How to prove $\|f*g\|_1 \leq \|f\|_1 \cdot \|g\|_1$ [duplicate]
Solution 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*}