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Thread: Identity Element for Convolution

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by mcbain666 View Post
    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.
    Last edited by Opalg; Aug 22nd 2008 at 11:46 PM. Reason: typo
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by mcbain666 View Post
    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.
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