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

#### by Orr Shalit

Recall Theorem 6 from Notes 3:

**Theorem 6:** *For every , the Fourier series of converges uniformly to . *

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 the Banach space with the sup norm. Let . Recall that for every integer we define the **th Fourier coefficient **of to be

The **th partial sum **of is defined to be the function

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

Note that

where .

**Definition 1: ***The sequence of functions is called the Dirichlet kernel. *

**Definition 2: ***If , then the convolution of with is the function *

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 , the partial sums are given by *

**Proposition 4:** *For all , *

**Proof:** This follows from multiplying both sides by and using trig identities.

#### 2. Divergence of Fourier series

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

**Theorem 5: ***Let be the Banach space with the sup norm. Then there exists a function whose Fourier series diverges at . *

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 be as above. *

*Given ,**the set of functions in whose Fourier series diverges at is generic.**Given any sequence in ,**the set of functions in whose Fourier series diverges at for all is generic.*

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

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

For every , let be the linear functional

**Lemma 7:** *For all , is bounded, and . *

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

**Proof of Lemma 7: **The straightforward estimate

gives , and in particular is bounded. Let be a sequence of continuous periodic functions of norm less than converging in the norm to the function . Then , while . This gives .

We shall now show that diverges to infinity when . For this we make the estimates: