)\,dt\\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi}F_N'(t)g(x_0-t)\,dt. \end{aligned}\]Because $F_N$ is $2\pi$-periodic,
\[\int_{-\pi}^{\pi}F_N'(t)\,dt =F_N(\pi)-F_N(-\pi)=0.\]Therefore,
\[\sigma_N(g)'(x_0) =\frac{1}{2\pi}\int_{-\pi}^{\pi}F_N'(t)\bigl(g(x_0-t)-g(x_0)\bigr)\,dt.\]Differentiability at $x_0$ implies that $\vert g(x_0-t)-g(x_0)\vert \leq C\vert t\vert $ for sufficiently small $\vert t\vert $. Since $g$ is bounded, increasing $C$ makes this bound valid for all $\vert t\vert \leq\pi$. Hence
\[\left\vert \sigma_N(g)'(x_0)\right\vert \leq C\int_{-\pi}^{\pi}\vert F_N'(t)\vert \,\vert t\vert \,dt.\]We use the following kernel bounds to estimate this integral.
Kernel bounds
There is a constant $A>0$, independent of $N$ and $t$, such that
\[\vert F_N'(t)\vert \leq AN^2, \qquad \vert F_N'(t)\vert \leq\frac{A}{\vert t\vert ^2} \quad(0<\vert t\vert \leq\pi).\]Proof of the first bound. The finite Fourier expansion of the Fejér kernel is
\[F_N(t)=\sum_{m=-N}^{N}a_m e^{imt}, \qquad a_m:=1-\frac{\vert m\vert }{N}.\]For $-N\leq m\leq N$, these coefficients satisfy $0\leq a_m\leq 1$. Differentiating the finite sum gives
\[F_N'(t)=\sum_{m=-N}^{N}im a_m e^{imt}.\]Since $\vert e^{imt}\vert =1$, the triangle inequality yields
\[\begin{aligned} \left\vert F_N'(t)\right\vert &\leq\sum_{m=-N}^{N}\vert m\vert a_m\\ &\leq\sum_{m=-N}^{N}\vert m\vert\\ &=N(N+1)\\ &\leq 2N^2. \end{aligned}\]Proof of the second bound. For $0<\vert t\vert\leq\pi$, use the explicit formula
\[F_N(t)=\frac{1}{N}\frac{\sin^2(Nt/2)}{\sin^2(t/2)}.\]Differentiating with the product and chain rules gives
\[\begin{aligned} F_N'(t) &=\frac{\sin(Nt/2)\cos(Nt/2)}{\sin^2(t/2)}\\ &\quad-\frac{\sin^2(Nt/2)\cos(t/2)}{N\sin^3(t/2)}. \end{aligned}\]We use $\vert\cos u\vert\leq 1$ and the estimates
\[\sin^2(Nt/2) \leq\vert\sin(Nt/2)\vert \leq\frac{N\vert t\vert}{2},\]and
\[\vert\sin(t/2)\vert\geq c\vert t\vert, \qquad c:=\frac{1}{\pi}.\]The first estimate uses $\vert\sin u\vert\leq\min{1,\vert u\vert}$, and the second holds for $\vert t\vert\leq\pi$. Thus,
\[\begin{aligned} \left\vert F_N'(t)\right\vert &\leq\frac{1}{c^2\vert t\vert^2} +\frac{1}{N}\frac{N\vert t\vert/2}{c^3\vert t\vert^3}\\ &=\left(\frac{1}{c^2}+\frac{1}{2c^3}\right) \frac{1}{\vert t\vert^2}. \end{aligned}\]The factor $1/N$ in the second term cancels the $N$ from the sine estimate. Choosing $A$ large enough proves both bounds with the same constant, independent of $N$ and $t$.
Splitting the integral at $\vert t\vert =1/N$, we obtain, for $N\geq 2$,
\[\begin{aligned} \left\vert \sigma_N(g)'(x_0)\right\vert &\leq C\int_{\vert t\vert \leq 1/N}\vert F_N'(t)\vert \,\vert t\vert \,dt\\ &\quad+C\int_{1/N<\vert t\vert \leq\pi}\vert F_N'(t)\vert \,\vert t\vert \,dt\\ &\leq CAN^2\int_{\vert t\vert \leq 1/N}\vert t\vert \,dt\\ &\quad+CA\int_{1/N<\vert t\vert \leq\pi}\frac{1}{\vert t\vert }\,dt\\ &=O(1)+O(\log N)\\ &=O(\log N). \end{aligned}\]This proves the lemma. $\square$
To finish the proof of the theorem, suppose that $f_\alpha$ is differentiable at some point $x_0$. The lemma gives
\[\begin{aligned} \Delta_N(f_\alpha)'(x_0) &=2\sigma_{2N}(f_\alpha)'(x_0)-\sigma_N(f_\alpha)'(x_0)\\ &=O(\log N). \end{aligned}\]Consequently,
\[\left\vert \Delta_{2N}(f_\alpha)'(x_0)-\Delta_N(f_\alpha)'(x_0)\right\vert =O(\log N).\]But for $N=2^{n-1}$, the same quantity equals $(2N)^{1-\alpha}$. Since $1-\alpha>0$,
\[\frac{(2N)^{1-\alpha}}{\log N}\longrightarrow\infty,\]which is a contradiction. Thus, $f_\alpha$ is nowhere differentiable. $\square$
References
- Elias M. Stein and Rami Shakarchi, Fourier Analysis: An Introduction.
- Math 139 Fourier Analysis Notes.