## Category: Fun stuff

### A proof of Holder’s inequality

One of the parts of this blog that I am most proud of is my series of “Souvenirs” post, where I report about my favorite new finds in conferences. In July I went to a big conference (IWOTA 2016 in St. Louis) that I was looking forward to going to for a long time, but I did not write anything after I returned. It’s not that there was nothing to report – there was a lot, it was great conference. I was just too busy with other things.

Why am I so busy? Besides being the father of seven people (most of them “kids”) and preparing for next year, I am in the last stages of writing a book, partly based on the lecture notes on “Advanced Analysis” that appeared in this blog, and on lecture notes that evolved from that. (When it will be ready I will tell you about it, you can be sure). I want to share here and now one small excerpt from it (thanks to Guy Salomon for helping me finesse it!)

Working on the final touches to the book, I decided to include a proof of Holder’s inequality in it, but I did not want to copy a proof from somewhere.  So I came up with the following proof, which I think is new (and out of curiosity I am asking you to please tell me if you have seen it before). The lazy idea of the proof is to use the fact that we already know – thanks to Cauchy-Schwarz – that the inequality holds in the $p =2$ case, and to try to show how the general case follows from that.

In other words, instead of bringing you fancy souvenirs from St. Louis, I got you this little snack from the nearby mall (really, the proof crystallized in my head when my daughter, my dog and I were sitting and waiting on a bench in the mall until other members of our family finish shopping).

Definition. Two extended real numbers $p,q \in [1, \infty]$ are said to be conjugate exponents if $\frac{1}{p} + \frac{1}{q} = 1$.

If $p=1$ then we understand this to mean that $q = \infty$, and vice versa.

For any (finite or infinite) sequence $x_1, x_2, x_3, \ldots$, and and any $p \in [1,\infty]$, we denote $\|x\|_p =\big(\sum |x_k|^p \big)^{1/p}$.

Theorem (Holder’s inequality): Let $p,q \in [1, \infty]$ be conjugate exponents.
Then for any two (finite or infinite) sequences $x_1, x_2, \ldots$ and $y_1, y_2, \ldots$ $\sum_k |x_k y_k| \leq \|x\|_p \|y\|_q.$

Proof. The heart of the matter is to prove the inequality for finite sequences. Pushing the result to infinite sequences does not require any clever idea, and is left to the reader (no offense).
Therefore, we need to prove that for every $x = (x_k)_{k=1}^n$ and $y = (y_k)_{k=1}^n$ in $\mathbb{C}^n$,

(HI) $\sum |x_ky_k| \leq \big(\sum |x_k|^p \big)^{1/p} \big( \sum |y_k|^q \big)^{1/q}$.

The case $p=1$ (or $p=\infty$) is immediate. The right hand side of (HI) is continuous in $p$ when $x$ and $y$ are held fixed, so it enough to verify the inequality for a dense set of values of $p$ in $(1,\infty)$.

Define $S = \Big\{\frac{1}{p} \in (0,1) \Big| p$ satisfies  (HI)  for all $x,y \in \mathbb{C}^n \Big\}$.

Now our task reduces to that of showing that $S$ is dense in $(0,1)$. By the Cauchy-Schwarz inequality, we know that $\frac{1}{2} \in S$. Also, the roles of $p$ and $q$ are interchangeable, so $\frac{1}{p} \in S$ if and only if $1 - \frac{1}{p} \in S$.

Set $a = \frac{q}{2p+q}$ ( $a$ is chosen to be the solution to $2ap = (1-a)q$, we will use this soon). Now, if $\frac{1}{p} \in S$, we apply (HI) to the sequences $(|x_k| |y_k|^{a})_k$ and $(|y_k|^{1-a})_k$, and then we use the Cauchy-Schwarz inequality, to obtain $\sum |x_k y_k| = \sum|x_k||y_k|^a |y_k|^{1-a}$ $\leq \Big(\sum |x_k|^p |y_k|^{ap} \Big)^{1/p}\Big(\sum |y_k|^{(1-a)q} \Big)^{1/q}$ $\leq \Big((\sum |x_k|^{2p})^{1/2} (\sum |y_k|^{2ap})^{1/2} \Big)^{1/p}\Big(\sum |y_k|^{(1-a)q} \Big)^{1/q}$ $= \Big(\sum |x_k|^{p'} \Big)^{1/p'} \Big(\sum|y_k|^{q'} \Big)^{1/q'}$

where $\frac{1}{p'} = \frac{1}{2p}$ and $\frac{1}{q'} = \frac{1}{2p} + \frac{1}{q}$. Therefore, if $s = \frac{1}{p} \in S$, then $\frac{s}{2} = \frac{1}{2p} \in S$; and if $s = \frac{1}{q} \in S$, then $\frac{s+1}{2} = \frac{1}{2}\frac{1}{q}+\frac{1}{2} = \frac{1}{q} + \frac{1}{2}\frac{1}{p}$ is also in $S$.

Since $\frac{1}{2}$ is known to be in $S$, it follows that $\frac{1}{4}$ and $\frac{3}{4}$ are also in $S$, and continuing by induction we see that for every $n \in \mathbb{N}$ and $m \in \{1,2, \ldots, 2^n-1\}$, the fraction $\frac{m}{2^n}$ is in $S$. Hence $S$ is dense in $(0,1)$, and the proof is complete.

### Where have all the functional equations gone (the end of the story and the lessons I’ve learned)

This will be the last of this series of posts on my love affair with functional equations (here are links to parts one, two and three).

#### 1. A simple solution of the functional equation

In the previous posts, I told of how I came to know of the functional equations

(*) $f(t) = f\left(\frac{t+1}{2}\right) + f \left( \frac{t-1}{2}\right) \,\, , \,\, t \in [-1,1]$

and more generally

(**) $f(t) = f(\delta_1(t)) + f(\delta_2(t)) \,\, , \,\, t \in [-1,1]$

(where $\delta_1$ and $\delta_2$ satisfy some additional conditions) and my long journey to discover that these equations have, and now I will give it away… Read the rest of this entry »

### Where have all the functional equations gone (part II)

I’ll start off exactly where I stopped in the previous post: I will tell you my solution to the problem my PDEs lecturer (and later master’s thesis advisor) Paneah gave us:

Problem: Find all continuously differentiable solutions to the following functional equation:

(FE) $f(t) = f\left(\frac{t+1}{2} \right) + f \left(\frac{t-1}{2} \right) \,\, , \,\, t \in [-1,1] .$

Before writing a solution, let me say that I think it is a fun exercise for undergraduate students, and only calculus is required for solving it, so if you want to try it now is your chance.

### A sneaky proof of the maximum modulus principle

The April 2013 issue of the American Mathematical Monthly has just appeared, and with it my small note “A Sneaky Proof of the Maximum Modulus Principle”. Here is a link to the current issue on the journal’s website, and here is a link to a version of the paper on my homepage. As the title suggest, the note contains a new proof — which I find extremely cool — for the maximum modulus principle from the theory of complex variables. The cool part is that the proof is based on some basic linear algebra. The note is short and very easy, and I am not going to say anything more about the proof, except that it relates to some of my “real” research (the way in which it relates can be understood by reading the Note and its references).

I am writing this post not only to publicize this note, but also to record somewhere my explanation why I have been behaving in a sneaky fashion. Indeed, this is the first paper that I wrote which I did not post on the Arxiv. Why?

Unlike research journals, the American Mathematical Monthly is a journal which has, if I am not mistaken, actual subscribers. I mean real people, some of them perhaps old school (like myself), and I could see them waiting to receive their copy in the mailbox, and then when the new issue finally arrives they gently open the envelope — or perhaps they tear it open, depending on their custom — after which they sit down and browse through the fresh issue. I could believe that there are such persons (for I myself am such a person) that do not look at the online version of the journal even though they have access, because that would spoil their fun with the paper copy which is to arrive a few days later.

Now I wouldn’t like to spoil a small pleasure of a subscriber, somewhere out there. So I did not post the Note on the Arxiv, lest it pop up on somebody’s mailing list. “Oh, this I have already seen…”. I shall not be resposible for such spoilers! So I decided to keep my note relatively secret, putting it on my homepage, but putting off the Arxiv until the journal really gets published and all the physical copies are safely in the mailboxes of all subscribers. I made this decision about a year ago from now, and to tell the truth I felt that a year is a terribly long time to wait. In the end, this year appears much much shorter from this end than from the other one.

(I guess that it does not matter much if I put it on the Arxiv now: in the meanwhile I discovered that google scholar has managed to figure out that such a note exists on somebody’s webpage. Probably I will post it on the Arxiv, for the sake of all things being in good order).