## 2009年2月27日 星期五

### 圓周率日活動

3 月 14 日是圓周率日，也是數學資料庫的生日！如此大日子，我們安排了一連串的活動，大家萬勿錯過！
• 當天上午有數學資料庫協辦的培正數學邀請賽決賽（詳見比賽網頁）。
• 當天晚上是數學資料庫的週年晚宴暨生日會，將有一系列的抽獎和遊戲活動（詳情快將在此公佈）。

From "What is Mathematical Research" to "A Characterization Problem in Graph Theory"

As a graduate student in Mathematics, the most common question I have ever been asked is "What are there to be researched in Mathematics?". In this seminar, I will give the audience a taste of mathematical research by sharing my experience of working on a characterization problem in Graph Theory -- the characterization of planar maps with order dimension at most 2.

The content of this seminar is conceivable for secondary students.

## 2009年2月25日 星期三

### 小心解讀數字

100001-200000　　　470
200001-300000　　　2200
300001-400000　　　3840
400001-600000　　　5270
600001-900000　　　5990
900001 或以上　　　　6000

## 2009年2月10日 星期二

### 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 $f_n$ converges pointwisely to a function f and if the derivatives $\nabla f_n$ converges uniformly to a function g (say all convergence are on some open subset of $\mathbb{R}^n$, for the conclusions are local anyway), then the limit f is differentiable and $\nabla f = g$.

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 $\nabla f_n$ 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 $f_n$ is a sequence of differentiable functions in $\mathbb{R}^n$ that converges pointwisely to a function f and suppose their derivatives $\nabla f_n$ converges uniformly to a function g. Then to argue differentiability at $x_0$, consider the sequence of functions

$F_n(x):=\begin{cases}\frac{f_n(x)-f_n(x_0)-\nabla f_n(x_0)\cdot(x-x_0)}{|x-x_0|}&\quad\text{if x \ne x_0}\\0&\quad\text{if x = x_0}\end{cases}$

It is easy to verify that $F_n$ is a sequence of continuous functions of x, and that they are Cauchy in $L^{\infty}$. But by assumption $F_n$ converges pointwisely to

$F(x):=\begin{cases}\frac{f(x)-f(x_0)-g(x_0)\cdot (x-x_0)}{|x-x_0|}&\quad\text{if x \ne x_0}\\0&\quad\text{if x = x_0}\end{cases}$

Since the uniform limit of a sequence of continuous functions is continuous, F is continuous at $x_0$, so f is differentiable at $x_0$ and $\nabla f(x_0) = g(x_0)$. This completes the proof.