### Advanced Analysis, Notes 10: Banach spaces (application: divergence of Fourier series)

Recall Theorem 6 from Notes 3:

Theorem 6: For every $f \in C_{per}([0,1]) \cap C^1([0,1])$, the Fourier series of $f$ converges uniformly to $f$

It is natural to ask how much can we weaken the assumptions of the theorem and still have uniform convergence, or how much can we weaken and still have pointwise convergence. Does the Fourier series of a continuous (and periodic) function always converge? In this post we will use the principle of uniform boundedness to see that the answer to this question is a very big NO.

Once again, we begin with some analytical preparations.

#### 1. The Dirichlet kernel

Denote by $X$ the Banach space $C_{per}([0,1])$ with the sup norm. Let $f \in X$. Recall that for every integer $n$ we define the $n$th Fourier coefficient of $f$ to be $\hat{f}(n) = \int_0^1 f(x) e^{-2\pi i nx} dx .$

The $N$th partial sum of $f$ is defined to be the function $S_N(f)(x) = \sum_{n=-N}^N \hat{f}(n) e^{2\pi i nx} .$

The most natural questions one asks in harmonic analysis regarding pointwise convergence is whether the sequence of functions $S_N(f)$ converges pointwise or almost everywhere to $f$. It is convenient to have an alternative representation for the partial sums $S_N(f)$.

Note that $S_N(f)(x) = \sum_{n=-N}^N \left(\int_0^1 f(t) e^{-2\pi i nt} dt \right) e^{2\pi i nx} = \int_0^1 f(t) D(x-t) dt,$

where $D_N(z) = \sum_{n=-N}^N e^{2\pi i nz}$.

Definition 1: The sequence of functions $\{D_N\}$ is called the Dirichlet kernel

Definition 2: If $g,h \in X$, then the convolution of $g$ with $h$ is the function $g * h (x) = \int_0^1 g(t) h(x-t) dt = \int_0^1 g(x-t)h(t) dt .$

The first equality above is the definition, while the second is a simple observation using change of variables. We reached the following result.

Proposition 3: For every $f \in X$, the partial sums are given by $S_N(f) = f * D_N .$

Proposition 4: For all $N$ $D_N(x) = \frac{\sin(2\pi (N+1/2)x)}{\sin(\pi x)} .$

Proof: This follows from multiplying both sides by $\sin(\pi x)$ and using trig identities.

#### 2. Divergence of Fourier series

Recall the category theoretic vocabulary that we introduced in the previous notes.

Theorem 5: Let $X$ be the Banach space $C_{per}([0,1])$ with the sup norm. Then there exists a function $f \in X$ whose Fourier series diverges at $0$

Theorem 5 can also be proved in a constructive manner, by writing down an example. We will obtain the following much stronger, albeit non-constructive, result, which implies Theorem 5 immediately.

Theorem 6: Let $X$ be as above.

1. Given $x_0 \in [0,1]$the set of functions in $X$ whose Fourier series diverges at $x_0$ is generic.
2. Given any sequence $x_1, x_2, \ldots$ in $[0,1]$the set of functions in $X$ whose Fourier series diverges at $x_i$ for all $i$ is generic.

In particular, we see that  for any countable dense subset $S$ of the interval, there is a continuous and periodic function whose Fourier series diverges on that $S$.  Things cannot, however, get much worse, because the Fourier series of any $L^2$ function converges almost everywhere.

Proof: A countable intersection of generic sets is generic, hence it suffices to prove 1. Also one may assume that $x_0 = 0$ (yes, it is obvious, but make sure you really know how to fix this).

For every $N$, let $F_N$ be the linear functional $F_N(f) = S_N(f)(0) = \int_0^1 f(t) D_N(t) dt .$

Lemma 7: For all $N$, $F_N$ is bounded, and $\|F_N\| \rightarrow \infty$

Assuming the lemma for now, let us prove the theorem. Let $A \subseteq X$ be the set of functions whose Fourier series converges at $0$. If $A$ was of the second category, the principle of uniform boundedness would have implied that $\sup_N \|F_N\| < \infty$, contradicting the lemma. Hence $A$ is of the first category, thus its complement is generic (by definition of “generic”).

Proof of Lemma 7: The straightforward estimate $\|F_N(f)\| \leq \int|f(t) D_N(t)| dt \leq \|f\| \int |D_N(t)| dt$

gives $\|F_N\| \leq \|D_N\|_1$, and in particular $F_N$ is bounded. Let $\{f_n\}$ be a sequence of continuous periodic functions of norm less than $1$ converging in the $L^1$ norm to the function $sign (D_N)$. Then $F_N(f_n) \rightarrow \|D_N\|_1$, while $|F_N(f_n)| \leq \|F_N\|$. This gives $\|F_N\| = \|D_N\|_1$.

We shall now show that $\|D_N\|_1$ diverges to infinity when $N \rightarrow \infty$. For this we make the estimates: $\|D_N\|_1 \geq \int_0^1 \frac{|\sin(2\pi(N+1/2)x)|}{\pi x} dx = \int_0^{2\pi(N+1/2)} \frac{|\sin(u)|}{\pi u} du$ $\geq \frac{1}{\pi}\sum_{k=1}^{N} \int_{2 \pi (k-1)}^ {2 \pi k} \frac{|\sin u|}{u} du$ $\geq \frac{1}{2\pi^2}\int_0^1 |\sin u | du \sum_{k=1}^N \frac{1}{k} \geq C \log N \rightarrow \infty.$