## 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.