TITLE: Hölder inequality and interpolation
QUESTION [5 upvotes]: I was seeking the other day for intuition about the Hölder inequality and more precisely the geometric intuition behind the relation $\frac{1}{p} + \frac{1}{q} = 1$.
I found this interpretation of the Hölder inequality.
It says that we are trying to find an estimate of :
$$ \int f^a g^b$$ with the knowledge of $\int f^p$ and $\int g^q$.
For $(a, b) = (1,1)$, we need the relationship $\frac{1}{p} + \frac{1}{q} = 1$ so that the points : $(p, 0), (1,1), (0, q)$ lie on the same line. We can know use interpolation to have the Hölder inequality which give an estimate at the point $(1, 1)$.
But I feel like an argument is missing here. Why the estimate is using the geometric mean of the  function $f$ and $g$ ?  Why we don't have something like (using arithmetic mean) :
$$
\int \mid fg \mid \leq \frac{1}{p} \int f^p + \frac{1}{q}\int g^q \;\;?
$$
Moreover the article is saying we can generalize this approach to approximate $\int f^ag^b$ but in this case we need three points such that $(a,b)$ is in the convex hull of these three points. Why for $(1, 1)$ two points are needed while for general $(a, b)$ we need three points ?

REPLY [3 votes]: In my opinion, the correct answer to the question has been already given by John Dawkins: thus the following one is simply an illustration of his answer based on the proof of Hölder's inequality given in reference [1], §4.1, pp. 100-101, the best I've seen in print.
The proof given by Vol'pert and Hudjiaev starts by giving a geometric proof of Young's product inequality based on the following picture ([1], p. 100, Figure 1),

where
$$
\eta=\xi^{p-1} \iff \xi = \eta^{q-1}.
$$
As correctly stated by Dawkins, $\eta=\eta(\xi)$ and $\xi=\xi(\eta)$ are inverse functions and this is the main reason for which we have the relation $\frac{1}{p}+\frac{1}{q}=1$: indeed
$$
(u^{p-1})^{q-1}= (u^{q-1})^{p-1}= u \iff (p-1)(q-1)=1\iff \frac{1}{p}+\frac{1}{q}=1.
$$
The proof then goes as follows: calling respectively $A_1$ and $A_2$ the areas of the green and pink plane regions shown, we have
$$
\begin{split}
A_1 &=\int\limits_0^a \xi^{p-1} \mathrm{d}\xi = \frac{a^p}{p}\\
A_2 &=\int\limits_0^b \eta^{q-1} \mathrm{d}\eta = \frac{b^q}{q}\\
\end{split}\implies A_1+A_2 \ge ab \iff  \frac{a^p}{p} + \frac{b^q}{q} \ge ab \label{1}\tag{Y}
$$
since the rectangle whose sides are $[0, a]$ and $[0, b]$ is evidently contained in the union of the two shown plane regions.
Now, in order to estimate the product of the two functions $f(x)$ and $g(x)$ by using \eqref{1}, it is only needed a wise choice of $a$ and $b$:
$$
\begin{split}
a &= \frac{|f(x)|}{\|f\|_p}\\
b &= \frac{|g(x)|}{\|g\|_q}
\end{split}
$$
Then we have
$$
|f(x)g(x)|\le \frac{\|f\|_p^{1-p} \|g\|_q}{p} |f(x)|^p  + \frac{\|g\|_q^{1-q} \|f\|_p}{q} |g(x)|^q 
$$
and integrating both members of this inequality, one gets the sought for result.
Reference
[1] Aizik Isaakovich Vol'pert and Sergei Ivanovich Hudjaev (1985), Analysis in classes of discontinuous functions and equations of mathematical physics, Mechanics: analysis, 8, Dordrecht–Boston–Lancaster: Martinus Nijhoff Publishers, pp. xviii+678, ISBN 90-247-3109-7, MR 0785938, Zbl 0564.46025.