Saturday, October 27, 2012

On convolutions

Somebody on MathOverlow  asked for some intuition behind the operation of convolution of two functions. Here is my take on this.  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$


Suppose we are given a function $f:\bR\to \bR$. Discretize  the real axis and think of it as  the collection of point $\Lambda_\hbar:=\hbar \bZ$, where $\hbar>0$ is a small number.  We can then approximate $f$ with its restriction  $f^\hbar:=f|_{\Lambda_\hbar}$. This  is   determined by its generating function, i.e., the   formal power series $\newcommand{\ii}{\boldsymbol{i}}$

$$G^\hbar_f(t)=\sum_{n\in\bZ}f(n\hbar)t^n\in \bR[[t,t^{-1}]]. $$

Then

$$G^\hbar_{f_0\ast f_1}(t)= G^\hbar_{f_0}(t)\cdot G^\hbar_{f_1}(t).\tag{1} \label{1} $$

Observe that if  we   set $t=e^{-\ii\xi \hbar}$, then

$$G^\hbar_f(t)=\sum_{x\in\Lambda_\hbar} f(x) e^{-\ii \xi x}. $$

Moreover

$$ \hbar G^\hbar_f(e^{-\ii\xi \hbar})=\sum _{n\in \bZ} \hbar f(n\hbar) e^{-\ii\xi(n\hbar)}, \tag{2}\label{2}$$

and the expression in the right hand sum is  a "Riemann sum"  approximating

$$\int_{\bR} f(x)^{-\ii\xi x} dx. $$

Above we recognize the Fourier transform of $f$. If we let $\hbar\to 0$  in (\ref{2}) and we use (\ref{1})  we obtain the wellknown fact that the Fourier transform  maps the convolution to the usual pointwise product of functions. (The fact that this rather careless passing to the limit  can be rigorous is what the Poisson formula is all about.)

The above argument shows that we can regard $\hbar G_f^\hbar(1)$ as an approximation for $\int_{\bR} f(x) dx$.





Denote by $\delta(x)$ the Delta function concentrated at $0$. The Delta function concentrated at $x_0$ is then $\delta(x-x_0)$. What could be the generating function of $\delta(x)$, $G_\delta^\hbar$?  First, we know that $\delta(x)=0$, $\forall x\neq 0$ so that


$$G_\delta^\hbar(t) =ct^0=c. $$

The constant $c$ can be determined from the equality 

$$ 1= \int_{\bR} \delta(x) dx=\hbar G_\delta^\hbar(1)=\hbar c$$

Hence $\hbar G_\delta^\hbar(1)=1$.  Similarly

$$ G^\hbar_{\delta(\cdot-n\hbar)} =\frac{1}{\hbar} t^n. $$

In particular, the discretization $\delta^\hbar(x-n\hbar)$ of $\delta(x-n\hbar)$ is the  function $\Lambda_\hbar\to \bR$ with value $\frac{1}{\hbar}$ at $x=n\hbar$ and $0$  elsewhere.

Putting together all of the above we obtain an equivalemn description for the  generating functon af a function $f:\Lambda_\hbar\to\bR$. More precisely

$$ G^\hbar_f(t)=\hbar\sum_{\lambda\in\Lambda_\hbar}f(\lambda) G^\hbar_{\delta(\cdot-\lambda)}(t). $$
In other words

$$f^\hbar= \hbar\sum_{\lambda\in\Lambda_\hbar} f(\lambda)\delta^\hbar_\lambda,\;\;\delta^\hbar_\lambda(\cdot):=\delta^\hbar(\cdot-\lambda). \tag{3}\label{3}$$


The last  equality  suggests an interpretation for the generating function as an algebraic encoding of the fact that $f:\Lambda_\hbar\to\bR$ is a superposition of $\delta$ functions  concentrated along the points of the lattice $\Lambda_\hbar$. The  factor $\hbar$ in (\ref{3}) is a discretization of the infinitesimal $dx$, which indicates that $\hbar\delta^\hbar_\lambda$  should be viewed as a measure.    Observe that

$$(\hbar\delta^\hbar_\lambda)\ast (\hbar\delta^\hbar_\mu)=\hbar\delta^\hbar_{\lambda+\mu}. \tag{4}\label{4}$$
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