# 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 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 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 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 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 , §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 (, 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

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