Inequality for the p norm of a convolution Let $1\le p < \infty$. Let $g \in L^p(\Bbb R^n)$, and $f \in L^1(\Bbb R^n)$. 
Show: $||f*g||_p \le ||f||_1||g||_p$, where $*$ indicates the convolution defined by 
$$ f*g (x) := \int f(x-t) g(t)dt.$$
Here is a relevant hint, apparently:
Let $q$ be the conjugate of $p$. Note that:
$$|f*g| \le \int [|g(t)||f(x-t)|^{1/p}]|f(x-t)|^{1/q}dt,$$
then apply Holder's inequality.
I did exactly as the hint suggested, and ended up with what looked like a meaningless inequality compared to the final answer. I even took the $p$ power of both sides. 
Any help?
 A: You have - don't forget the $x$ -
$$\lvert (f\ast g)(x)\rvert \leqslant \int \bigl[\lvert g(t)\rvert\cdot\lvert f(x-t)\rvert^{1/p}\bigr]\cdot \lvert f(x-t)\rvert^{1/q}\,dt.$$
Applying Hölder's inequality to that yields
$$\begin{align}
\lvert (f\ast g)(x)\rvert &\leqslant \left(\int \lvert g(t)\rvert^p \lvert f(x-t)\rvert\,dt\right)^{1/p}\cdot \left(\int \lvert f(x-t)\rvert\,dt\right)^{1/q}\\
&= \lVert f\rVert_1^{1/q}\cdot\left(\int \lvert g(t)\rvert^p \lvert f(x-t)\rvert\,dt\right)^{1/p}.
\end{align}$$
Raising to the $p$-th power,
$$\lvert (f\ast g)(x)\rvert^p \leqslant \lVert f\rVert_1^{p/q}\cdot \int \lvert g(t)\rvert^p\lvert f(x-t)\rvert\,dt,$$
and integrating with respect to $x$:
$$\begin{align}
\int \lvert (f\ast g)(x)\rvert^p\,dx &\leqslant \lVert f\rVert_1^{p/q}\cdot \iint \lvert g(t)\rvert^p\lvert f(x-t)\rvert\,dt\,dx\\
&=  \lVert f\rVert_1^{p/q}\cdot \iint \lvert g(t)\rvert^p\lvert f(x-t)\rvert\,dx\,dt\tag{Fubini}\\
&= \lVert f\rVert_1^{p/q} \lVert f\rVert_1\lVert g\rVert_p^p.
\end{align}$$
Taking the $p$-th root then yields
$$\lVert f\ast g\rVert_p \leqslant \lVert f\rVert_1^{1/q+1/p}\cdot \lVert g\rVert_p,$$
which, since $\frac1q + \frac1p = 1$, is the desired result.
