# Hölder inequality and interpolation

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 ?

• The geometric mean is $\le$ the arithmetic mean, so what you suggest is a weaker inequality. Jan 18, 2021 at 14:01
• @MartinR I am aware of that but I am just wondering why it should be true with the arithmetic mean. Why the interpolation is using the geometric mean ? Jan 18, 2021 at 14:06
• 1 "but I am just wondering why it should be true with the arithmetic mean"? I cannot understand this sentence. can you try to say it in different words? 2 "We can know use interpolation to have the Hölder inequality" if this is the interpolation I am thinking of, then it is equivalent to Hölder, so I don't know what you are doing? 3 If you downvoted the answer below, can you explain why? That answer mirrors what was my first instinct Jan 24, 2021 at 6:55

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).

• Magnificent answer, really! My only point was, maybe you could add some of the applications of the fact that this inequality is multiplicative : for example, I believe the multiplicative nature of the inequality shows that multiplication by $f \in L^p$ is a well defined operator from $L^q$ to $L^1$ Jan 25, 2021 at 6:13

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.

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.

• I'm curious about the downvote. Jan 24, 2021 at 16:11

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 $$\frac{1}{p} + \frac{1}{q}=1$$.