7
$\begingroup$

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.)

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Gyu Eun Lee Dec 15 '12 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.