Generalization of Hölder inequality On this wiki page, one can see that 
$$\|fg\|_r\leq \|f\|_p\|g\|_q,$$
when
$$\frac{1}{r} = \frac{1}{p} + \frac{1}{q}, \ p,q\in(0,\infty].$$
I have two questions,


*

*Is the statement true?

*If so, is it stated in some references? I need it for my thesis.

 A: $\newcommand{\eq}{=}\newcommand{\plus}{+}$
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.
A: Hint: Consider $p' = \frac{r}{p}$, $q' = \frac{r}{q}$, $f' = f^r$, $g' = g^r$.
A: 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
$$
\|fg\|_r\le\|f\|_p\|g\|_q
$$
A: For $ p,q\in (0,1)$ I don't believe that it is correct, since the conjugate $q$ of $p$ would not fit.
However, for $p, q \in [1, \infty)$ I made a proof, I hope it is helpful to you.
We know that:
a) Let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite, complete measure space.
\begin{equation*}
    L^P(x,\mu)=\{f:X\rightarrow \mathbb{C}\hspace{0.1cm} \text{ measurable} \mid \int \limits_X \lvert f \lvert^P d\mu < +\infty\} ,\hspace{0.3cm}1\leq P < \infty
\end{equation*}
\begin{gather*}
\text{and also} \hspace{0.1cm} {\Vert{f}\Vert}_{P} =(\int \limits_X \lvert f \lvert^P d\mu)^{1/P}.  
\end{gather*}
b) Let $1\leq p \leq \infty$ and $q$ its conjugate, i.e., such that it satisfies $\dfrac{1}{p} +\dfrac{1}{q}=1$. 
If $f \in L^p (\mu)$ and $g \in L^q (\mu)$, then $ f.g \in L^1(\mu)$ and it holds that:
\begin{equation*}
\int \limits_X \lvert f.g \lvert d\mu  \leqslant \Vert f \Vert_p \Vert g\Vert_q \text{ (Holder's Inequality).}   
\end{equation*}
Now:
Since $\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{r}$ $\Rightarrow$ $\dfrac{1}{p/r}+\dfrac{1}{q/r}=1$ $\Rightarrow$ $\dfrac{q}{r}$ is the conjugate of $\dfrac{p}{r}$    ...............($\alpha$)
Now let's prove that $f^r \in L^{p/r}$ and that $g^r \in L^{q/r}$.
With effect:
$\int \limits_X \lvert f^r \lvert^{p/r} d\mu=\int \limits_X \lvert f \lvert^{r(\dfrac{p}{r})} d\mu=\int \limits_X \lvert f \lvert^{p} d\mu< \infty $ , is convergent since $f\in L^p$. So $f^r \in L^{p/r}$ .....($\phi$) 
$\int \limits_X \lvert g^r \lvert^{q/r} d\mu=\int \limits_X \lvert g \lvert^{r(\dfrac{q}{r})} d\mu=\int \limits_X \lvert g \lvert^{q} d\mu< \infty $ , is convergent since $g\in L^q$. So $g^r \in L^{q/r}$....($\gamma$)
Knowing ($\alpha$), ($\phi$) and ($\gamma$), we can apply Holder's inequality.
Applying it, we have:
\begin{equation*}
\int \limits_X \lvert f^r.g^r \lvert d\mu  \leqslant (\int \limits_X \lvert f^r \lvert^{p/r} d\mu)^{r/p}.(\int \limits_X \lvert g^r \lvert^{q/r} d\mu)^{r/q} , \\
\end{equation*}
\begin{equation*}
\int \limits_X \lvert f.g \lvert^r d\mu=\int \limits_X \lvert f^r.g^r \lvert d\mu \leqslant (\int \limits_X \lvert f^r \lvert^{p/r} d\mu)^{r/p}.(\int \limits_X \lvert g^r \lvert^{q/r} d\mu)^{r/q}= (\int \limits_X \lvert f \lvert^{r.\dfrac{p}{r}} d\mu)^{r/p}.(\int \limits_X \lvert g \lvert^{r.\dfrac{q}{r}} d\mu)^{r/q},
\end{equation*}
\begin{equation*}
\int \limits_X \lvert f.g \lvert^r d\mu \leqslant (\int \limits_X \lvert f \lvert^{p} d\mu)^{r/p}.(\int \limits_X \lvert g \lvert^{q} d\mu)^{r/q}.
\end{equation*}
Raising both sides to $1/r$, we have:
\begin{equation*}
    (\int \limits_X \lvert f.g \lvert^r d\mu)^{1/r} \leqslant \left[ (\int \limits_X \lvert f \lvert^{p} d\mu)^{r/p}.(\int \limits_X \lvert g \lvert^{q} d\mu)^{r/q}\right] ^ {1/r},
\end{equation*}
\begin{equation*}
(\int \limits_X \lvert f.g \lvert^r d\mu)^{1/r} \leqslant \left(\int \limits_X\lvert f \lvert^{p} d\mu \right)^{{\dfrac{r}{p}}{\dfrac{1}{r}}}  \left(\int \limits_X\lvert g \lvert^{q} d\mu \right)^{{\dfrac{r}{q}}{\dfrac{1}{r}}},
\end{equation*}
\begin{equation*}
     \lVert{fg}\lVert_{r}=(\int \limits_X \lvert f.g \lvert^r d\mu)^{1/r} \leqslant \left(\int \limits_X \lvert f \lvert^{p} d\mu \right)^{1/p}.\left(\int \limits_X \lvert g \lvert^{q} d\mu \right)^{1/q}=\lVert{f}\lVert_{p}.\lVert{g}\lVert_{q},
\end{equation*}
\begin{equation*}
    \lVert fg \lVert_{r} \leqslant \lVert f \lVert_{p}.\lVert g\lVert_{q}.
\end{equation*}
