We are going to prove the two first mentioned divergence results. Both du Bois-Reymond's and Kolmogorov's proofs are good examples of a general method to construct counterexamples in analysis: Take a set of functions each of which is ''closer'' to being a counter example than the previous ones, and use these functions to form a series that is pathological enough to give the desired counter example. This is also for example the method one usually uses to construct a nowhere differentiable continuous function. However, for du Bois-Reymond's result, we will apply the Banach-Steinhaus theorem, as it is perhaps clearer and emphasizes another important method to show the existence of counterexamples: Suppose there is no counterexample in the function space under consideration, and show that the lack of bad functions leads to too good properties for the space to be true. Another application of that principle would be the proof that there exists a sequence $(a_n)_{n=-\infty}^{\infty}$ converging to zero that is not the sequence of Fourier coefficients of any $L^1$-function (by the Riemann-Lebesgue lemma, the condition for $(a_n)$ is necessary). A drawback of such proofs is that they give no explicit example, but in practice, the constructive proofs are in many cases also not very explicit.
A pathological continuous function
The reason behind the existence of pathological Fourier series is, as mentioned in the earlier post, that the Dirichlet kernels $D_n$ are not a good family of kernels. More precisely,
Theorem 1. We have
\[\begin{eqnarray}\int_{-\frac{1}{2}}^{\frac{1}{2}}|D_N(x)|dx\gg \log N\end{eqnarray}\]
for all $N$, so $(D_N)_{N=1}^{\infty}$ violates the second condition for a good family of kernels (we use the notation $f\ll g$ for $f=O(g)$).
Proof. Using $|\sin a|\leq a$, one obtains
\[\begin{eqnarray}\int_{-\frac{1}{2}}^{\frac{1}{2}}|D_N(x)|dx&\geq& \int_{-\frac{1}{2}}^{\frac{1}{2}}\left|\frac{\sin((2N+1)\pi x)}{\pi x}\right|dx\\ &=&\int_{-\frac{2N+1}{2}}^{\frac{2N+1}{2}}\left|\frac{\sin(\pi y)}{\pi y}\right|\\&>&\sum_{k=-N}^{N} \int_{\frac{k}{2}-\frac{1}{4}}^{\frac{k}{2}+\frac{1}{4}}\left|\frac{\sin(\pi y)}{\pi y}\right|\\&\gg&\sum_{k=1}^{N} \frac{1}{k+\frac{1}{2}}\gg \log N\end{eqnarray}\]
by comparison with the harmonic series. ■
We recall that the Fourier partial sums are given by the convolution $S_N(f,x)=(f*D_N)(x)$, so we want to study the linear operators $S_N:C_{\#}([0,1])\to C_{\#}([0,1])$, where $C_{\#}([0, 1])$ is the space of continuous $1$-periodic functions on $[0,1].$ It is well-known, and not difficult to prove, that this space is a Banach space (the norm used is of course the sup norm). As we are getting close to a situation where the Banach-Steinhaus theorem, or the uniform boundedness principle, applies, we now formulate it.
Theorem 2 (Banach-Steinhaus). Let $(T_j)_{j \in J}$ be a collection of bounded linear operators from a Banach space $B$ to a normed vector space $X$. Then exactly one of the following holds:
(i) $\sup_{j\in J}\|T_j\|_{B\to X}<\infty$ (the operator norms are uniformly bounded)
(ii) There exists $f\in B$ such that $\sup_{j\in J}\|T_j f\|_X=\infty$ (the operators are unbounded at a fixed point).
This Theorem is a rather direct consequence of the Baire category theorem about complete metric spaces. Now we are ready to formulate the first divergence result.
Theorem 3. There exists a continuous $1$-periodic function whose Fourier series diverges at a point.
Proof. We consider the linear operators $S_N(f)(0):C_{\#}([0,1])\to \mathbb{C}$ (the $N$th Fourier partial sum at the origin). The operators $S_N$ are bounded from $C_{\#}([0,1])$ to $\mathbb{C}$ since $|S_N(f)(0)\|\leq \|f\|_{\infty}\|D_N\|_1$. We may therefore apply the Banach-Steinhaus theorem. We will see that the operator norms of $S_N(\cdot)(0)$ are unbounded, so condition (ii) holds in the Banach-Steinhaus theorem, and this gives the desired function. If $g(x)=\text{sign}(D_N(x))$ (this is a piecewise continuous function), then
\[\begin{eqnarray}S_N(g)(0)=\int_{0}^{1} g(x)D_N(x)dx=\|D_N\|_1.\end{eqnarray}\]
Now take a sequence $(f_n)$ of continuous $1$-periodic functions that approach $g$ uniformly and are bounded by $1$ in absolute value. Then$|S_N(f_n)(0)|\to \|D_N\|_1$ as $n\to \infty.$ We conclude that the operator norm of $S_N(\cdot)(0)$ is $\|D_N\|_1$, which is unbounded, giving a contradiction. ■
A pathological integrable function
We now turn to Kolmogorov's example on the pointwise divergence of $L^1$-functions. Surprisingly, I could not find online lecture notes containing a proof of it (there are probably such lecture notes, but they are not very easy to find). Almost all the lecture notes I found mentioned the result but said that the proof is beyond the scope of the notes, or too long, or something along those lines. I then decided to take a look at Kolmogorov's original article Une série de Fourier–Lebesgue divergente presque partout, and the proof given there uses no difficult theorems; it is of comparable difficulty to de Bois-Reymond's example of a continuous function whose Fourier series diverges at a point, but for some reason the latter proof is usually presented and the former is not. On the other hand, Kolmogorov's proof is five pages and involves choosing carefully a large collection of parameters but I don't think that should not be too long a proof for most lecture notes on Fourier analysis, especially given the importance of the result.
The proof is in two parts. We first show that if there exists a sequence $(\varphi_n)$ of sufficiently pathological integrable functions, then a suitable infinite linear combination of them is a desired function. After that, we construct those $\varphi_n$.
What do we want from the functions $\varphi_n$? First of all, they should be close to having a divergent Fourier series; the function $\varphi_n$ should have a Fourier partial sum that is large as a function of $n$ in a subset of $[0,1]$ with large measure (this measure should approach $1$). Moreover, we want some control over their linear combinations, so we require that the functions are nonnegative an that the $L^1$-norms are constant. We formulate these requirements more precisely.
Lemma 4. There exists a sequence $(\varphi_n)$ of integrable functions on $[0,1]$ such that
(i) $\varphi_n\geq 0, \|\varphi_n\|_1=2$
(ii) The Fourier partial sums of $\varphi_n$ are uniformly bounded.
(iii)There exists a sequence $(M_n)$ tending to infinity such that the $q_n$th Fourier partial sum of $\varphi_n$ has absolute value greater than $M_n$ in a set $E_n\subset[0,1]$, where $m(E_n)\to 1$ as $n\to \infty$ ($m(\cdot)$ is the Lebesgue measure and $q_n$ is just a suitable number).
Assume for now that this has been proved. We apply this to prove
Theorem 5. There exists an $L^1$-function whose Fourier series diverges almost everywhere.
Proof. We choose
\[\begin{eqnarray}\Phi(x)=\sum_{k=1}^{\infty} \frac{1}{\sqrt{M_{n_k}}}\varphi_{n_k}(x),\end{eqnarray}\]
where the $n_k$ are chosen suitably, and show that this function has an almost everywhere divergent Fourier series. Choose $n_1<n_2<...$ inductively in the following way.
(1) $M_{n_k}\geq 2^{2k}$
(2) All the Fourier partial sums of $\varphi_{n_i}, i<k,$ are bounded by $\frac{1}{2}\sqrt{M_{n_k}}$ in absolute value
(3) We have $\frac{1}{2^k}\sqrt{M_{n_k}}\geq q_{n_i}$ for all $i<k.$
Clearly this can be done by taking $n_k$ to be very large in terms of $n_1,...,n_{k-1}$. Now
\[\begin{eqnarray}\|\Phi\|_1&\leq& \sum_{k=1}^{\infty} \frac{1}{\sqrt{M_{n_k}}}\|\varphi_{n_k}\|_1&\leq&\sum_{k=1}^{\infty}\frac{2}{2^k}=2,\end{eqnarray}\]
so $\Phi\in L^1$, and in particular $\Phi$ is finite almost everywhere. Moreover, we can integrate termwise to get
\[\begin{eqnarray}\hat{\Phi}(N)=\sum_{k=1}^{\infty} \frac{1}{\sqrt{M_{n_k}}}\widehat{\varphi_{n_k}}(N),\end{eqnarray}\]
and therefore
\[\begin{eqnarray}S_N(\Phi,x)=\sum_{k=1}^{\infty} \frac{1}{\sqrt{M_{n_k}}}S_N(\varphi_{n_k},x).\end{eqnarray}\]
This is allowed since a series with nonnegative terms can be integrated termwise. We take $N=q_{n_r}$ for some $r$, and $x\in E_{n_r}$. Then by (iii), the $r$th term in the previous sum is at least $\sqrt{M_{n_r}}$ in absolute value. The sum of the terms with $k<r$ can be estimated as follows.
\[\begin{eqnarray}\left|\sum_{k=1}^{r-1} \frac{1}{\sqrt{M_{n_k}}}S_{q_{n_r}}(\varphi_{n_k},x)\right|&<& \sum_{k=1}^{\infty}\frac{\frac{1}{2}\sqrt{M_{n_r}}}{\sqrt{M_{n_k}}}\\&\leq &\sum_{k=1}^{\infty}\frac{\sqrt{M_{n_r}}}{2^{k+1}}= \frac{1}{2}\sqrt{M_{n_r}},\end{eqnarray}\]
owing to (2) and (1). Furthermore, the contribution of the terms with $k>r$ can be bounded by
\[\begin{eqnarray}\left|\sum_{k=r+1}^{\infty}\frac{1}{\sqrt{M_{n_k}}}S_{q_{n_r}}(\varphi_{n_k},x)\right|&\leq&\sum_{k=r+1}^{\infty}\frac{2\cdot(2q_{n_k}+1)}{\sqrt{M_{n_k}}}\\&\leq& \sum_{k=1}^{\infty}\frac{6}{2^k}=6.\end{eqnarray}\]
using (3) and noticing that $|S_N(f,x)|\leq (2N+1)\max_n |\hat{f}(n)|\leq (2N+1)\|f\|_1$.
We conclude that for any $r$, the function $\Phi(x)$ has a Fourier partial sum that is larger in absolute value than $\frac{1}{2}\sqrt{M_{r}}-6$ in the set $E_{n_r}$. If $E=\limsup_r E_{n_r}$ (the set of those $x$ belonging to infinitely may $E_{n_r}$), then $E$ has measure $1$ and for $x\in E$ the Fourier partial sums of $\Phi$ at $x$ are unbounded. ■
We are now left with proving the lemma. How could one construct such functions $\varphi_n$? It would be nice if the $\varphi_n$ were simple (that is, finite linear combinations of characteristic functions), and even better if they were piecewise constant. In addition, one could define the functions $\varphi_n$ recursively on different intervals so that each step of the recursion we would add something to the function. Due to Gibbs phenomenon, one may expect suitable partial sums to be large near the points of discontinuity.
Proof of Lemma 4. We are going to choose a fast growing sequence of integers $m_1,m_2,...$. We set $m_1=n$, and define the $m_i$ inductively. When $m_k$ has been defined for some $1\leq k\leq n$, let $\varphi_n(x)=\frac{m_k^2}{n}$ for $x$ on the interval $I_k=(\frac{2k}{2n+1}-\frac{1}{m_k^2},\frac{2k}{2n+1}+\frac{1}{m_k^2})$. Outside these intervals, set $\varphi_n(x)=0.$ Then $\varphi_n\geq 0$ trivially, and the integral of $\varphi_n$ is 2, as the length of $I_k$ is $\frac{2}{m_k^2}$. Hence it remains to prove (iii).
The $N$th Fourier partial sum is
\[\begin{eqnarray}S_N(\varphi_n,x)=\int_{0}^1 \varphi_n(t)\frac{\sin((2N+1)\pi (t-x))}{\sin(\pi (t-x))}dt=\int_{0}^1 \varphi_n(t)D_N(t-x)dt.\end{eqnarray}\]
We define $m_k$ in terms of $m_1,...,m_{k-1}$ in such a way that
\[\begin{eqnarray}\left|\int_{\bigcup_{i<k} I_i}\frac{\sin((2m_k+1)\pi (t-x))}{\sin(\pi (t-x))}dt\right|<1\end{eqnarray}\]
for all \[\begin{eqnarray}x\in J_k= \left[\frac{2(k-1)}{2n+1}+\frac{2}{n^2},\frac{2k}{2n+1}-\frac{2}{n^2}\right].\end{eqnarray}\]
Such a number $m_k$ exists since the integral converges to $0$ as $m_k\to \infty$ by the Riemann-Lebesgue lemma (notice that $t$ and $x$ belong to different intervals, so there is no problem with integrability). In particular, we may choose $2m_k+1$ to be divisble by $2n+1$.
We have now defined the sequence $(m_k)$, and hence $\varphi_n$ for all $n$. In particular, the integral producing the partial sum, taken over $\bigcup_{i<k}I_i$, is by construction less than $1$ in absolute value. If we show that the corresponding integral over $\bigcup_{k\leq i\leq n} I_i$ is large in a set $E_n$ whose measure tends to $1$, we are done.
Let $r\geq k$, $x\in J_k$, and write the $m_k$th partial sum integral over $I_r$ as
\[\begin{eqnarray}\int_{I_r}\frac{m_r^2}{n}\cdot D_{m_k}(\frac{2r}{2n+1}-x)dt+\int_{I_r}\frac{m_r^2}{n}\left(D_{m_k}(t-x)-D_{m_k}(\frac{2r}{2n+1}-x)\right)dt.\end{eqnarray}\]
The first integrand is just the integral of a constant, while the second integrand, the difference of the Dirichlet kernels is by the mean value theorem at most
\[\begin{eqnarray}\frac{m_k^2}{n}\cdot\frac{1}{m_k^2}\left|\max_t\frac{d}{dt}\frac{\sin((2m_k+1)\pi t)}{\sin(\pi t)}\right|\leq \frac{8\pi}{n},\end{eqnarray}\]
where we used $|t-\frac{2r}{2n+1}|\leq \frac{1}{m_k^2}$. Hence the integral over $I_r$ is
\[\begin{eqnarray}\frac{2}{n} \frac{\sin((2m_k+1)\pi(\frac{2r}{2n+1}-x))}{\sin(\pi(\frac{2r}{2n+1}-x))}+O\left(\frac{1}{n}\right),\end{eqnarray}\]
where the constant in the big $O$ is at most $8\pi$. Since $|\frac{2r}{2n+1}-x|\leq \frac{2}{n^2}$, we have by $\sin y \geq \frac{2}{\pi y}$ ($y\in [0,\frac{\pi}{2}]$) the inequality
\[\begin{eqnarray}\frac{1}{\sin(\pi(\frac{2r}{2n+1}-x))}>\frac{1}{\sin(\pi(\frac{2r}{2n+1}-\frac{2(k-1)}{2n+1})}\geq \frac{n}{r-k+1}.\end{eqnarray}\]
It follows that for $x\in J_k$,
\[\begin{eqnarray}|S_{m_k}(\varphi_n,x)|\geq \sum_{r=1}^{n}\frac{|\sin((2m_k+1)\pi(\frac{2r}{2n+1}-x))|}{r-k+1} -5.\end{eqnarray}\]
We assumed that $m_k$ is divisible by $2n+1$, so $\sin((2m_k+1)\pi(\frac{2r}{2n+1}-x))=\sin((2m_k+1)\pi(\frac{2k}{2n+1}-x))$ for all $r$. Therefore
\[\begin{eqnarray}|S_{m_k}(\varphi_n,x)|&\geq& \left|\sin((2m_k+1)\pi(\frac{2k}{2n+1}-x))\right|\sum_{r=1}^{n-k}\frac{1}{r}\\&\geq& \frac{1}{2}(\log n)\left|\sin((2m_k+1)\pi(\frac{2k}{2n+1}-x))\right|\end{eqnarray}\]
for $n-k\geq \sqrt{n}$.
Finally, choose $M_n=\sqrt{\log n}.$ Then $\varphi_n(x)>M_n$ for those $x\in \bigcup_{k\leq n-\sqrt{n}} J_k$ that satisfy
\[\begin{eqnarray}\left|\sin((2m_k+1)\pi(\frac{2k}{2n+1}-x))\right|\geq \frac{2}{\sqrt{\log n}}.\end{eqnarray}\]
Denote this set by $E_n$. Since the measure of $\displaystyle\bigcup_{k>n-\sqrt{n}} J_k$ is trivially bounded by $\sqrt{n}\cdot(\frac{1}{2n+1}+\frac{4}{n^2})$, the measure of the points not belonging to any $J_k$ with $k\leq n-\sqrt{n}$ is $O(\frac{1}{\sqrt{n}})$.
On the other hand, for any $k\leq n-\sqrt{n}$, the measure of $x\in J_k$ not satisfying the desired inequality is not greater than $\frac{4}{\pi n\sqrt{\log n}}$, because if the integer part of $(2m_k+1)|\frac{2k}{2n+1}-x|$ is fixed, the measure of such points is at most $\frac{4}{\pi\sqrt{\log n}(2m_k+1)}$, and the number of possible integer parts is at most $\frac{2m_k+1}{n}$. We conclude that the measure of $x\in [0,1]$ that do not satisfy the inequality is $O(\frac{1}{\sqrt{\log n}})$, so $m(E_n)\to 1 $ as $n\to \infty.$ The proof of the Lemma is now finished, and this also completes the proof of the Theorem. ■
Kolmogorov's example actually gives a stronger divergence result for free.
Theorem 6. There is an $L^1$-function whose Fourier series diverges in measure almost everywhere.
Proof. Consider the function $\Phi(x)$ again. If its partial sums $S_N(\Phi,x)$ converged in measure in some set $F$, then every subsequence would have a pointwise almost everywhere in $F$ convergent sub-subsequence (This is well-known). In particular, $S_{q_{n_k}}(\Phi,x)$ would have an almost everywhere in $F$ converging subsequence. However, it is unbounded almost everywhere, and this is a contradiction. ■
