Young's Inequality for Convolutions

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Let $p, q, r \in \R_{\ge 1}$ satisfy:

$1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$

Let $L^p \left({\R^n}\right)$, $L^q \left({\R^n}\right)$, and $L^r \left({\R^n}\right)$ be Lebesgue spaces with seminorms $\left\Vert{\cdot}\right\Vert_p$, $\left\Vert{\cdot}\right\Vert_q$, and $\left\Vert{\cdot}\right\Vert_r$ respectively.

Let $f \in L^p \left({\R^n}\right)$ and $g \in L^q \left({\R^n}\right)$.

Then the convolution $f * g$ is in $L^r \left({\R^n}\right)$ and the following inequality is satisfied:

$\left\Vert{f * g}\right\Vert_r \le \left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q$

Corollary 1

Let $f: \R^n \to \R$ be a Lebesgue integrable function.

Let $p \in \R_{\ge 1}$.

Let $g: \R^n \to R$ be a Lebesgue $p$-integrable function.

Let $\left\Vert{f}\right\Vert_p$ denote the $p$-seminorm of $f$.

Then the convolution $f * g$ of $f$ and $g$ satisfies:

$\left\Vert{f * g}\right\Vert_p \le \left\Vert{f}\right\Vert_1 \cdot \left\Vert{g}\right\Vert_p$

and hence is also Lebesgue $p$-integrable.

Corollary 2

Let $f, g: \R^n \to \R$ be Lebesgue integrable functions.

Then their convolution $f * g$ is also Lebesgue integrable, and:

$\left\Vert{f * g}\right\Vert \le \left\Vert{f}\right\Vert \, \left\Vert{g}\right\Vert$


We begin by seeking to bound $\left\vert{\left({f * g}\right) \left({x}\right)}\right\vert$:

\(\displaystyle \left({f * g}\right) \left({x}\right)\) \(=\) \(\displaystyle \int f \left({x - y}\right) g \left({y}\right) \ \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \left\vert{ \left({f * g}\right) \left({x}\right) }\right\vert\) \(\le\) \(\displaystyle \int \left\vert{ f \left({x - y}\right)}\right\vert \cdot \left\vert{g \left({y}\right)}\right\vert \ \mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int \left\vert{ f \left({x - y}\right) }\right\vert^{1 + p/r - p/r} \cdot \left\vert{ g \left({y}\right) }\right\vert^{1 + q/r - q/r} ~\mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int \left\vert{ f \left({x - y}\right)}\right\vert ^{p/r} \cdot \left\vert{g \left({y}\right)}\right\vert ^{q/r} \cdot \left\vert{ f \left({x - y}\right)}\right\vert ^{1-p/r} \cdot \left\vert{g \left({y}\right)}\right\vert ^{1-q/r} ~\mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int \left({\left\vert{ f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right)^{1/r} \cdot \left\vert{ f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} \cdot \left\vert{ g \left({y}\right)}\right\vert^{ (r-q)/r} ~\mathrm d y\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \underset{I} {\underbrace{ \left\Vert{ \left({ \left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q }\right)^{1/r} }\right\Vert_r} } \cdot \underset{II}{\underbrace{\left\Vert{\left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} }\right\Vert_{\textstyle \frac{pr}{r-p} } } } \cdot \underset{III}{\underbrace{\left\Vert{\left\vert{ g \left({y}\right)}\right\vert^{(r-q)/r} }\right\Vert_{\textstyle \frac{qr}{r-q} } } }\) $\quad$ $\quad$

where the last inequality is via the Generalized Hölder Inequality applied to three functions.

Note that the relation of conjugate exponents in the Generalized Hölder Inequality is satisfied:

\(\displaystyle \frac 1 r + \frac {r-p} {p r} + \frac {r-q} {q r}\) \(=\) \(\displaystyle \frac 1 r + \frac 1 p - \frac 1 r + \frac 1 q - \frac 1 r\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 p + \frac 1 q - \frac 1 r\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ by hypothesis on $p$, $q$, $r$ $\quad$

We now analyze terms $I$, $II$, and $III$ individually:

\((I):\quad\) \(\displaystyle \left\Vert{ \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert ^q}\right)^{1/r} }\right\Vert_r\) \(=\) \(\displaystyle \left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert ^q}\right)^{\textstyle \frac 1 r \times r} \ \mathrm d y}\right)^{1/r}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int \left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q \ \mathrm d y}\right)^{1/r}\) $\quad$ $\quad$
\((II):\quad\) \(\displaystyle \left\Vert{\left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} }\right\Vert_{\textstyle \frac {p r} {r - p} }\) \(=\) \(\displaystyle \left({\int \left\vert{f \left({x - y}\right)}\right\vert^{\textstyle \frac {r - p} r \times \frac {p r} {r - p} } \ \mathrm d y}\right)^{\textstyle\frac {r - p} {p r} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int \left\vert{f \left({x - y}\right)}\right\vert^p \ \mathrm d y}\right)^{\textstyle\frac 1 p \times \frac {r-p} r}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{f}\right\Vert_p}^{\left({r - p}\right) /r}\) $\quad$ $\quad$
\((III):\quad\) \(\displaystyle \left\Vert{\left\vert{g \left({y}\right)}\right\vert^{(r-q)/r} }\right\Vert_{\textstyle\frac{qr}{r-q} }\) \(=\) \(\displaystyle \left({\int \left\vert{g \left({y}\right)}\right\vert^{\textstyle \frac{r-q} r \times \frac{q r}{r-q} } \ \mathrm d y}\right)^{\textstyle \frac{r-q}{qr} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int \left\vert{g \left({y}\right)}\right\vert^{\textstyle \frac{r-q} r \times \frac{q r}{r-q} } \ \mathrm d y}\right)^{\textstyle \frac{r-q}{qr} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\int \left\vert{g \left({y}\right)}\right\vert^q \ \mathrm d y}\right)^{\textstyle \frac 1 q \times \frac{r-q} r}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{g}\right\Vert_q}^{(r-q)/r}\) $\quad$ $\quad$

With these preliminary calculations out of the way, we turn to the main proof:

\(\displaystyle {\left\Vert{f * g}\right\Vert_r}^r\) \(=\) \(\displaystyle \int \vert \left({f * g}\right) \left({x}\right)\vert ^r ~\mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \int \left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right)^{1/r} \cdot \left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} \cdot \left\vert{g \left({y}\right)}\right\vert^{(r-q)/r} \ \mathrm d y}\right)^r \mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int \left({\left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right) \ \mathrm d y}\right)^{1/r} \cdot {\left\Vert{f}\right\Vert_p}^{\left({r - p}\right) /r} \cdot {\left\Vert{g}\right\Vert_q}^{(r-q)/r} }\right)^r \ \mathrm d x\) $\quad$ See $I$, $II$, $III$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int \left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right) }\right\vert^q}\right) \ \mathrm d y}\right) \cdot {\left\Vert{f}\right\Vert_p}^{r-p} \cdot {\left\Vert{g}\right\Vert_q}^{r-q} \ \mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{f}\right\Vert_p}^{r-p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \iint \left\vert{g \left({y}\right)}\right\vert^q \left\vert{f \left({x - y}\right)}\right\vert^p \ \mathrm d y \ \mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{f}\right\Vert_p}^{r-p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \int \vert g \left({y}\right)\vert^q \left({\int \vert f \left({x - y}\right)\vert^p \ \mathrm d x}\right) \ \mathrm d y\) $\quad$ Fubini's Theorem $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{f}\right\Vert_p}^{r-p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \int \left\vert{ g \left({y}\right) }\right\vert^q \ \mathrm d y \int \left\vert{f \left({x}\right)}\right\vert^p \ \mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{f}\right\Vert_p}^{r - p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \ {\left\Vert{g}\right\Vert_q}^q \ {\left\Vert{f}\right\Vert_p}^p\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\left\Vert{f}\right\Vert_p}^r \ {\left\Vert{g}\right\Vert_q}^r\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left\Vert{f * g}\right\Vert_r\) \(\le\) \(\displaystyle \left\Vert{f}\right\Vert_p \ \left\Vert{g}\right\Vert_q\) $\quad$ $\quad$


Source of Name

This entry was named for William Henry Young.