Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 23 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.

share|cite|improve this answer

We can use a generalized AM-GM inequality to deduce that if $1/p+1/q+1/r=1$, then


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


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.