Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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 $$ (f*g)(x)=\prod_{j=1}^k\int_{\mathbb{R}}e^{-a(x_j-y_j)^2}e^{-by_j^2}dy=\prod_{j=1}^k\frac{\sqrt{\pi}}{\sqrt{a+b}}e^{-\frac{abx^2}{a+b}} $$ $$ =\left(\frac{\pi}{a+b} \right)^\frac{k}{2}e^{-\frac{a}{a+b}\|x\|^2}. $$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.