Theorem: Young's Inequality for Convolutions

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

Proof:
We begin by seeking to bound $\size {\map {\paren {f * g} } x}$:
{{begin-eqn}}
{{eqn | l = \map {\paren {f * g} } x 
      | r = \int \map f {x - y} \map g y \rd y
}}
{{eqn | ll= \leadsto
      | l = \size {\map {\paren {f * g} } x}
      | o = \le
      | r = \int \size {\map f {x - y} } \cdot \size {\map g y} \rd y
}}
{{eqn | r = \int \size {\map f {x - y} }^{1 + p/r - p/r} \cdot \size {\map g y}^{1 + q/r - q/r} \rd y
}}
{{eqn | r = \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
}}
{{eqn | r = \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
}}
{{eqn | o = \le
      | r = \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} } } }
}}  
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \frac 1 r + \frac {r - p} {p r} + \frac {r - q} {q r} 
      | r = \frac 1 r + \frac 1 p - \frac 1 r + \frac 1 q - \frac 1 r 
}}
{{eqn | r = \frac 1 p + \frac 1 q - \frac 1 r
}}
{{eqn | r = 1
      | c = {{hypothesis}} on $p$, $q$, $r$
}}
{{end-eqn}}
We now analyze terms $(1)$, $(2)$ and $(3)$ in turn:
{{begin-eqn}}
{{eqn | n = 1
      | l = \norm {\paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} }_r
      | r = \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} 
}}
{{eqn | r = \paren {\int \size {\map f {x - y} }^p \cdot \size {\map g y}^q \rd y}^{1/r}
}}
{{eqn | n = 2
      | l = \norm {\size {\map f {x - y} }^{\paren {r - p} /r } }_{\textstyle \frac {p r} {r - p} }
      | r = \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} } 
}}
{{eqn | r = \paren {\int \size {\map f {x - y} }^p \rd y}^{\textstyle\frac 1 p \times \frac {r - p} r}
}}
{{eqn | r = {\norm f_p}^{\paren {r - p} / r}
}}
{{eqn | n = 3
      | l = \norm {\size {\map g y}^{\paren {r - q} / r} }_{\textstyle\frac {q r} {r - q} }
      | r = \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} }
}}
{{eqn | r = \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} }
}}
{{eqn | r = \paren {\int \size {\map g y}^q \rd y}^{\textstyle \frac 1 q \times \frac {r - q} r}
}}
{{eqn | r = {\norm g_q}^{\paren {r - q} / r}
}}
{{end-eqn}}
Therefore we have
{{begin-eqn}}
{{eqn | l = \size {\map {\paren {f * g} } x}
      | o = \le
      | r = \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}
}}
{{end-eqn}}
With these preliminary calculations out of the way, we turn to the main proof:
{{begin-eqn}}
{{eqn | l = {\norm {f * g}_r}^r 
      | r = \int \size {\map {\paren {f * g} } x}^r \rd x
}}
{{eqn | o = \le
      | r = \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  
}}
{{eqn | r = \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
}}
{{eqn | r = {\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
}}
{{eqn | r = {\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 
      | c = Fubini's Theorem
}}
{{eqn | r = {\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
}}
{{eqn | r = {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \ {\norm g_q}^q \ {\norm f_p}^p
}}
{{eqn | r = {\norm f_p}^r \ {\norm g_q}^r
}}
{{eqn | ll= \leadsto
      | l = \norm {f * g}_r 
      | o = \le
      | r = \norm f_p \norm g_q
}}
{{end-eqn}}
{{qed}}
{{Namedfor|William Henry Young|cat = Young, W.H.}}
Category:Young's Inequality for Convolutions
Category:Measure Theory
Category:Inequalities
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