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 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 are said to be conjugate exponents if
.
If then we understand this to mean that
, and vice versa.
For any (finite or infinite) sequence , and and any
, we denote
.
Theorem (Holder’s inequality): Let be conjugate exponents.
Then for any two (finite or infinite) sequences and
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 and
in
,
(HI) .
The case (or
) is immediate. The right hand side of (HI) is continuous in
when
and
are held fixed, so it enough to verify the inequality for a dense set of values of
in
.
Define
satisfies (HI) for all
.
Now our task reduces to that of showing that is dense in
. By the Cauchy-Schwarz inequality, we know that
. Also, the roles of
and
are interchangeable, so
if and only if
.
Set (
is chosen to be the solution to
, we will use this soon). Now, if
, we apply (HI) to the sequences
and
, and then we use the Cauchy-Schwarz inequality, to obtain
where and
. Therefore, if
, then
; and if
, then
is also in
.
Since is known to be in
, it follows that
and
are also in
, and continuing by induction we see that for every
and
, the fraction
is in
. Hence
is dense in
, and the proof is complete.
The idea may be put thus:
In fact, H\”older’s inequality is equivalent to the convexity of the multivariate function:
Indeed, for two
-vectors, taken as
and
, we have
and
And we find that H\”older’s inequality for some fixed
is equivalent to the convexity inequality for 
for the particular value
.
And for a continuous function to be convex it suffices that the inequality holds (for all vectors
,
) for some particular
, say
.
Thanks, Eliahu! That’s an interesting comment.