Hölder Inequality Application I am trying to follow a step in a proof which uses the Hölder Inequality.
The setting is:  $f\in L^{p}(\mathbb{R}^{d})$, $g\in L^{1}(\mathbb{R}^{d})$, with $1\leq p\leq \infty$.
The claim is that:  By the Hölder Inequality, for $1\leq p < \infty$,
$$\left|\int_{\mathbb{R}^{d}}|f(x-y)g(y)|dy\right|^{p} \leq \left(\int_{\mathbb{R}^{d}}|f(x-y)|^{p}|g(y)|dy\right)\left(\int_{\mathbb{R}^{d}}|g(y)|dy\right)^{p-1}$$
But I am having a hard time seeing which Hölder Conjugates were used.
Is the Minkowski Inequality also used here?  Thank you!
 A: Note that when $\frac{1}{p}+\frac{1}{q}=1$, $q=\frac{p}{p-1}$. Therefore,
$$
\small\begin{align}
\left|\int_{\mathbb{R}^n}|f(x-y)g(y)|\mathrm{d}y\right|^p
&\le\left(\int_{\mathbb{R}^n}\left(|f(x-y)||g(y)|^{1/p}\right)\left(|g(y)|^{1-1/p}\right)\mathrm{d}y\right)^p\\
&\le\left(\left(\int_{\mathbb{R}^n}\left(|f(x-y)||g(y)|^{1/p}\right)^p\mathrm{d}y\right)^{1/p}\left(\int_{\mathbb{R}^n}\left(|g(y)|^{1-1/p}\right)^{\frac{p}{p-1}}\mathrm{d}y\right)^{\frac{p-1}{p}}\right)^p\\
&=\left(\int_{\mathbb{R}^n}|f(x-y)|^p|g(y)|\mathrm{d}y\right)\left(\int_{\mathbb{R}^n}|g(y)|\mathrm{d}y\right)^{p-1}
\end{align}
$$

Another approach:
Let $G=\int_{\mathbb{R}^n}|g(y)|\mathrm{d}y$. Then Jensen's inequality yields
$$
\begin{align}
\left|\int_{\mathbb{R}^n}|f(x-y)g(y)|\mathrm{d}y\right|^p
&=\left|G\int_{\mathbb{R}^n}\left|f(x-y)\right|\frac{|g(y)|}{G}\mathrm{d}y\right|^p\\
&\le G^p\int_{\mathbb{R}^n}\left|f(x-y)\right|^p\frac{|g(y)|}{G}\mathrm{d}y\\
&=\left(\int_{\mathbb{R}^n}\left|f(x-y)\right|^p|g(y)|\mathrm{d}y\right)\left(\int_{\mathbb{R}^n}|g(y)|\mathrm{d}y\right)^{p-1}
\end{align}
$$
A: I think a conceptual manner of understanding this situation is to define a positive measure $\mu$ on $\mathbb{R}^n$ by the rule $d\mu = \left|g\right|dy$, where $dy$ is Lebesgue measure, i.e., by the rule $\mu(E)=\int_{E} \left|g(y)\right| dy$, where $E$ is a Lebesgue measurable subset of $\mathbb{R}^n$. (In the language of measure theory, $\left|g\right|$ is the Radon-Nikodym derivative $\frac{d\mu}{dy}$.)
If $x\in\mathbb{R}^n$ is fixed, then let us apply Hölder's inequality to the functions $y\to \left|f(x-y)\right|$ and $y\to 1$ on $\mathbb{R}^n$ with respect to the measure $\mu$ and with the exponents $p$ and $p'$:
$$
\begin{eqnarray*}

\int_{\mathbb{R}^n} \left|f(x-y)g(y)\right| dy

&=& \int_{\mathbb{R}^n} \left|f(x-y)\right|d\mu(y)

\\ &\leq&
\left(\int_{\mathbb{R}^n} \left|f(x-y)\right|^{p}d\mu(y)\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}^n} 1^{p'}d\mu(y)\right)^{\frac{1}{p'}}


\\ &=& 
\left(\int_{\mathbb{R}^n} \left|f(x-y)\right|^{p}\left|g(y)\right|dy\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}^n}\left|g(y)\right|dy\right)^{\frac{1}{p'}}.
\end{eqnarray*}
$$
You should justify the equalities in the computation above (the inequality is simply a special case of Hölder's inequality). If we take the $p$th power of both sides of the inequality in the above computation, then we deduce the desired inequality.
The technique that I have used to answer your question is an important one; e.g., it is used in at least one proof of Young's inequality on convolutions (if I remember correctly).
I hope this helps!
