I was doing some problems from Rudin's Principles of Mathematical Analysis and came across a problem in which he asks you to prove Hölder's inequality via Young's inequality:

If $u$ and $v$ are nonnegative real numbers, and $p$ and $q$ are positive real numbers such that $\displaystyle \frac{1}{p}+\frac{1}{q}=1$, then $\displaystyle uv \leq \frac{1}{p}u^p+\frac{1}{q}v^q$.

I'm familiar with the proof using convexity of the $\log$ function and Jensen's inequality, but Rudin hasn't defined the $\log$ function by chapter $6$ (where this problem originates) and hasn't done anything with convexity. Usually he gives everything necessary for a problem before he poses one, so this seems to be something of an omission. Perhaps he wants us to read Chapter $8$ to learn about $\log$ and prove Jensen's inequality before attacking this problem? But then why put it in Chapter $6$?

My question: is there a proof of Young's inequality that does not use convexity of $\log$ or something similar? If one exists, can it be done using only the material from chapters $1-6$ of Principles?

(For clarity, chapter 1-6 essentially cover the real number system, metric space topology, sequences and series, continuity, differentiability, and the Riemann-Stieltjes integral.)


1 Answer 1


First note that we have $$ab \leq \int_0^{a} f(x) dx + \int_0^{b} f^{-1}(x) dx $$ for any strictly increasing integrable function $f(x)$. The geometric interpretation is from looking at the area of the rectangle with coordinates $(0,0)$,$(a,0)$,$(a,b)$ and $(0,b)$ and comparing it with the areas given by the integrals. From the image it is also clear that the equality hold only when $b=f(a)$.

To get the Young's inequality, choose $f(x) = x^{p-1}$.

I have added the following picture for clarity. enter image description here

The image was made using grapher and some post processing was done using LaTeXiT and preview on Mac OSX.

  • $\begingroup$ Ah. Now that I think about it, I remember doing this proof in Spivak's Calculus...and it was called Young's inequality there! Thanks! $\endgroup$ Dec 15, 2012 at 6:32
  • $\begingroup$ How does it work if $f(0)\ne0$? If $f(0)>0$ then $f^{-1}$ is negative near 0, and the integral on the right contains negative terms that does not seems to be straightforward to consider in your drawings. If $f(0)<0$ the same problem occurs with some parts of the left integral that is negative. $\endgroup$ Jan 14 at 5:09

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