Du-Bois- Reymond Function.

Question

Please give me an example of a continuous function whose Fourier series diverges at a point. (given by Du Bois-Reymond). thanks.

Answer 1

If you're satisfied with a theoretical way of constructing such a function, the idea being to pile "bad" functions...

Definitions: Let $f:\mathbb{R}\to\mathbb{C}$ be a $2\pi$-periodic function.

• $f\in C(\mathbb{T})$ if $f$ is continuous
• $f\in L^1(\mathbb{T})$ if $f$ is measurable and $\displaystyle\int_0^{2\pi}|f(t)|\,dt<\infty$
• $\displaystyle\|f\|_{\infty}:=\sup_t|f(t)|$

Note that $C(\mathbb{T})\subset L^1(\mathbb{T})$.

If $f\in L^1(\mathbb{T})$ then its Fourier coefficients are defined by $$ \hat{f}(k) := \frac{1}{2\pi}\int_0^{2\pi}f(t)e^{-ikt}\,dt\quad(k\in\mathbb{Z}) $$ The Fourier series of $f$ is then the expression $$ \sum_{k=-\infty}^{\infty}\hat{f}(k)e^{ikt} $$ Its symmetric partial sums are noted $$ s_nf(x) := \sum_{k=-n}^n\hat{f}(k)e^{ikx}\quad(n\geq0) $$

Theorem (Du Bois Reymond): There is a $f\in C(\mathbb{T})$ such that $$ \sup_n|s_nf(0)|=\infty $$ In particular, the Fourier series of $f$ diverges at $x=0$.

We need preliminary results in order to prove this.

Lemma 1: Let $f \in L^1(\mathbb{T})$. Then $$ s_n f(x) = \frac{1}{2\pi} \int_0^{2\pi} f(x-t) D_n(t) \,dt $$

where $\begin{aligned}[t] D_n(t) :&= \sum_{k=-n}^n e^{ikt} \\ &= \frac{\sin( (n+1/2) t )}{\sin(t/2)} \quad (\textit{Dirichlet kernel}) \end{aligned}$

Proof: We have \begin{align*} s_nf(x) &= \sum_{k=-n}^n \hat{f}(k) e^{ikx} \\ &= \sum_{k=-n}^n \left( \frac{1}{2\pi} \int_0^{2\pi} f(t) e^{-ikt} \,dt \right) e^{ikx} \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(t) \sum_{k=-n}^n e^{ik(x-t)} \,dt \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(x-t) \sum_{k=-n}^n e^{ikt} \,dt \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(x-t) D_n(t) \,dt \end{align*}

where \begin{align*} D_n(t) &= \sum_{k=-n}^n e^{ikt} \\ &= e^{-int} \frac{e^{ i(2n+1)t }-1}{e^{it}-1} \\ &= \frac{e^{ i(n+1)t }-e^{-int}}{e^{it/2} \left( e^{it/2}-e^{-it/2} \right)} \\ &= \frac{e^{ i(n+1/2)t }-e^{ -i(n+1/2)t }}{\left( e^{it/2}-e^{-it/2} \right)} \\ &= \frac{\sin( (n+1/2)t )}{\sin(t/2)} \end{align*}

Lemma 2: $\displaystyle\int_{-\pi}^{\pi}|D_n(t)|\,dt\to\infty$ as $n\to\infty$

Proof: This has been asked here.

Lemma 3: $|s_nf(0)|\leq(2n+1)\|f\|_{\infty}$

Proof: $$ |s_nf(0)|=\bigg|\sum_{k=-n}^n\hat{f}(k)\bigg|\leq\sum_{k=-n}^n|\hat{f}(k)|\leq(2n+1)\|f\|_{\infty} $$

Lemma 4: For all $\epsilon>0$ and all $k>0$ there exists $f\in C(\mathbb{T})$ and $n\geq1$ such that $$ \|f\|_{\infty}\leq\epsilon\quad\text{and}\quad|s_nf(0)|\geq k $$

Proof: Let's take $$ f_n(t):=\frac{\epsilon \overline{D_n(-t)}}{1+|D_n(-t)|} $$ where $n$ is to be chosen. Then $f_n\in C(\mathbb{T})$ and $\|f_n\|_{\infty}\leq\epsilon$. Finally, \begin{align} s_nf_n(0) &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f_n(0-t)D_n(t)\,dt\\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\epsilon |D_n(t)|^2}{1+|D_n(t)|}\,dt\\ &=\epsilon\bigg(\underbrace{\frac{1}{2\pi}\int_{-\pi}^{\pi}|D_n(t)|\,dt}_{\xrightarrow[(n \to \infty)]{}\,\infty} - \underbrace{\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{|D_n(t)|}{1+|D_n(t)|}\,dt}_{\leq1}\bigg)\\ &\to\infty\quad(n\to\infty) \end{align} Choosing $n$ large enough we have $|s_nf_n(0)|\geq k$.

Proof of the theorem

By lemma 4 we can recursively choose $f_k\in C(\mathbb{T})$ and positive integers $n_k$ such that $$ \|f_k\|_{\infty}\leq2^{-k}\min_{1\leq j\leq k-1}\left(\frac{1}{2n_j+1}\right)\quad(\leq2^{-1}\text{ if }k=1) $$ and $$ |s_{n_k}f_k(0)|\geq k $$ Also, we might assume that $$ \bigg|s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j+f_k\bigg)(0)\bigg|\geq\bigg|s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j-f_k\bigg)(0)\bigg| $$ since if it's false we can replace $f_k$ by $-f_k$. This has as an effect that \begin{align} \bigg|s_{n_k}\bigg(\sum_{j=1}^{k}f_j\bigg)(0)\bigg| &\geq \frac{1}{2}\bigg|s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j+f_k\bigg)(0)\bigg|+\frac{1}{2}\bigg|s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j-f_k\bigg)(0)\bigg|\\ &=\frac{1}{2}\bigg|s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j+f_k\bigg)(0)\bigg|+\frac{1}{2}\bigg|-s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j-f_k\bigg)(0)\bigg|\\ &\geq\frac{1}{2}\bigg|s_{n_k}\bigg(\sum_{j=1}^{k-1}f_j+f_k-\sum_{j=1}^{k-1}f_j+f_k\bigg)(0)\bigg|\\ &=|s_{n_k}f_k(0)|\\ &\geq k \end{align} where the first inequality comes from the fact that if $A\geq B$ then $A\geq\frac{A+B}{2}$.

Now let $\displaystyle f:=\sum_{j\geq1}f_j$.

Since $\|f_j\|_{\infty}\leq2^{-j}$ for all $j$, this series converges uniformly and hence $f\in C(\mathbb{T})$.

Also, for all $k$, $$ s_{n_k}f(0) = s_{n_k}\bigg(\sum_{j=1}^kf_j\bigg)(0)+s_{n_k}\bigg(\sum_{j=k+1}^{\infty}f_j\bigg)(0) $$

But we've seen that $$ \bigg|s_{n_k}\bigg(\sum_{j=1}^{k}f_j\bigg)(0)\bigg| \geq k $$ and by lemma 3 we have \begin{align} \bigg|s_{n_k}\bigg(\sum_{j=k+1}^{\infty}f_j\bigg)(0)\bigg| &\leq (2n_k+1)\bigg\|\sum_{j=k+1}^{\infty}f_j\bigg\|_{\infty}\\ &\leq (2n_k+1)\sum_{j=k+1}^{\infty}\|f_j\|_{\infty}\\ &\leq (2n_k+1)\sum_{j=k+1}^{\infty}\frac{2^{-j}}{2n_k+1}\\ &=\sum_{j=k+1}^{\infty}2^{-j}\\ &\leq1 \end{align} Hence $|s_{n_k}f(0)|\geq k-1$ which $\to\infty$ as $k\to\infty$.