Du-Bois- Reymond Function. Please give me an example of a continuous function whose Fourier series diverges at a point. (given by Du Bois-Reymond).
thanks.
 A: 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 $k\to\infty$.
