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Let $p , q , r$ be three real numbers in $[1 , + \infty]$ such that $\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$. If $f \in L^p({\mathbb{R}}^n)$, $g \in L^q({\mathbb{R}}^n)$ and $r < \infty$, I have to prove that $$ {|(f*g)(x)|}^r \leq {\|f\|}_p^{r - p} {\|g\|}_q^{r - q} \int_{{\mathbb{R}}^n} {|f(y)|}^p {|g(x - y)|}^q dy $$ and my indication is to use Hölder inequality for three functions; my intuition sais that $f$ and $g$ are two of that functions but I have to obtain the third. Can you help me with this problem please? Thank you very much.

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Let us define $$\alpha = \left(\frac{1}{p} - \frac{1}{r}\right)^{-1}, \quad \beta = \left(\frac{1}{q} - \frac{1}{r}\right)^{-1}.$$ Notice that they are such that $$ \frac{1}{\alpha} + \frac{1}{\beta} + \frac 1 r = 1. $$ We have: $$\begin{align} |(f*g)(x)| &\leq \int_{\mathbb R^n} |f(y)|\,|g(x-y)|\,dy \\ &= \int_{\mathbb R^n} \left(|f(y)|^{1-\frac{p}{r}}\right) \, \left(|g(x-y)|^{1 - \frac{q}{r}}\right)\, \left(|f(y)|^{p} \, |g(x-y)|^{q}\right)^{\frac{1}{r}}\,dy \\ &= \int_{\mathbb R^n} \left(|f(y)|^{\frac{p}{\alpha}}\right) \, \left(|g(x-y)|^{\frac{q}{\beta}}\right)\, \left(|f(y)|^{p} \, |g(x-y)|^{q}\right)^{\frac{1}{r}}\,dy \end{align}$$ Now apply apply Holder's inequality with exponents $\alpha, \beta, r$ to obtain: $$\begin{align} |(f*g)(x)| &\leq \|f\|_p^{\frac{p}{\alpha}} \, \|g\|_q^{\frac{q}{\beta}} \, \left(\int_{\mathbb R^n} |f(y)|^{p} \, |g(x-y)|^{q}\,dy\right)^{\frac{1}{r}}. \end{align}$$ from which the statement follows by raising both sides to the power $r$ and simplifying.

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    I can't see why $$ \frac{1}{\alpha} + \frac{1}{\beta} + \frac 1 r = 1. $$2017-01-27
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    Sorry, I can already see the equality.2017-01-27
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    But I think that if $p > r$ or $q > r$ there is a problem; for example, if $p > r$, then $\frac{1}{p} < \frac{1}{r}$, so $\alpha < 0$.2017-01-27
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    In fact, $\alpha$ is always positive, because $1/p>1/r$. Indeed, $1/p = 1/r + (1-1/q)$ and $(1 - 1/q) > 0$.2017-01-27
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    Sorry again, it can't happen because $$ \frac{1}{p} - \frac{1}{r} = 1 - \frac{1}{q} $$ and $$ 1 - \frac{1}{q} \geq 0 $$ because $q \in [1 , + \infty]$.2017-01-27
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    I understand everything. Good answer and thank you very much. Like +1.2017-01-27
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    Thanks, and you're welcome :)2017-01-27