How can we show $\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}$?

Question

Let $p\ge1$, $f\in L^p(\mathbb R^d)$ and $g\in L^1(\mathbb R^d)$. I've read that the inequality $$\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}\tag1$$ would follow from applying the Minkowski inequality to $$F(x,y):=g(y)f(x-y)$$ and using that $$\left\|\lambda_yf\right\|_{L^p}=\left\|f\right\|_{L^p}\tag2,$$ where $(\lambda_yf)(x):=f(x-y)$.

However, I absolutely don't get why $(1)$ does hold. How can we show it

The "usual" version of Young's inequality, as stated on Wikipdia, is clear to me.

Answer 1

$\|\int f(x-y)g(y)dy\|_p \leq \int \|f(x-y)g(y)\|_p dy$ where the norm inside the integral is w.r.t. the variable $x$. Since $g(y)$ is a constant we get $\|\int f(x-y)g(y)dy\|_p \leq \int |g(y)| \|f(x-y)\|_p dy=\|f\|_p \int |g(y)|dy=\|f\|_p \|g\|_1$.