A proof of Holder’s inequality

by Orr Shalit

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.