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From Rudin, Real and Complex Analysis, 1st edition, Chapter 7, Problem 4

Suppose $1\le p\le \infty$, $f\in L^{1}(\mathbb{R}^{1})$, $g\in L^{p}(\mathbb{R}^{1})$. Show that the the integral defining $f*g$ exists for almost all $x$, that $f*g\in L^{p}$, and that $|f*g|_{p}\le |f|_{1}|g|_{p}$.

Show that the equality can hold if $p=1$ and if $p=\infty$, and find conditions under which this happens. Assume $1<p<\infty$, and equality holds, then either $f=0$ a.e or $g=0$ a.e.

My thoughts:

By Fubini's theorem we can prove $$|f*g|_{p}^{p}\le |f|_1^{p} |g|_p^{p}$$ So the first claim is okay. But I do not know how to show when the equality holds.

So it suffice to prove $$\int\left(\int f(x-y)g(y)dy\right)^{p}dx<\left(\int |f|dx\right)^{p}\left(\int |g|^{p}dy\right)$$

By Fubini's theorem we have \begin{align}\int\left(\int f(x-y)g(y)dy\right)^{p}dx&\le \int (\int |f(x-y)|^{p}|g(y)|^{p}dy)dx\\&=\left(\int |g(y)|^p dy)(\int |f(x-y)| dx\right)^{p-1}\int |f(x-y)|dy\\&=\int |g(y)^p|dy (\int |f(t)|dt)^{p} \end{align} using Lebesgue measure's translation invariance. Since the first inequality is just from absolute value, we may assume $f,g$ are non-negative without losing any generality. But I do not see which one of the last few equalities is just an inequality.

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Here's how: look at each inequality that you used in the proof of $\|f*g\|_p \le \|f\|_1\|g\|_p$ separately, and ask yourself: when does equality hold here? –  user53153 Jan 5 '13 at 6:20
    
Let me expand the details and you may comment. –  Bombyx mori Jan 5 '13 at 6:23
    
Now updated with the "proof". –  Bombyx mori Jan 5 '13 at 6:31
    
Your comments after the proof are exactly backwards. Equalities are equalities, there is nothing to discuss there. They cannot turn into strict inequalities. We must find out when the first inequality turns into equality. No, it's not "just from absolute value". And no, you can't say "non-negative without losing any generality" when considering the equality cases. –  user53153 Jan 5 '13 at 6:35
    
I see. So the first inequality is the problem. Yes I am at a loss when (and why) it is true. –  Bombyx mori Jan 5 '13 at 6:38

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