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.

Let $u$ and $v$ be two $L^1(\mathbb{R})$ functions such that $\|u\|_{L^1} \le \|v\|_{L^1}$ and $f$ is non-negative $L^1(\mathbb{R})$ with non-negative inverse Fourier transform. Is it true that for the convolution $\|u*f\| \le \|v*f\|$? If not, maybe someone know additional condition that will give the last inequality.

share|improve this question
    
In what norm is the desired inequality? –  Glen Wheeler Feb 12 '11 at 22:43

1 Answer 1

If you additionally assume that the functions are in $L^2$ you could use Plancherel to obtain $\|u \ast v\|_2 = \|\hat{u} \hat{f}\|_2$. So now $\hat{f} \geq 0$ so the multiplication operator is order preserving.

I assume you mean the $L^2$ norm in your desired inequality (natural for convolutions).

share|improve this answer
    
Sorry I forgot to put the $L^1$ norm of the convolution. The desired inequality is $\|u*f\|_1 \le \|v*f\|_1$. Thank you for your reply. –  user6997 Feb 12 '11 at 23:56
    
Just one little detail, why is $\hat{f}\geq0$? –  AD. Nov 14 '13 at 20:30
    
Also, as I see it $L^1$ is more natural since it is closed under convolution. –  AD. Nov 14 '13 at 20:33

Your Answer

 
discard

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