$f*g$ denotes the convolution of $f$ and $g$.
Let $f,g \in L^1(\mathbb{R}^n)$. How can I proof that $||f*g||_1 \leq ||f||_1 \cdot ||g||_1$? (Where $||\,.||_1$ is the $L^1$-norm.)
I tried using the Hölder inequality like this:
$||f*g||_1 = \int |f(x-y)| \cdot |g(y)| \,\,dy \leq \int |f(x-y)| \,dy \cdot (\int |g(y)|^q\,dy)^{\frac{1}{q}}$
In the left integral I can simply use that the Lesbegue measure is a Haar measure, so also left-translation-invariant, and I get $||f||_1$.
But this doesn't work since it requires $\frac{1}{q}+ 1=1$ since I chose $p=1$, but such $q$ doesn't exist. Also I wouldn't know how to continue there, since $(\int |g(y)|^q\,dy)^{\frac{1}{q}}$ is not $||g||_1$ but $||g||_q$ and I'm not sure how to get from there to $||g||_1$.
\begin{align*} \|f\ast g\|_{1}&=\int|(f\ast g)(y)|dy\\ &=\int\left|\int f(y-x)g(x)dx\right|dy\\ &\leq\int\int|f(y-x)||g(x)|dxdy\\ &=\int\int|f(y-x)||g(x)|dydx\\ &=\int|g(x)|\int|f(y-x)|dydx\\ &=\int|g(x)|\int|f(y)|dydx\\ &=\int|g(x)|\|f\|_{1}dx\\ &=\|f\|_{1}\int|g(x)|dx\\ &=\|f\|_{1}\|g\|_{1}. \end{align*}