Two Analysis Problems

Recently I heard two problems in analysis, both I think are interesting, and they do not require too deep knowledge in analysis, which is the kind of questions I like most. Share here.

1) Suppose converges. Also, for each positive integer k, it is known that (just to avoid confusion, allow me clarify here that jk means "j times k"). Prove that for all positive integers i.

2) S contains all elements such that for any , there exists a rational number (where p,q are positive integers) satisfying . Prove that S is uncountable.

A beautiful solution

Suppose you want to prove that the sum of 1/(m^2 + n^2) diverges as m,n ranges over all positive integers. What do you do?

Here's a very beautiful solution, from one of my students in an undergraduate complex analysis class:

Every prime of the form 4k+1 is expressible as the sum of two squares. Hence the previous sum is bounded below by the sum of 1/p, where p ranges over all primes that are congruent to 1 mod 4. The latter sum diverges. Q.E.D.

AM-GM 不等式的幾個證明（六）

看見 Kahoo 給了 AM-GM 不等式這麼多個精彩的證明，不禁想再來一個用微分和凸函数 (convex functions) 的。

這裏我們只需留意對任意實數 x 和 a，



這不等式可以用很多不同的方法來證明，其中要想清楚當中的概念的，比較好的方法，是留意 是一凸函數，故其導數是遞增的，用中值定理即可推出上述不等式。（也可以不失一般性假設，從而用一元微積分或 的 Taylor 展開求得此不等式。）

有了這不等式以後，對正數 ，



這裏 a 可以是任何實數。把上列的不等式平均起來，得到

------(1)

取 ，則



從而 AM-GM 不等式得證。事實上不等式 (1) 的右邊是一關於 a 的函數，左邊則跟 a 無關，所以我們會取的 a 使得 (1) 的右邊達至最大值，而這正是我們上面這樣取 a 的原因。

這證明的一個好處，是我們可以用同樣的辦法證明以下這個 AM-GM 不等式的推廣：對任意 ，和任意正數 使得 ，有



從上面的討論可知，其實 AM-GM 不等式是函數 的 convexity 的反映。對任何的凸函數 f，我們都有 Jensen 不等式，詳見 wikipedia。

Differentiability and uniform convergence

Today I came across a problem in elementary analysis. It is quite interesting, so I thought I'd share it here.

The problem is whether the following statement is true:

If a sequence of differentiable functions converges pointwisely to a function f and if the derivatives converges uniformly to a function g (say all convergence are on some open subset of , for the conclusions are local anyway), then the limit f is differentiable and .

When n = 1 this is a standard theorem in mathematical analysis (see e.g. Rudin's Principle of Mathematical Analysis). However, I have not seen this explicitly stated for higher dimensions in any book, and if one just copies say the proof in Rudin's book line-by-line, one only concludes that the directional derivatives of f exist in every possible direction everywhere. Since in higher dimensions this is not quite sufficient to conclude that f is differentiable, I struggled for a while on whether the result is true as stated. (By the way, the result is trivial if one assumes all are continuously differentiable rather than just differentiable.)

My friends and I finally managed to give a positive answer to the above question, and I thought I'd present the proof here. (The proof is just a simple variant of say the proof in Rudin.)

Suppose is a sequence of differentiable functions in that converges pointwisely to a function f and suppose their derivatives converges uniformly to a function g. Then to argue differentiability at , consider the sequence of functions



It is easy to verify that is a sequence of continuous functions of x, and that they are Cauchy in . But by assumption converges pointwisely to



Since the uniform limit of a sequence of continuous functions is continuous, F is continuous at , so f is differentiable at and . This completes the proof.

Analytic function on a disk

Suppose f is a holomorphic function on the unit disk that is continuous up to the boundary. If f vanishes on an arc of the boundary circle, show that it is identically zero.

This is a standard question from complex analysis. I just learn today that there is a really cute solution to it. You may want to think about it before continuing.

The classical method is to use Schwartz reflection principle and argue that you can analytically continue the function outside the disk a little bit; then the continued function vanishes on a segment, so it must vanish identically.

The cute solution that I was referring to is the following: take (finitely many) copies of f, rotate each of them suitably and multiply the rotated functions altogether. Then the product is going to vanish identically on the boundary of the unit disk, and of course the product is holomorphic inside the disk. Hence the product is identically zero, and thus f is identically zero. (Just argue that all derivatives of f vanishes at the origin by differentiating the product.)

也談一談重排

剛剛看了johnmayhk寫的關於重排（Part 1、Part 2、Part 3）的文章，碰巧最近看probability theory時也遇到一個關於重排的小問題，在這裏說說。

在probability theory裏有一個結果是這樣的（這些還是用英文寫比較好…… ）：

If X, Y are independent random variables and discrete,



Range of X是countable的。若重排後計算出來的並不一樣，在數學上是不容許的。

要解決這個問題，可以用Dirichlet Rearrangement Theorem：

If and converges absolutely, then every rearrangement converges to the same sum.

因為是absolutely convergent的（怎樣證明？這個作為大學的分析課考試的一條基本題還挺不錯的），所以不論怎樣把它重排，它的極限值都是唯一的。


P.S.1：（這裏不想解釋independent, random variable, discrete, converges absolutely, rearrangement等用語；有興趣者自行查wiki吧！）

P.S.2：最後中國在奧運奪得 51金。