Convolution and good kernels

March 26, 2012

At the end of my last post, A motivational problem for the study of Fourier series, I said that I would completely solve the Dirichlet problem in the disk in my next post. However, I decided to introduce some of the concepts needed first. In this post, I will define some basic concepts and state some of their basic properties. First, the convolution of periodic functions and how it is directly related to Fourier series, then the notion of a good kernel (approximation to the identity), as treated in Stein & Shakarchi’s book. In my next post, I will treat the notion of Cesàro summability, which is another way of assigning a sum to an infinite series, and then show how these things all tie together, giving a quick and easy proof of Fejér’s theorem, which, in particular, implies Weierstrass’s approximation theorem. So I will resume my study of the Dirichlet problem in the disk in 2 posts.

Definition: Let $f$ and $g$ be two $2\pi$ -periodic (Riemann-)integrable functions on $\textbf{R}$ . Then their convolution $f*g$ is defined on $[-\pi, \pi]$ as

$(f*g)(x) = \displaystyle\frac{1}{2\pi} \int_{-\pi}^{\pi} f(y)g(x-y)dy.$

Since $f$ and $g$ are periodic, it is clear from exercice 1 that we can also write

$(f*g)(x) = \displaystyle\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x-y)g(y)dy.$

The convolution can be viewed as a “weighted average” or as the amount of overlap between the two functions as one is shifted over another (see Wolfram). There is also a nice intuitive explanation in terms of the Dirac distribution on Wikipedia. More importantly, here are some algebraic properties of the convolution :

Proposition: Let $f$ , $g$ and $h$ be $2\pi$ -periodic integrable functions on $\textbf{R}$ . Then :

$f*(g+h) = (f*g) + (f*h)$

$(cf)*g = c(f*g) = f*(cg)$ for any $c \in \textbf{C}$

$f*g = g*f$

$(f*g)*h = f*(g*h)$

$f*g$ is continuous

$\widehat{f*g}(n) = \hat{f}(n)\hat{g}(n).$

Where $\hat{f}(n)$ and $\hat{g}(n)$ respectively denote the $n$ th Fourier coefficient of $f$ and $g.$ The first two points are a consequence of the linearity of the integral. To prove the other points, one can proceed as follows: first prove the property in the case where $f$ and $g$ are continuous. It is fairly straightforward to do, using Fubini’s theorem to interchange the order of integration freely and using a change of variable when needed. For point 5, one needs to see that $g$ must be uniformly continuous on all of $mathbf{R}$ . Once the continuous case is settled, the idea is to approximate $f$ and $g$ by continuous functions and use uniform convergence arguments to conclude.

It is also of interest to note that the first four points of the last proposition tell us that the integrable functions on the circle have the structure of an algebra (without identity) with the product being the convolution. Point 6 tells us that taking the Fourier coefficients is an algebra homomorphism from the convolution product algebra to the pointwise product algebra of functions on the circle.

Let’s see why the convolution is of immediate interest for the study of Fourier series. Let the $N^{th}$ Dirichlet kernel $D_N$ be defined as

$D_N(x) = \displaystyle\sum_{n = -N}^N e^{inx}.$

Then, for $f$ a function on the circle, the Fourier series of $f$ can be expressed as

$S_N(f)(x) = \displaystyle\sum_{n=-N}^N \hat{f}(n)e^{inx}$

$= \displaystyle\sum_{n=-N}^N \left(\frac{1}{2\pi}\int_{-\pi}^{\pi} f(y)e^{-iny}dy\right) e^{inx}$

$= \displaystyle\frac{1}{2\pi} \int_{-\pi}^{\pi} f(y) \left(\sum_{n=-N}^N e^{in(x-y)}\right) dy$

$= (f*D_N)(x).$

So studying Fourier series is the same as studying the convolution $f*D_N$ .

Definition (from Stein’s book): A family of kernels $\{K_n(x)\}_{n \in \textbf{N}^*}$ on the circle is a family of good kernels if it satisfies the following properties:

For all $n \geq 1,$ we have $\frac{1}{2\pi}\int_{-\pi}^{\pi}K_n(x)dx = 1.$
There exists $M >0$ such that for all $n \geq 1$ , $\int_{-\pi}^{\pi}|K_n(x)|dx \leq M$ .
For every $\delta > 0$ , $\int_{\delta \leq |x| \leq \pi}|K_n(x)|dx \rightarrow 0, \quad$ as $n \rightarrow \infty$ .

So a family of good kernels on the circle can be viewed as a sequence of functions on $[-\pi, \pi]$ having constant area under their curve, peaking at the origin. At the limit, we get the Dirac delta function (which is not a function). property 2 assures us that the first property isn’t just a consequence of the $K_n$ ‘s being symmetric with respect to the $x$ axis. Viewed as weight distributions, we can interpret property 1 as saying that $K_n$ assigns unit mass to the whole circle and property 3 as saying that this mass concentrates near the origin as $n \rightarrow \infty$ . With this in mind, the next theorem shouldn’t come as a big surprise :

Theorem (Stein’s book, p.49): Let $\{K_n\}_{n \in \textbf{N}}$ be a family of good kernels, and $f$ an integrable function on the circle. Then

$\displaystyle\lim_{n\rightarrow \infty} (f*K_n)(x) = f(x)$

whenever $f$ is continuous at $x$ . If $f$ is continuous everywhere, then the above limit is uniform.

Proof: Suppose $f$ is continuous at $x$ and let $M$ be like the $M$ of property 2 in the definition of good kernels. For $\epsilon > 0$ , choose $\delta$ so that $|f(x-y) - f(x)| < \epsilon \pi / M$ whenever $|y| < \delta.$ The first property of good kernels let us write

$(f*K_n)(x) - f(x) = \displaystyle\frac{1}{2\pi} \int_{-\pi}^{\pi}K_n(y)f(x-y)dy - f(x)$

$= \displaystyle\frac{1}{2\pi} \int_{-\pi}^{\pi}K_n(y)[f(x-y) - f(x)]dy.$

We thus get

$|(f*K_n)(x) - f(x)| \leq \displaystyle\frac{1}{2\pi} \int_{|y| < \delta}|K_n(y)||f(x-y)-f(x)|dy$

$+ \displaystyle\frac{1}{2\pi} \int_{\delta \leq |y| \leq \pi} |K_n(y)||(f(x-y)-f(x)|dy$

$\leq \displaystyle\frac{\epsilon \pi}{2\pi M} \int_{-\pi}^{\pi} |K_n(y)|dy + \displaystyle\frac{2B}{2\pi} \int_{\delta \leq |y| \leq \pi} |K_n(y)|dy$

where $B$ is a bound for $|f|$ (which is necessarily bounded since it is periodic). Now, from the second property of good kernels, we can deduce that the first term is less than $\epsilon / 2$ and by the third property, for $n$ large enough the second term will also be less than $\epsilon / 2$ . Therefore, we showed that for $n$ large enough, we have

$|(f*K_n)(x) - f(x)| < \epsilon$

hence the first assertion is proved. But if $f$ is continuous everywhere, then it is uniformly continuous (being defined on a compact set). so $\delta$ can be chosen independently of $x$ . This proves that $f*K_n \rightarrow f$ uniformly if $f$ is continuous everywhere.

QED

Recall that the (convolution) algebra of integrable functions on the circle does not have an identity. Still, this last result gives us something that resemble an identity and the family $\{K_n\}$ is sometimes referred to as an approximation to the identity. Viewing the convolutions as weighted averages, the result is intuitive because as $n \rightarrow \infty,$ the value $f(x)$ is assigned the full mass.

With this powerful result, we see that if $\{D_N\}$ would be a family of good kernels, then the Fourier series of $f$ would converge to $f(x)$ whenever $f$ would be continuous at $x.$ Unforunately, this is not how it is (the Fourier series of a continuous function can actually diverge). Indeed, it can be shown that $\int_{-\pi}^{\pi}|D_N(x)|dx \geq c \log N$ as $N \rightarrow \infty$ for some $c \in \textbf{R},$ so the Dirichlet kernels don’t satisfy property 2 of the definition of good kernels. Rest assured though, not all is lost. Being sums of exponentials, $\{D_N\}$ actually satisfies property 1 of the definition.

From → Analysis, Fourier analysis

7 Comments

sabrina permalink

Thank you to give us the opportunity of see this posts.
I don’t understand only because the dirichlet kernel satisfies property 1 of the definition.
thank you.

Reply
- J-F Arbour permalink
  
  You are right, the Dirichlet kernels don’t form a family of good kernels. However, when you take the average of Dirichlet kernels, you get a family of good kernels called the Fejér kernels. The next post after this one, called Cesàro summability and Fejér’s theorem, adresses this result.
  
  Thank you for your comment, I’m glad this is of use to some people!
  
  Reply
  - sabrina permalink
    
    I’m sorry, I have another question for you about this post.
    I can’t write a proof of “the Dirichlet kernels don’t satisfy property 2”.
    I would be grateful if you show me how to begin this proof to make me an idea of this.
    Thank you.
    Sabrina
alicialalal permalink

Amazingly clear and succint introduction. It would be even better if you post some graphs to help us to understand clearer

Reply

Convolution and good kernels

Trackbacks & Pingbacks

Leave a Reply Cancel reply

Search

Categories

Tags

Convolution and good kernels

Share this:

Like this:

Related

Trackbacks & Pingbacks

Leave a Reply Cancel reply

Search

Categories

Tags