There are questions that concerns me when I read the following proof regarding the generalized Holder inequality :
Let $U$ be a subset of $\mathbb{R}$. Let $1 < p, q, r < \infty$ with $p^{-1} + q^{-1} + r^{-1} = 1$. Let $f \in L^p(U), g \in L^q(U)$ and $h \in L^r(U)$. Then $$||fgh||_1 \leq ||f||_p||g||_q||h||_r.$$
Assume that we have the original version of Holder inequality : Let $1 < p,q < \infty$. For $f \in L^p(U)$ and $g \in L^q(U)$, $$||fg||_1 \leq ||f||_p||g||_q.$$
$\textbf{Proof}$ Let $s = (1/p + 1/q)^{-1}.$ Then $1/s + 1/r = 1.$ Then apply the original Holder inequlity gives $$\int_U (fg)h dx \leq ||h||_r (\int_U (fg)^s)^{1/s}.$$ Then apply Holder again to $(fg)^s$ to get the result.
$\textbf{Question}$ My confusion is that when $s$ is set. The next step is to apply the original Holder inequality to $(fg)$ and $h$. Clearly, $h \in L^r$. But how do we know that $fg \in L^s$ ? Is it trivial to see that $fg \in L^s$ ?? I try to verify this, but not quite successful.
(Note if $fg$ is NOT in $L^s$, then its integrate is $\infty$. How can $\infty \leq ||f||_p||g||_q$ which suppose to be a finite number!! )
You can verify this using Holder's inequality: if $1 \le p,q,s < \infty$ and $\dfrac 1p + \dfrac 1q = \dfrac 1s$, then $f \in L^p$ and $g \in L^q$ implies $fg \in L^s$.
The result is still true in the case either $p = \infty$ or $q = \infty$ but the proof is slightly different from what follows.
As long as $s < \infty$ you have $\dfrac sp + \dfrac sq = 1$, so that a routine application of Holder's inequality gives you $$ \int |fg|^s = \int |f|^s |g|^s \le \left( \int (|f|^s)^{p/s} \right)^{s/p} \left( \int (|g|^s)^{q/s} \right)^{s/q}$$ which easily rearranges to $$ \left( \int |fg|^s \right)^{1/s} \le \left( \int |f|^p \right)^{1/p} \left( \int |g|^q \right)^{1/q}.$$