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Let $p,q,r \in (1,\infty)$ with $1/p+1/q+1/r=1$. Prove that for every functions $f \in L^p(\mathbb{R})$, $g \in L^q(\mathbb{R})$,and $h \in L^r(\mathbb{R})$ $\int_{\mathbb{R}} |fgh|\leq \|f\|_p\centerdot \|g\|_q \centerdot\|h\|_r.$

I was going to use Hölder's inequality by letting $1/p+1/q= 1/(pq/p+q)$ and WLOG let $p so that $L_q(\mathbb{R})\subseteq L_p(\mathbb{R})$, but I cannot use this inclusion because $\mathbb{R}$ does not have finite measure.

Would you please help me if you have any other method to approach this problem?

2 Answers 2

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The rough idea is to show a series of inequalities: \int|fgh|\leq\|fg\|_{p'}\|h\|_r\leq\|f\|_p\|g\|_q\|h\|_r where p'=\frac{pq}{p+q} or \frac{1}{p'}=\frac{1}{p}+\frac{1}{q} or 1=\frac{1}{p/p'}+\frac{1}{q/p'}.

First we show that \|fg\|_{p'}\leq \|f\|_p\|g\|_q. This is easy since \|fg\|_{p'}=\left(\int|fg|^{p'}\right)^{\frac{1}{p'}}\leq(\|f^{p'}\|_{p/p'}\|g^{p'}\|_{q/p'})^{\frac{1}{p'}}=\|f\|_p\|g\|_q, where we apply the Holder's inequality (it is permissible since $|f|\in L^p(\mathbb{R})$, thus $|f|^{p'}\in L^{p/p'}(\mathbb{R})$). As a result, |fg|\in L^{p'}(\mathbb{R}). Apply Holder's inequality again, we get the first inequality in far above. Hope this will help you.

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    @SuryakantShrivastava Maybe you also need to note that the function $f(x)=x^{\frac{1}{p'}}$ is strictly increasing on $[0,\infty)$, and the integral of non-negative measurable function is also non-nagetive.2018-03-30
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We can use a generalized AM-GM inequality to deduce that if $1/p+1/q+1/r=1$, then

$abc\le\frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r}$

for nonnegative $a,b,c$. Let $a=|f(x)|/\|f\|_p,\,b=|g(x)|/\|g\|_q,\,c=|h(x)|/\|h\|_r$, and then integrate both sides of the inequality over $\mathbb{R}$ to obtain

$\frac{\|fgh\|_1}{\|f\|_p\|g\|_q\|h\|_r}\le\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1.$

Multiply out and you have Holder's inequality for three functions.

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    @Dragonite, by the very first inequality and then integrating, you can get the integrability of $fgh$.2018-04-20