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I have to prove that if $f$, $g$ $\in L^1(\mathbb{R^n})$ then $\operatorname{dom}\left(f*g\right)$ is a set of full measure and: $\left\|f*g\right\|_{L^{1}} \le \left\|f\right\|_{L^1} \left\|g\right\|_{L^1}$

Any help would be appreciated.

user127.0.0.1
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luka5z
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3 Answers3

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Start with $\|f*g\|_{L^1}$ and write it up as a double integral, then interchange the order of integration and see what you can get out of it. I recommend assuming they are both nonnegative first, to give you a feeling for the problem. And don't try to go for too much rigor on the first pass; just try to find out where you're going. Once you have the inequality you seek, go back and try to justify the steps rigourously.

Harald Hanche-Olsen
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$\left\|f*g\right\|_{L^{1}}=\int_{R^n}|\int_{R^n}f(x-y)g(y)dy|dx \le \int_{R^n}\int_{R^n}|f(x-y)g(y)|dydx = $ $\int_{R^n}|g(y)|\int_{R^n}|f(x-y)|dxdy=\int_{R^n}|g(y)|\int_{R^n}|f(t)|dtdy = \left\|g\right\|_{L^{1}} \left\|f\right\|_{L^{1}}$

luka5z
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  • That is correct. You only need to quote the appropriate theorems to justify your calculation. – Harald Hanche-Olsen Jan 17 '14 at 05:55
  • I used Change of Variables Theorem, Fubini's Theorem and one of the integral inequalities – luka5z Jan 17 '14 at 07:55
  • Good. Did you take care to convince yourself that the integral defining $f*g(x)$ actually converges for almost all $x$? (It's not hard, but it needs doing. Hint: Tonelli.) – Harald Hanche-Olsen Jan 17 '14 at 08:04
  • But we know that both $ \left|g\right|{L^{1}}$, $\left|f\right|{L^{1}}$ are finite and their product is majorizing $ \left|fg\right|_{L^{1}}$ so isn't $fg$ convergent form that? – luka5z Jan 17 '14 at 08:16
  • You know that $\int\int|f(x-y)g(y)|,dx,dy<\infty$ by direct calculation. By Tonelli, $\int\int|f(x-y)g(y)|,dy,dx<\infty$ as well. So $\int|f(x-y)g(y)|,dy<\infty$ for almost all $x$, and thus $\int f(x-y)g(y),dy$ exists for almoas all $x$. Before you know this, you are strictly speaking not allowed to begin your calculation, as you don't know that $f*g$ is defined a.e. – Harald Hanche-Olsen Jan 17 '14 at 14:53
  • So Tonelli theorem says that if some nonnegative $f$ is in $L_1\mathbb(X)$ and $\mathbb{X}$ is sigma finite measure space, then we can transform our integral into iterate integrals. But how does this help us? I found Tonelli theorem on Wikipedia. – luka5z Jan 17 '14 at 22:12
  • More importantly, just like Fubini you can use it to interchange the order of integration. But whereas Fubini needs the function to be integrable, Tonelli does not. Instead, Tonelli works for nonnegative functions, integrable or not. The classic use of the pair is to use Tonelli on the absolute value of a function to show it is integrable, to justify the use of Fubini. – Harald Hanche-Olsen Jan 18 '14 at 11:03
  • Thank you! Now, everything is clear to me! – luka5z Jan 18 '14 at 11:17
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It might be edifying to consider that the convolution product turns the Banach space $L^{1}(\mathbb{R}^{n})$ into a commutative and associative algebra, for which your desired inequality holds. That is, $[L^{1}(\mathbb{R}^{n}),*]$ is a commutative Banach algebra. More commonly, we consider the convolution kernel $K$. That is, we suppose $K(x) \in L^{1}(\mathbb{R}^{n})$. Then the linear mapping $ f \rightarrow K*f$ is a bounded map on $L^{1}(\mathbb{R}^{n})$ with operator norm $ \leq \left\Vert K \right\Vert_{L^{1}}$, i.e. $\left\Vert K*f \right\Vert_{L^{1}} \leq \left\Vert K \right\Vert_{L^{1}}\left\Vert f \right\Vert_{L^{1}}$.

user 3462
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