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To be more specific, prove that $L^1(\mathbb{R}^n)$ with multiplication defined by convolution: $$ (f\cdot g)(x)=\int_\mathbb{R^n}f(x-y)g(y)dy $$ is a Banach algebra. All the properties of Banach algebra are easily proved except the last one: $$ \|f\cdot g\|\le \|f\|\|g\|. $$ Could anyone give me a hint?

Actually I found a proof on the web, but I don't see why the last equality holds.

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What about Fubini? – 1015 Feb 13 '13 at 18:52
I edited my question. Now I found a proof on the web, but I don't see why the last equality holds. – user60610 Feb 13 '13 at 19:02
This is a special case of Young's inequality for convolutions by the way, with $p = q = r = 1$. – anonymous Feb 13 '13 at 19:04
up vote 1 down vote accepted

$$ \int\int |f(x-y)g(y)|\;dx\;dy = \int\left[\int|f(x-y)|\;dx\right]|g(y)|\;dy = \int\left[\int|f(x)|\;dx\right]|g(y)|\;dy = \left[\int|f(x)|\;dx\right]\;\left[\int |g(y)| \;dy\right] = \|f\|\;\|g\| $$

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I can see that by variable substitution $\int |f(x-y)|dx$ reduces to $\int |f(x)|dx$. But why isn't $\int |f(x-y)|dx$ a function of $y$? I am confused. – user60610 Feb 13 '13 at 19:05
It is. But you can just shift it since you integrate over all of $\mathbb R$. – anonymous Feb 13 '13 at 19:06
But I remember in the calculation of conditional density, by integrating over one variable $\int_{\mathbb{R}} f(x,y)dy=f_X(x)$ is a function of $x$. What is the difference here? – user60610 Feb 13 '13 at 19:09
Yes, $\int |f(x-y)|dx$ is a function of $y$. But a calculation shows it is a constant function. – GEdgar Feb 13 '13 at 19:10
I managed to get the conclusion that $\int |f(x-y)|dx$ is a constant by using the substitution $t=x-y$. Is there a more enlightening way of doing this? Moreover, I can also conclude that $\int |f(x-y)|dx=\int |f(x-y)|dy$, right? – user60610 Feb 13 '13 at 19:12

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