Convolution inequality

Question

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.

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).