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!

  • $\begingroup$ I think my answer to your question is fairly conceptual; have you read it? $\endgroup$ Nov 18, 2011 at 2:11

2 Answers 2


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 $$ \begin{align} \left|\int_{\mathbb{R}^n}|f(x-y)g(y)|\mathrm{d}y\right|^p &=\left|G\int_{\mathbb{R}^n}\left|f(x-y)\right|\frac{|g(y)|}{G}\mathrm{d}y\right|^p\\ &\le G^p\int_{\mathbb{R}^n}\left|f(x-y)\right|^p\frac{|g(y)|}{G}\mathrm{d}y\\ &=\left(\int_{\mathbb{R}^n}\left|f(x-y)\right|^p|g(y)|\mathrm{d}y\right)\left(\int_{\mathbb{R}^n}|g(y)|\mathrm{d}y\right)^{p-1} \end{align} $$

  • $\begingroup$ +1 I just figured out the exponents. You beat me to it. =) $\endgroup$
    – Srivatsan
    Nov 17, 2011 at 2:49
  • 1
    $\begingroup$ Neat as usual when you do $p,q$-gymnastics :) $\endgroup$
    – t.b.
    Nov 17, 2011 at 2:51
  • 3
    $\begingroup$ Oops, I just noticed that I proved this for $\mathbb{R}^n$, not $\mathbb{R}^d$! ;-) $\endgroup$
    – robjohn
    Nov 17, 2011 at 3:21

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!

  • $\begingroup$ I was just looking back at my answer again, and I was thinking that we could use $g$ as a weight. Then looking at your answer, I saw that is just what you did (+1). $\endgroup$
    – robjohn
    Feb 4, 2012 at 21:08

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