# Thread: Identity Element for Convolution

1. ## Identity Element for Convolution

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

2. 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.

3. 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...

4. 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.

5. Thanks