Identity Element for Convolution

**mcbain666** · August 22nd 2008, 08:08 AM

Hi,
can anyone please explain to me why convolution in $L^1$ does not have an identity element?
i.e.
there is no $u \in L^1(R^n)$ such that

$u * g = g$ $\forall g \in L^1(R^n)$

Thanks

**Opalg** · August 22nd 2008, 11:13 AM

Originally Posted by mcbain666

Hi,
can anyone please explain to me why convolution in $L^1$ does not have an identity element?
i.e.
there is no $u \in L^1(R^n)$ such that

$u * g = g$ $\forall g \in L^1(R^n)$

An informal, "physicist's", explanation would go like this. If g*u=g for all g then $g(x) = \int_{\mathbb R^n}g(x-t)u(t)\,dt$ . But you would not expect the value of g at x to depend on the values of g at points other than x itself. Therefore the function u must be zero everywhere except at 0, where it has to be so infinitely huge that the value of the integral is g(x). In other words, u has to be the Dirac delta function. But this is not an element of $L^1(\mathbb R^n)$ (and in fact is not a function at all).

If you want a more mathematical argument, then you could try something like this. let $\varepsilon>0$ and define f by $f(x) = \begin{cases}1&\text{if }|x|<\varepsilon,\\0&\text{otherwise}.\end{cases}$ Then $f(x) = (f*u)(x) = \int_{\mathbb R^n}f(x-t)u(t)\,dt = \int_{|x-t|<\varepsilon}u(t)\,dt$ . When |x|>ε that gives $\int_{|x-t|<\varepsilon}u(t)\,dt = 0$ . You can probably believe (and you may even be able to prove) that this implies that u(t)=0 whenever |x|>ε. Since ε is arbitrary, it follows that u is zero everywhere except at 0.

However, $L^1(\mathbb R^n)$ does have what is called an approximate identity. That is a sequence of elements $u_k$ of norm 1 such that $f*u_k\to f$ (in the $L^1$ -norm) as k→∞, for all $f\in L^1(\mathbb R^n)$ . For example, when n=1 you can take $u_k(x) = {\textstyle\frac k{\sqrt{\pi}}}e^{-k^2x^2}$ . This gives a sequence of functions that have increasingly large sharp spikes at the origin and are very small elsewhere.

**mcbain666** · August 22nd 2008, 11:32 AM

Thanks,
both explanations are pretty convincing.
I've been trying with characteristic functions but haven't managed to prove the claim.
By the way, if
$\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ where $p,q,r \in (1,\infty)$ ,
how could one prove that if
$f \in L^p, g \in L^q$
then
$f * g \in L^r$
?

I've been trying with Holder but can't seem to get anywhere, I really should need a good functional analysis book...

Thanks again for your help.

**Opalg** · August 22nd 2008, 12:27 PM

Originally Posted by mcbain666

By the way, if
$\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ where $p,q,r \in (1,\infty)$ ,
how could one prove that if
$f \in L^p, g \in L^q$
then
$f * g \in L^r$
?

I believe this is usually proved by using interpolation theorems, as discussed here.

By the way, if you want to ask a new question, you should really start a new thread. That makes it easier for other people to access it.

**mcbain666** · August 22nd 2008, 02:22 PM

Thanks

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