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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!! )

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  • $\begingroup$ Could you double-check your definition of $r$ and $s$? There are inconsistencies as currently written. $\endgroup$ – Alex R. Mar 12 at 22:15
  • $\begingroup$ Sorry, there is a typo which says $$1/p + 1/q + 1/s = 1.$$ I correct it to be $$1/p + 1/q + 1/r = 1.$$ $\endgroup$ – user117375 Mar 12 at 22:21
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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 $$ \left( \int |fg|^s \right)^{1/s} \le \left( \int |f|^p \right)^{1/p} \left( \int |g|^q \right)^{1/q}.$$

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  • $\begingroup$ You mean $s \neq \infty$ ? $\endgroup$ – user117375 Mar 12 at 22:34
  • $\begingroup$ Argh. Yes I do. $\endgroup$ – Umberto P. Mar 13 at 0:26
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Holder's inequality in its original form sets no constraints on $f,g$, except that they are measurable, so:

$$\|fg\|_1\leq \|f\|_p\|g\|_q,$$

regardless of whether-or-not $fg\in L^1, f\in L^p, g\in L^q$. Recall that one way of proving Holder's inequality is through Young's inequality: $|fg|\leq \frac{|f|^p}{p}+\frac{|q|^q}{q}$, so that upon integration, the $l.h.s.$ is finite whenever the r.h.s. is finite. Conversely, if the l.h.s. is infinite, then at least one term on the r.h.s. is infinite.

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