[**Update**: *Thanks to Andreas for pointing out that I may have been a little sloppy in stating the *__maximum modulus principle__! The version below is an updated and correct one. Also, Andreas pointed out an excellent post on "amplification, arbitrage and the tensor power trick" (by Terry Tao) in which the "tricks" discussed are indeed very useful and far more powerful generalizations of the "method" of E. Landau discussed in this post. *The Landau method mentioned here, it seems, is just one of the many examples of the "tensor power trick".*]

The *maximum modulus principle* states that if (where ) is a holomorphic function, then attains its maximal value on any compact on the boundary of . (If attains its maximal value anywhere in the interior of , then is a constant. However, we will not bother about this part of the theorem in this post.)

*Problems and Theorems in Analysis II*, by Polya and Szego, provides a short proof of the “inequality part” of the principle. The proof by E. Landau employs Cauchy’s integral formula, and the technique is very interesting and useful indeed. The proof is as follows.

From Cauchy’s integral formula, we have

,

for every in the interior of .

Now, suppose on . Then,

,

where the constant depends only on the curve and on the position of , and is independent of the specific choices of . Now, this rough estimate can be significantly improved by applying the same argument to , where , to obtain

, or .

By allowing to go to infinity, we get , which is what we set out to prove.

Polya/Szego mention that the proof shows that *a rough estimate may sometimes be transformed into a sharper estimate by making appropriate use of the generality for which the original estimate is valid.*

I will follow this up with, maybe, a couple of problems/solutions to demonstrate the effectiveness of this useful technique.

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March 10, 2008 at 8:23 pm

Andreas D.Dear Vishal,

Polya/Szegö’s comment is quite interesting and Terence Tao had a wonderful article in September 2007 on what he colourfully calls “arbitrage” and which extends that comment ( I would be curious to know if Terence knew about theirs remark).Here is the link:

http://terrytao.wordpress.com/2007/09/05/amplification-arbitrage-and-the-tensor-power-trick/

You should be a little more careful in your statement of the theorem: if f is defined only on D, you are not allowed to speak about the values of |f| on the boundary of D. You might want to assume that f can be continuously extended to the closure of D . You should also suppose closure(D) compact , so that |f| is guaranteed to have a maximum.

This is an interesting post and your choice of

Polya/Szegö confirms your excellent mathematical taste.

Cheers, A.D.

March 11, 2008 at 8:16 am

VishalDear Andreas,

First, thank you very much for providing the link to that excellent post by Dr Tao. Indeed, the Landau method seems to me an example of the powerful “tensor power trick.” That post will keep me busy for a long time as it contains a wealth of useful techniques for establishing stronger estimates from weaker ones.

Second, indeed, I was sloppy in stating the principle when I first wrote the post. I have corrected the same now.

Lastly, there is a certain kind of joy in reading mathematical books (or literature, in general) by top-class mathematicians who not only know their trade well but also know how to communicate their ideas in a way that almost reads like a fantastic novel! The analogy I have provided is crude but I am sure you know what I mean.

March 22, 2008 at 7:27 pm

JosephI’m sorry, but I don’t understand your proof. Is K the lenght of the curve L?

Also, I understand how to use the tensor powers trick you describe, but how can you conclude that f obtains its maximum on the boundry of D? I’m just unclear in those details.

Best,

Joseph

September 19, 2013 at 10:00 pm

maths14Hello Joseph. You can simply think that K is an upper-bound of the length of the curve L. Also Vishal indicated that he just showed the inequality part of the maximal modulus principle.

March 24, 2008 at 3:28 am

Jennyebi am gonna show this to my friend, guy