Hi,

can anyone please explain to me why convolution in does not have an identity element?

i.e.

there is no such that

Thanks

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- Aug 22nd 2008, 08:08 AMmcbain666Identity Element for Convolution
Hi,

can anyone please explain to me why convolution in does not have an identity element?

i.e.

there is no such that

Thanks - Aug 22nd 2008, 11:13 AMOpalg
An informal, "physicist's", explanation would go like this. If g*u=g for all g then . 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 (and in fact is not a function at all).

If you want a more mathematical argument, then you could try something like this. let and define f by Then . When |x|>ε that gives . 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, does have what is called an*approximate identity*. That is a sequence of elements of norm 1 such that (in the -norm) as k→∞, for all . For example, when n=1 you can take . This gives a sequence of functions that have increasingly large sharp spikes at the origin and are very small elsewhere. - Aug 22nd 2008, 11:32 AMmcbain666
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

where ,

how could one prove that if

then

?

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. - Aug 22nd 2008, 12:27 PMOpalg
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. - Aug 22nd 2008, 02:22 PMmcbain666
Thanks