Looking for a specific norm for a bound of the difference of two convolutions I want to show a bound like the following,
$||g*f-g*s||\leq C ||g|| \text{ } ||f-s||$, where f and s are bounded and $g$ can be in $L^{p}(D)$, for $1\leq p<\infty$ and D a bounded domain. In  which norm could such a bound hold?
Thanks for your help!
 A: If $g \in L^p(\mathbb{R})$ has compact support, we can write:
$$
\lVert g*f - g*s \rVert_{L^\infty(\mathbb{R})} \leq \mathrm{esssup}_{x \in \mathbb{R}} \int_D \lvert g(x-y) \rvert \lvert f(y) - s(y) \rvert~\mathrm{d}y \leq 
$$
$$
\lVert f - s \rVert_{L^\infty(D)} \mathrm{esssup}_{x \in \mathbb{R}}\int_D \lvert g(x-y) \rvert~\mathrm{d}y
$$
Use the transformation formula:
$$
\lVert f - s \rVert_{L^\infty(D)} \mathrm{esssup}_{x \in \mathbb{R}}\int_D \lvert g(x-y) \rvert~\mathrm{d}y \leq \lVert f - s \rVert_{L^\infty(D)}  \mathrm{esssup}_{x \in \mathbb{R}} \int_{\mathbb{R}} \lvert g(x-y) \rvert~\mathrm{d}y = 
\lVert f - s \rVert_{L^\infty(D)}  \mathrm{esssup}_{x \in \mathbb{R}} \int_{\mathbb{R}} \lvert g(t) \rvert~\mathrm{d}t = \lVert f - s \rVert_{L^\infty(D)} \lVert g \rVert_{L^1(\mathbb{R})} = \lVert f - s \rVert_{L^\infty(D)} \lVert g \rVert_{L^1(\mathrm{supp}(g))}
$$
Furthermore use Hölder's inequality:
$$
\lVert f - s \rVert_{L^\infty(D)}\lVert g \rVert_{L^1(\mathrm{supp}(g))} \leq C \lVert f - s \rVert_{L^\infty(D)} \lVert g \rVert_{L^p(\mathrm{supp}(g))}
$$
where $C := \lvert \mathrm{supp}(g) \rvert^{1-\frac{1}{p}}$. Therefore:
$$
\lVert g*f - g*s \rVert_{L^\infty(\mathbb{R})} \leq C \lVert f - s \rVert_{L^\infty(D)} \lVert g \rVert_{L^p(\mathrm{supp}(g))}
$$
