# Understanding Completeness of Fourier Basis

1. Oct 18, 2013

### stephenkeiths

So the other day in class my teacher gave a proof for the completeness of $\phi_n(x) = \frac{1}{\sqrt{2\pi}}e^{inx}$ in $L^2([-\pi,\pi])$. And I'm trying to convince my self I understand it at least a little. He defined Frejer's Kernel
$K_n(x) = \frac{1}{2\pi(n+1)} \sum_{k=-n}^{n}(n+1-|k|)e^{ikx}$
He claimed without showing that it could be written as
$K_n(x) = \frac{1}{2\pi(n+1)} \frac{sin^2((n+1)\frac{x}{2})}{sin^2(x/2)}$
He showed it was a summability kernel, and then claimed that any summability kernel approaches the identity. And used this to prove that the only vector orthogonal to all $\phi_n$ is the zero vector. I understand why the last step completes the proof, I'm just looking for some more insight into the rest of the proof.

What I'd like to be able to do, is understand where these definitions come from. What is this kernel and why is it so useful? For instance, if i was trying to prove completeness of the equivalent sines and cosines what would be different? Would the kernel look any different?

Also, in my class on Fourier series, we learned that $sin(nx)$ and $cos(nx)$ are complete on their own, in $L^2([0,\pi])$. How does this relate? How could I prove this? Would I have to use a new kernel?

2. Oct 19, 2013

### R136a1

One of the major problems in the theory of Fourier series is that of pointswise convergence. That is, given a function $f$ and an $x\in [-\pi,\pi]$, when can we guarantee that

$$f(x) = \sum_{n=-\infty}^{+\infty} c_n e^{inx}$$

There have been inventend various conditions, such as differentiability, which make it work. It is however very curious that continuity is not sufficient.

However, there are other types of convergence which behave much better than pointswise convergence. One of these types is called Cesaro convergence. Define

$$s_N(x)= \sum_{n=-N}^{N} c_n e^{inx}$$

Usual pointswise convergence cares about $\lim_{N\rightarrow +\infty} s_N(x)$. Cesaro convergence only requires converges of the averages. That is, it demands that

$$\lim_{N\rightarrow +\infty} \frac{1}{N} \sum_{k=-N}^N s_k(x)$$

For example, take $s_N(x) = (-1)^N$, then this sequence does not converge, but it does Cesaro converge to $0$. On the other hand, if you have usual pointswise convergence, then you always have Cesaro convergence. One of the major theorems in Fourier analysis is the Fejer theorem which states that any Fourier series Cesaro converges to the function provided it is continuous.

How does all this relate to the Fejer kernels? Let us introduce the concept of a convolution product. Given any two integrable functions $f$ and $g$, we define

$$(f*g)(x) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x-t)g(t)dt$$

One of the major advantages of the convolution product is that it relates to ordinary product in the sense that if $c_n$ and $d_n$ are the respective Fourier coefficients of $f$ and $g$, then the coefficients of $f*g$ are $c_nd_n$.
Here, the important property is that if you take the function $e_k(x) = e^{ikx}$, then
$$(e_k*f)(x) = c_k e_k$$
So the partial Fourier series of $f$ can be written as

$$s_n = \sum_{k = -n}^n c_k e_k = \left(\sum_{k=-n}^n e_k\right)*f$$

So the Fourier series can be seen as the convolution with one function. This function is called the Dirichlet kernel, $D_n = \sum_{k=-n}^n e_k$ and it controls pointswise convergence.
Cesaro convergence is

$$\frac{1}{N} \sum_{n=-N}^N s_n = \left(\frac{1}{N}\sum_{n=-N}^N \sum_{k = - n}^n e_k\right)*f$$

The somewhat complicated function $K_N = \frac{1}{N}\sum_{n=-N}^N \sum_{k = - n}^n e_k$ is exactly the Fejer kernel and can be checked to have the form you describe above.

So this Fejer kernel controls Cesaro convergence. It has a few properties that make it very well behaved, these properties are:

• $K_n\geq 0$
• $\frac{1}{2\pi}\int_{-\pi}^\pi K_n(t)dt = 1$
• For any $\delta$, we have $\int_{\delta\leq |t|\leq \pi} K_n(t)dt\rightarrow 0$

The Dirichlet kernel for pointswise convergence does not share these properties and this explains why pointswise convergence has worse properties.

The completeness of the sine and cosines can easily be proven from the completeness of the exponential functions, just use the relations

$$\sin(x) = \frac{ e^{ix} - e{-ix}}{2i},~\cos(x) = \frac{e^{ix} + e^{-ix}}{2},~e^{ix} = \cos(x) + i\sin(x)$$

3. Oct 19, 2013

### stephenkeiths

First of all, thank you for the very informative post; it was very enlightening.

Why do $sin(nx)$ form a complete basis in $L^2([0,\pi])$?

What is the essential difference? Does it have to do with odd and even functions?

Thanks again