# Understanding generalized Holder inequality proof

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!! )

• Could you double-check your definition of $r$ and $s$? There are inconsistencies as currently written. – Alex R. Mar 12 at 22:15
• Sorry, there is a typo which says $$1/p + 1/q + 1/s = 1.$$ I correct it to be $$1/p + 1/q + 1/r = 1.$$ – user117375 Mar 12 at 22:21

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 $$\left( \int |fg|^s \right)^{1/s} \le \left( \int |f|^p \right)^{1/p} \left( \int |g|^q \right)^{1/q}.$$

• You mean $s \neq \infty$ ? – user117375 Mar 12 at 22:34
• Argh. Yes I do. – Umberto P. Mar 13 at 0:26

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.