Is there a geometric intuition for Hölder's inequality?

I am referring to $||fg||_1 \le ||f||_p ||f||_q $, when $\frac{1}{p}+\frac{1}{q}=1$.

For $p=q=2$ this is just the Cauchy-Schwarz inequality, for which I have geometric intution: The projection of a vector along a direction shortens its length.

My question is if there is a similar geometric interpretation for Hölder's inequality. I am aware of the scaling argument, which shows the inequality can only hold when $\frac{1}{p}+\frac{1}{q}=1$; but why should we expect this to be true when the $p,q$ are conjugates? Perhaps there is some physical interpretation?

Note that I am looking for intuition, not necessarily a formal proof. Hölder's inequality can be proved using Young's inequality, for which a beautiful intuition is given here.

In my perspective, even though this gives intuition to a component in the proof of Hölder's inequality, this does not really give an intuition for the inequality itself.

(But maybe I am wrong? does the actual integration have "true content" in it, or is Hölder really nothing but Young's inequality in disguise? Part of the confusion is that the intuition for Young's inequality is based on integration, so if we only rely on that, the intuition for Hölder should be some sort of "double integration"... )

To see that the geometric intuition of Young's and Hölder's inequalities are somewhat different, we can look at $p=q=2$:

In that case, Young's inequality is just the standard AM-GM inequality for two variables. This inequality can be interpreted geometrically. Although here one can also view this as "projection only shortens", the scenario is a little different than the one in the Cauchy-Schwarz inequality. (At least the the reasons behind the "equality cases" seem slightly different to me).


1 Answer 1


(Before getting to intuition: you've mention the proof by Young's inequality, but there is another nice proof that is a direct application of Jensen's inequality. Assume $\|f\|_p=\|g\|_q=1$ and $f,g>0.$ Then $(\int fg)^p=[\int g^q(fg^{1-q})]^p\leq \int g^q(fg^{1-q})^p=\int f^p=1.$ The inequality comes from the convexity of $x^p$ and probability measure $d\mu=g^qdx.$)

In any Banach space $V$ there is an inequality $|\langle x,f\rangle|\leq \|x\|_V \|f\|_{V^*}.$ This is almost a triviality, but it is a reflection of the geometrical fact that unit balls are convex. Vectors in the dual space $V^*$ represent hyperplanes that support the unit ball in $V.$

From the viewpoint of the geometry of Banach spaces, the intuition behind Hölder's inequality is that the dual space of $L^p$ has an integral representation as $L^q.$ If you take a point $f_0$ on the $L^p$ unit sphere, the supporting hyperplane is $\{f\mid\int fg=1\}$ where $g=\overline{f_0}|f_0|^{p-2},$ or simply $g=f_0^{p-1}$ with the assumption $f_0>0.$ For $\|f\|_p=1$ we get $\int fg\leq 1,$ which is Hölder's inequality.

Even if you had no idea what the dual space of $L^p$ might be, it would be natural to derive the convexity inequality $f^p\geq f_0^p+pf_0^{p-1}(f-f_0)$ - this is the equation for the supporting hyperplane at $x=f_0$ on the curve $x\mapsto x^p.$ Integrating gives $\int f^p\geq \int f_0^p+p\int f_0^{p-1}(f-f_0),$ which is a type of supporting hyperplane equation in $L^p.$ Rearranging and setting $f_0=g^{1/(p-1)}$ gives $\int f^p+(p-1)\int g^q\geq p\int fg,$ which is Hölder's inequality, after scaling to get $\|f\|_p=\|g\|_q=1.$ This argument is just Young's inequality in disguise but I hope the appearance of $q$ seems quite natural here.

For all these kinds of inequalities it is worth reading https://terrytao.wordpress.com/2007/09/05/amplification-arbitrage-and-the-tensor-power-trick/ if you haven't already.

  • 1
    $\begingroup$ I am sorry, but I do not understand something: what happens in the inequality step in $(\int fg)^p=[\int g^q(fg_{1-q})]^p\leq \int g^q(fg_{1-q})^p=\int f^p=1$? I guess $g_{1-q}$ is supposed to be $g^{1-q}$. The inequality is supposed to be $[\int g^q(fg_{1-q})]^p\leq \int g^{pq}(fg^{1-q})^p$, no? $\endgroup$ Mar 18, 2018 at 13:22
  • $\begingroup$ @AsafShachar: Jensen's inequality for the probability measure $g^qdx$ gives $(\int g^q(\dots))^p\leq \int g^q(\dots)^p.$ (There's a more careful version at the bottom of the section en.wikipedia.org/wiki/… or in Lieb and Loss "Analysis") $\endgroup$
    – Dap
    Mar 18, 2018 at 14:59
  • 1
    $\begingroup$ I don't understand " but it is a reflection of the geometrical fact that unit balls are convex. " ? How are the two related ? I think it is due to the way $||f||_{V^*}$ is defined and the linearity of $f$ and the norm on $V$. $\endgroup$ Feb 12, 2021 at 7:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .