I'm having a silly time trying to prove the "simple" case of:
$g \in \|L\|_{1}$ and $f \in \|L\|_{\infty}$
Then it holds that:
$\| f \ast g \|_{L^\infty} \leq \|g\|_{L^1}\|f\|_{L^\infty}$
Where $\ast$ is the convolution. This is one of the edge cases in Minkowki's Inequality for convolutions when $p=1$.
So far I have only substituted the definition of infinity norm on both sides but it did not enlighten me at all.
Edit: here it is explicitly,
$\| f \ast g \|_{L^\infty} = inf\{B>0: \mu(\{x:| \int g(y)f(y^{-1}x) d\mu (y) |>B\})=0\}$
and on the other hand:
$\|g\|_{L^1}\|f\|_{L^\infty} = \int |g| \mu \cdot inf\{B>0: \mu(\{x:| \int f(y) d\mu (y) |>B\})=0\}$