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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 $p<q$ 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?

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2 Answers

up vote 16 down vote accepted

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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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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