Hölder inequality and interpolation

Question

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 ?

Answer 1

But you do have something like that: Young's inequality says that for $u,v\ge 0$ and conjugate exponents $p$ and $q$, $$ uv\le p^{-1}u^p+q^{-1}v^q. $$ (This can be proved with a picture once you notice that $u\mapsto u^{p-1}$ and $v\mapsto v^{q-1}$ (the derivatives of $u\mapsto p^{-1}u^p$ and $v\mapsto q^{-1}v^q$) are inverse functions.) Take $u=|f(x)|$ and $v = |g(x)|$ and integrate.

Answer 2

At first we look at the geometrical situation of the convex hull of three points in general position in $\mathbb{R^2}$.

Convex hull of three points with $(a,b)$ inside:

In OPs referenced article we have a triangle with points $(u_j,v_j),1\leq j\leq 3$ and a point $(a,b)$ inside the triangle as shown in the graphic below:

$\qquad\qquad\quad$

We can write $(a,b)$ as convex linear combination of $(u_j,v_j)$ according the graphic as follows: \begin{align*} \binom{a^{\prime}}{b^{\prime}}&=\binom{u_2}{v_2}+\lambda\left[\binom{u_3}{v_3}-\binom{u_2}{v_2}\right]\\ &=(1-\lambda)\binom{u_2}{v_2}+\lambda\binom{u_3}{v_3}\\ \\ \color{blue}{\binom{a}{b}}&=\binom{u_1}{v_1}+\mu\left[\binom{a^{\prime}}{b^{\prime}}-\binom{u_1}{v_1}\right]\\ &=(1-\mu)\binom{u_1}{v_1}+\mu\binom{a^{\prime}}{b^{\prime}}\\ &=(1-\mu)\binom{u_1}{v_1}+\mu\left[(1-\lambda)\binom{u_2}{v_2}+\lambda\binom{u_3}{v_3}\right]\\ &\,\,\color{blue}{=(1-\mu)\binom{u_1}{v_1}+\mu(1-\lambda)\binom{u_2}{v_2}+\lambda\mu\binom{u_3}{v_3}}\tag{1} \end{align*}

Note we have in (1) a convex combination of the points $(u_j,v_j)$ since \begin{align*} (1-\mu)+\mu(1-\lambda)+\lambda\mu=1 \end{align*}

Convex hull of three points with $(a,b)=(1,1)$ at the boundary:

Now we consider the triangle with the three points $(0,0), (p,0), (0,q)$, $p>1, q>1$ and the point $(a,b)=(1,1)$ at the line segment from $(p,0)$ to $(0,q)$.

$\qquad\qquad\qquad$

We calculate similarly as above: \begin{align*} \color{blue}{\binom{1}{1}}&=\binom{p}{0}+\nu\left[\binom{0}{q}-\binom{p}{0}\right]\\ &\,\,\color{blue}{=(1-\nu)\binom{p}{0}+\nu\binom{0}{q}} \end{align*}

We observe in this special case where $(a,b)=(1,1)$ is at the boundary of the convex hull, we need only two points $(p,0)$ and $(0,q)$ to obtain a convex combination of the triangle for $(1,1)$.

Hölder's inequality:

A few words to Hölder's inequality which might be helpful. We have for positive real numbers $x$ and $y$ and $p>1, q>1$ real numbers with $\frac{1}{p}+\frac{1}{q}=1$ the arithmetic-geometric means inequality: \begin{align*} x^{\frac{1}{p}}y^{\frac{1}{q}}\leq \frac{1}{p}x+\frac{1}{q}y\tag{2} \end{align*} We define for a $p$-integrable measurable function $f$: \begin{align*} N_p(f):=\left(\int|f|^p\,d\mu\right)^{\frac{1}{p}} \end{align*} and take \begin{align*} x:=\left(\frac{|f(\omega)|}{N_p(f)}\right)^p,\quad y:=\left(\frac{|g(\omega)|}{N_q(g)}\right)^q\tag{3} \end{align*}

We obtain from (2) and (3): \begin{align*} \frac{|f\,g|}{N_p(f)N_q(g)}&\leq \frac{1}{p\left(N_p(f)\right)^p}|f|^p+\frac{1}{q\left(N_q(g)\right)^q}|g|^q\tag{4}\\ \frac{1}{N_p(f)N_q(g)}\int|f\,g|\,d\mu &\leq \frac{1}{p\left(N_p(f)\right)^p}\int|f|^p\,d\mu+\frac{1}{q\left(N_q(g)\right)^q}\int|g|^q\,d\mu\\ &=\frac{1}{p}+\frac{1}{q}\\ &=1\\ \color{blue}{N_1(fg)}&\color{blue}{\leq N_p(f)N_q(g) }\tag{5} \end{align*} and (5) gives Hölder's inequality.

Note the transformation from the arithmetic mean in (4) to the multiplication of $N_p(f)$ with $N_q(g)$ in (5).

Answer 3

Focusing more on intuition, and landing on the geometry how integrals scale, let us check out a simple case.

Consider $\int_{0}^{r} 1 dx = r$. Applying Holder's inequality we would get on the opposite side $\left( \int_{0}^{r} 1 dx \right)^{\frac{1}{p}} \left( \int_{0}^{r} 1 dx \right)^{\frac{1}{q}}$. So, what can we say to relate $r$ to $r^{\frac{1}{p} + \frac{1}{q}}$?

Case 1: ($\frac{1}{p} + \frac{1}{q} = 1$)

Evidently if $\frac{1}{p} +\frac{1}{q} =1$, then we have the equality that $r = r^{\frac{1}{p} + \frac{1}{q}}$

Case 2: ($\frac{1}{p} + \frac{1}{q} < 1$)

If $\frac{1}{p} + \frac{1}{q} < 1$ then for $r < 1$ we have $r < r^{\frac{1}{p} +\frac{1}{q}}$ and for $r > 1$ we have $r > r^{\frac{1}{p} + \frac{1}{q}}$.

Case 3: ($\frac{1}{p} + \frac{1}{q} > 1$)

In this case, similar to Case 2, you can quickly rule at an inequality that holds for all values of $r$.

Conclusion: If an inequality like Holder's is true, the only possible option is when $\frac{1}{p} + \frac{1}{q}=1$.

Answer 4

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.