Hi,
can anyone please explain to me why convolution in $\displaystyle L^1 $ does not have an identity element?
i.e.
there is no $\displaystyle u \in L^1(R^n)$ such that
$\displaystyle u * g = g$ $\displaystyle \forall g \in L^1(R^n) $
Thanks
Hi,
can anyone please explain to me why convolution in $\displaystyle L^1 $ does not have an identity element?
i.e.
there is no $\displaystyle u \in L^1(R^n)$ such that
$\displaystyle u * g = g$ $\displaystyle \forall g \in L^1(R^n) $
Thanks
An informal, "physicist's", explanation would go like this. If g*u=g for all g then $\displaystyle g(x) = \int_{\mathbb R^n}g(x-t)u(t)\,dt$. 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 $\displaystyle L^1(\mathbb R^n)$ (and in fact is not a function at all).
If you want a more mathematical argument, then you could try something like this. let $\displaystyle \varepsilon>0$ and define f by $\displaystyle f(x) = \begin{cases}1&\text{if }|x|<\varepsilon,\\0&\text{otherwise}.\end{cases}$ Then $\displaystyle f(x) = (f*u)(x) = \int_{\mathbb R^n}f(x-t)u(t)\,dt = \int_{|x-t|<\varepsilon}u(t)\,dt$. When |x|>ε that gives $\displaystyle \int_{|x-t|<\varepsilon}u(t)\,dt = 0$. 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, $\displaystyle L^1(\mathbb R^n)$ does have what is called an approximate identity. That is a sequence of elements $\displaystyle u_k$ of norm 1 such that $\displaystyle f*u_k\to f$ (in the $\displaystyle L^1$-norm) as k→∞, for all $\displaystyle f\in L^1(\mathbb R^n)$. For example, when n=1 you can take $\displaystyle u_k(x) = {\textstyle\frac k{\sqrt{\pi}}}e^{-k^2x^2}$. This gives a sequence of functions that have increasingly large sharp spikes at the origin and are very small elsewhere.
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
$\displaystyle \frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r} $ where $\displaystyle p,q,r \in (1,\infty) $,
how could one prove that if
$\displaystyle f \in L^p, g \in L^q $
then
$\displaystyle 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.