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We have space $L^1(\mathbb{R}^k) $ and a set $C =$ {$f \in L^1 (\mathbb{R}^k) : f(x) = e^{-a\cdot ||x||^2}, a>0 $}. Prove that if $f, g \in C$, then $f * g \in C$ (convolution of functions). So the beginning of my solutions is: $$(f * g )(x) = \int\limits_{\mathbb{R}^k} f (x-y) g(y) dl_k(y) = \int\limits_{\mathbb{R}^k} e ^{-a||x-y||^2} e^ {-b ||y||^2} dl_k(y) $$ $$=\int\limits_{\mathbb{R}^k} e^{-a ((y_1 -x_1)^2 + ... + (y_k - x_k)^2) - b(y_1^2 +... y_k^2)} dl_k(y) $$ and...? I'm stuck here. Could somebody help?

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I don't think $dl_k$ is a standard notation for the Lebesgue measure. Is it? – 1015 Mar 22 '13 at 21:10
You had a typo: it is $y_j$'s in the last exponential, not $x_j$'s. I hope you don't mind if I edited that. – 1015 Mar 22 '13 at 21:13

Edit: This is false. But almost true. Up to multiplication by a constant.

Steps: Do the case $k=1$ first. Develop the quadratic in $y$. Complete the square. Change the variable appropriately to use $\int_\mathbb{R}e^{-t^2}dt=\sqrt{\pi}$. Then apply Fubini in the general $k$ case.

Details: For $k=1$, $f(x)=e^{-ax^2}$, and $g(x)=e^{-bx^2}$, we have $$ (f*g)(x)=\int_{\mathbb{R}}e^{-a(x-y)^2}e^{-by^2}dy=\int_{\mathbb{R}}\exp{\left(-(a+b)\left(y^2-\frac{2ax}{a+b}y+\frac{ax^2}{a+b}\right)\right)}dy $$ $$ =\int_{\mathbb{R}}\exp{\left(-(a+b)\left(\left(y-\frac{ax}{a+b}\right)^2+\frac{abx^2}{(a+b)^2}\right)\right)}dy $$ $$ =e^{-\frac{abx^2}{a+b}}\int_{\mathbb{R}}\exp{\left(-(a+b)\left(y-\frac{ax}{a+b}\right)^2\right)}dy $$ $$ =e^{-\frac{abx^2}{a+b}}\int_{\mathbb{R}}e^{-u^2}\frac{du}{\sqrt{a+b}}=\frac{\sqrt{\pi}}{\sqrt{a+b}}e^{-\frac{abx^2}{a+b}}. $$ Back to the general $k$ case, and applying Fubini, we obtain $$ (f*g)(x)=\prod_{j=1}^k\int_{\mathbb{R}}e^{-a(x_j-y_j)^2}e^{-by_j^2}dy=\prod_{j=1}^k\frac{\sqrt{\pi}}{\sqrt{a+b}}e^{-\frac{abx^2}{a+b}} $$ $$ =\left(\frac{\pi}{a+b} \right)^\frac{k}{2}e^{-\frac{a}{a+b}\|x\|^2}. $$

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In exponent we have $-ay^2 - ax^2 + 2ayx - by^2$. So t would be like $sqrt{a(x-y)^2 + by^2}$ but I don't obtain integral you wrote... – Anne Mar 22 '13 at 22:02
No, that's not the square completion with repect to $y$. Sorry, I have to go. Check on wikipedia or something for square completion. – 1015 Mar 22 '13 at 22:07
Well, ok, I got the completion. And we obtain this $\sqrt\pi$. But we should have something like $e^{-ax^2}$, which isn't a constant... I don't understand. – Anne Mar 22 '13 at 22:23
@Anne I gave the details. Please check my edit. And my computations. If I'm correct, the statement is false, although almost true. Are you sure you had the correct question? – 1015 Mar 23 '13 at 11:19

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