Generalization of Hölder inequality

Solution 1:

$\newcommand{\eq}{=}\newcommand{\plus}{+}$ It is an exercise in Zygmund and Wheeden Measure an Integral to prove the following

If $$\sum_{i=1}^k\frac1{p_i}=\frac1r,$$ then $$\|f_1\cdots f_k\|_r\leq \|f_1\|_{p_1}\cdots \|f_k\|_{p_k}.$$

There it's asked to assume that $p_i,r\geq 1$ though it's not necessary as the argument which follows will show.

Hereafter: $$[n]=\{1,\ldots,n\}.$$

First notice that

$$0\leq \frac1{p_j}\leq \sum_{i=1}^k\frac1{p_i}=\frac1r,$$

therefore

$$p_j\geq r\qquad \forall j\in [k].\tag{1}$$

We go by induction on $k$.

If $k=2$ we have

$$\begin{align*} \frac1{p_1}\plus\frac1{p_2} &\eq \frac1r\\ \frac1{p_1/r}\plus \frac1{p_2/r} &\eq 1, \end{align*}$$

by $(1)$, $p_1/r\geq 1$, so by Hölder's inequality, we have

$$\int \left|f_1\right|^r \left| f_2\right|^r\leq \left\| \left|f_1\right|^r\right\|_{p_1/r} \left\| \left|f_2\right|^r\right\|_{p_2/r}\eq \left(\int \left|f_1\right|^{p_1}\right)^{r/p_1}\left(\int \left|f_2\right|^{p_2}\right)^{r/p_2}.$$

Given that the map $t\mapsto t^{1/r}$ is increasing in $[0,\infty[$, we conclude

$$\left\| f_1f_2\right\|_r\leq \left\|f_1\right\|_{p_1} \left\|f_2\right\|_{p_2}.$$

Now suppose that the inequality holds for $k$. Let $p_i,r\in ]0,\infty]$ such that

$$\sum_{i\eq 1} ^{k\plus 1} \frac1{p_i} \eq \frac1r.$$

Let

$$\frac1{r'}\eq \sum_{i\eq 1} ^{k} \frac1{p_i}.$$

Thus

$$\begin{align*} \frac1r &\eq \frac1{r'} \plus \frac1{p_{k\plus1}}\\ 1 &\eq \frac1{r'/r}\plus \frac1{p_{k\plus 1}/r}. \end{align*}$$

Then, we have $r'/r\geq 1$. By what we have already done $$\begin{align*} \int \left|f_1\cdots f_k\right|^r \left| f_{k\plus 1}\right|^r &\leq \left\| \left| f_1\cdots f_k \right|^r \right\|_{r'/r} \left\| \left\| f_{k\plus 1} \right|^r \right\|_{p_{k\plus 1}/r} \\ &\eq \left( \int \left| f_1\cdots f_k \right|^{r'} \right)^{r/r'} \left( \int \left| f_{k\plus 1} \right|^{p_{k\plus 1}} \right)^{r/p_{k\plus 1}} \\ &\eq \left\| f_1\cdots f_k \right\|_{r'}^r \left\| f_{k\plus 1} \right\|_{p_{k\plus 1}}^r. \tag{2} \end{align*}$$

Using the induction hypothesis in $(2)$ we get

$$ \int \left| f_1\cdots f_{k\plus 1} \right|^r \leq \left( \left\| f_1 \right\|_{p_1} \cdots \left\| f_{k} \right\|_{p_{k}} \right)^r \left\| f_{k\plus 1} \right\|_{p_{k\plus 1}}^r. \tag{3}$$

Raising $(3)$ to $1/r$ we obtain the desired inequality.

Solution 2:

Hint: Consider $p' = \frac{r}{p}$, $q' = \frac{r}{q}$, $f' = f^r$, $g' = g^r$.

Solution 3:

It is a fairly common inequality.

Suppose that $\frac1p+\frac1q=\frac1r$, then $|f|^r\in L^{p/r}$ and $|g|^r\in L^{q/r}$ and $\frac{r}{p}+\frac{r}{q}=1$, so we can use the standard Hölder inequality to get $$ \begin{align} \int |f(x)g(x)|^r\,\mathrm{d}x &\le\left(\int\left(|f(x)|^r\right)^{p/r}\,\mathrm{d}x\right)^{r/p} \left(\int\left(|g(x)|^r\right)^{q/r}\,\mathrm{d}x\right)^{r/q} \end{align} $$ Raising to the $1/r$ power yields $$ \|fg\|_r\le\|f\|_p\|g\|_q $$