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.
Convolution inequality
2
$\begingroup$
analysis
fourier-analysis
harmonic-analysis
-
0In what norm is the desired inequality? – 2011-02-12
1 Answers
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).
-
0Also, as I see it $L^1$ is more natural since it is closed under convolution. – 2013-11-14