Recall Theorem 6 from Notes 3:
Theorem 6: For every $latex f in C_{per}([0,1]) cap C^1([0,1])$, the Fourier series of $latex f$ converges uniformly to $latex f$.
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 $latex X$ the Banach space $latex C_{per}([0,1])$ with the sup norm. Let $latex f in X$. Recall that for every integer $latex n$ we define the $latex n$th Fourier coefficient of $latex f$ to be
$latex hat{f}(n) = int_0^1 f(x) e^{-2pi i nx} dx .$
The $latex N$th partial sum of $latex f$ is defined to be the function
$latex S_N(f)(x) = sum_{n=-N}^N hat{f}(n) e^{2pi i nx} .$
The most natural questions one asks in harmonic analysis regarding pointwise convergence is whether the sequence of functions $latex S_N(f)$ converges pointwise or almost everywhere to $latex f$. It is convenient to have an alternative representation for the partial sums $latex S_N(f)$.
Note that
$latex S_N(f)(x) = sum_{n=-N}^N left(int_0^1 f(t) e^{-2pi i nt} dt right) e^{2pi i nx} = int_0^1 f(t) D(x-t) dt, $
where $latex D_N(z) = sum_{n=-N}^N e^{2pi i nz}$.
Definition 1: The sequence of functions $latex {D_N}$ is called the Dirichlet kernel.
Definition 2: If $latex g,h in X$, then the convolution of $latex g$ with $latex h$ is the function
$latex g * h (x) = int_0^1 g(t) h(x-t) dt = int_0^1 g(x-t)h(t) dt .$
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 $latex f in X$, the partial sums are given by
$latex S_N(f) = f * D_N . $
Proposition 4: For all $latex N$,
$latex D_N(x) = frac{sin(2pi (N+1/2)x)}{sin(pi x)} .$
Proof: This follows from multiplying both sides by $latex sin(pi x)$ and using trig identities.
2. Divergence of Fourier series
Recall the category theoretic vocabulary that we introduced in the previous notes.
Theorem 5: Let $latex X$ be the Banach space $latex C_{per}([0,1])$ with the sup norm. Then there exists a function $latex f in X$ whose Fourier series diverges at $latex 0$.
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 $latex X$ be as above.
- Given $latex x_0 in [0,1]$, the set of functions in $latex X$ whose Fourier series diverges at $latex x_0$ is generic.
- Given any sequence $latex x_1, x_2, ldots$ in $latex [0,1]$, the set of functions in $latex X$ whose Fourier series diverges at $latex x_i$ for all $latex i$ is generic.
In particular, we see that for any countable dense subset $latex S$ of the interval, there is a continuous and periodic function whose Fourier series diverges on that $latex S$. Things cannot, however, get much worse, because the Fourier series of any $latex L^2$ function converges almost everywhere.
Proof: A countable intersection of generic sets is generic, hence it suffices to prove 1. Also one may assume that $latex x_0 = 0$ (yes, it is obvious, but make sure you really know how to fix this).
For every $latex N$, let $latex F_N$ be the linear functional
$latex F_N(f) = S_N(f)(0) = int_0^1 f(t) D_N(t) dt .$
Lemma 7: For all $latex N$, $latex F_N$ is bounded, and $latex |F_N| rightarrow infty$.
Assuming the lemma for now, let us prove the theorem. Let $latex A subseteq X$ be the set of functions whose Fourier series converges at $latex 0$. If $latex A$ was of the second category, the principle of uniform boundedness would have implied that $latex sup_N |F_N| < infty$, contradicting the lemma. Hence $latex A$ is of the first category, thus its complement is generic (by definition of “generic”).
Proof of Lemma 7: The straightforward estimate
$latex |F_N(f)| leq int|f(t) D_N(t)| dt leq |f| int |D_N(t)| dt $
gives $latex |F_N| leq |D_N|_1$, and in particular $latex F_N$ is bounded. Let $latex {f_n}$ be a sequence of continuous periodic functions of norm less than $latex 1$ converging in the $latex L^1$ norm to the function $latex sign (D_N)$. Then $latex F_N(f_n) rightarrow |D_N|_1$, while $latex |F_N(f_n)| leq |F_N|$. This gives $latex |F_N| = |D_N|_1$.
We shall now show that $latex |D_N|_1$ diverges to infinity when $latex N rightarrow infty$. For this we make the estimates:
$latex |D_N|_1 geq int_0^1 frac{|sin(2pi(N+1/2)x)|}{pi x} dx = int_0^{2pi(N+1/2)} frac{|sin(u)|}{pi u} du$
$latex geq frac{1}{pi}sum_{k=1}^{N} int_{2 pi (k-1)}^ {2 pi k} frac{|sin u|}{u} du $
$latex geq frac{1}{2pi^2}int_0^1 |sin u | du sum_{k=1}^N frac{1}{k} geq C log N rightarrow infty.$