Young's Inequality for Convolutions

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Theorem

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

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

Let $\map {L^p} {\R^n}$, $\map {L^q} {\R^n}$, and $\map {L^r} {\R^n}$ be Lebesgue spaces with seminorms $\norm {\, \cdot \,}_p$, $\norm {\, \cdot \,}_q$, and $\norm {\, \cdot \,}_r$ respectively.

Let $f \in \map {L^p} {\R^n}$ and $g \in \map {L^q} {\R^n}$.

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

$\norm {f * g}_r \le \norm f_p \cdot \norm g_q$

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 $\norm f_p$ denote the $p$-seminorm of $f$.

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

$\norm {f * g}_p \le \norm f_1 \cdot \norm g_p$

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:

$\norm {f * g} \le \norm f \norm g$

Proof

We begin by seeking to bound $\size {\map {\paren {f * g} } x}$:

\(\ds \map {\paren {f * g} } x\)

\(=\)

\(\ds \int \map f {x - y} \map g y \rd y\)

\(\ds \leadsto \ \ \)

\(\ds \size {\map {\paren {f * g} } x}\)

\(\le\)

\(\ds \int \size {\map f {x - y} } \cdot \size {\map g y} \rd y\)

\(\ds \)

\(=\)

\(\ds \int \size {\map f {x - y} }^{1 + p/r - p/r} \cdot \size {\map g y}^{1 + q/r - q/r} \rd y\)

\(\ds \)

\(=\)

\(\ds \int \size {\map f {x - y} }^{p/r} \cdot \size {\map g y}^{q/r} \cdot \size {\map f {x - y} }^{1 - p / r} \cdot \size {\map g y}^{1-q/r} \rd y\)

\(\ds \)

\(=\)

\(\ds \int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} \cdot \size {\map f {x - y} }^{\paren {r - p} / r} \cdot \size {\map g y}^{\paren {r - q} / r} \rd y\)

\(\ds \)

\(\le\)

\(\ds \underset {(1)} {\underbrace {\norm {\paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} }_r} } \cdot \underset {(2)} {\underbrace {\norm {\size {\map f {x - y} }^{\paren {r - p} / r} }_{\textstyle \frac {p r} {r - p} } } } \cdot \underset {(3)} {\underbrace {\norm {\size {\map g y}^{\paren {r - q} / r} }_{\textstyle \frac {q r} {r - q} } } }\)

where the last inequality is via the Generalized Hölder's Inequality for Integrals applied to three functions.

Note that the relation of conjugate exponents in the Generalized Hölder's Inequality for Integrals is satisfied:

\(\ds \frac 1 r + \frac {r - p} {p r} + \frac {r - q} {q r}\)

\(=\)

\(\ds \frac 1 r + \frac 1 p - \frac 1 r + \frac 1 q - \frac 1 r\)

\(\ds \)

\(=\)

\(\ds \frac 1 p + \frac 1 q - \frac 1 r\)

\(\ds \)

\(=\)

\(\ds 1\)

by hypothesis on $p$, $q$, $r$

We now analyze terms $(1)$, $(2)$ and $(3)$ in turn:

\(\text {(1)}: \quad\)

\(\ds \norm {\paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} }_r\)

\(=\)

\(\ds \paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{\textstyle\frac 1 r \times r} \rd y}^{1/r}\)

\(\ds \)

\(=\)

\(\ds \paren {\int \size {\map f {x - y} }^p \cdot \size {\map g y}^q \rd y}^{1/r}\)

\(\text {(2)}: \quad\)

\(\ds \norm {\size {\map f {x - y} }^{\paren {r - p} /r } }_{\textstyle \frac {p r} {r - p} }\)

\(=\)

\(\ds \paren {\int \size {\map f {x - y} }^{\textstyle \frac {r - p} r \times \frac {p r} {r - p} } \rd y}^{\textstyle\frac {r - p} {p r} }\)

\(\ds \)

\(=\)

\(\ds \paren {\int \size {\map f {x - y} }^p \rd y}^{\textstyle\frac 1 p \times \frac {r - p} r}\)

\(\ds \)

\(=\)

\(\ds {\norm f_p}^{\paren {r - p} / r}\)

\(\text {(3)}: \quad\)

\(\ds \norm {\size {\map g y}^{\paren {r - q} / r} }_{\textstyle\frac {q r} {r - q} }\)

\(=\)

\(\ds \paren {\int \size {\map g y}^{\textstyle \frac {r - q} r \times \frac {q r} {r - q} } \rd y}^{\textstyle \frac {r - q} {q r} }\)

\(\ds \)

\(=\)

\(\ds \paren {\int \size {\map g y}^{\textstyle \frac {r - q} r \times \frac {q r} {r - q} } \rd y}^{\textstyle \frac {r - q} {q r} }\)

\(\ds \)

\(=\)

\(\ds \paren {\int \size {\map g y}^q \rd y}^{\textstyle \frac 1 q \times \frac {r - q} r}\)

\(\ds \)

\(=\)

\(\ds {\norm g_q}^{\paren {r - q} / r}\)

Therefore we have

\(\ds \size {\map {\paren {f * g} } x}\)

\(\le\)

\(\ds \paren {\int \size {\map f {x - y} }^p \cdot \size {\map g y}^q \rd y}^{1/r} {\norm f_p}^{\paren {r - p} / r} {\norm g_q}^{\paren {r - q} / r}\)

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

\(\ds {\norm {f * g}_r}^r\)

\(=\)

\(\ds \int \size {\map {\paren {f * g} } x}^r \rd x\)

\(\ds \)

\(\le\)

\(\ds \int \paren {\paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q} \rd y}^{1/r} \cdot {\norm f_p}^{\paren {r - p} / r} \cdot {\norm g_q}^{\paren {r - q} / r} }^r \rd x\)

\(\ds \)

\(=\)

\(\ds \int \paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q} \rd y} \cdot {\norm f_p}^{r - p} \cdot {\norm g_q}^{r - q} \rd x\)

\(\ds \)

\(=\)

\(\ds {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \iint \size {\map g y}^q \size {\map f {x - y} }^p \rd y \rd x\)

\(\ds \)

\(=\)

\(\ds {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \int \size {\map g y}^q \paren {\int \size {\map f {x - y} }^p \rd x} \rd y\)

Fubini's Theorem

\(\ds \)

\(=\)

\(\ds {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \int \size {\map g y}^q \rd y \int \size {\map f x}^p \rd x\)

\(\ds \)

\(=\)

\(\ds {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \ {\norm g_q}^q \ {\norm f_p}^p\)

\(\ds \)

\(=\)

\(\ds {\norm f_p}^r \ {\norm g_q}^r\)

\(\ds \leadsto \ \ \)

\(\ds \norm {f * g}_r\)

\(\le\)

\(\ds \norm f_p \norm g_q\)

$\blacksquare$

Source of Name

This entry was named for William Henry Young.