Suppose the functions $f_{n}(n=1,2, \ldots)$ are defined on the open interval $(0,1)$ and that $f_{n}$ tends uniformly on $(0,1)$ to a function $f$. If the $f_{n}$ are continuous, show that $f$ is continuous. If the $f_{n}$ are differentiable, show by example that $f$ need not be differentiable.

Assume now that each $f_{n}$ is differentiable and the derivatives $f_{n}^{\prime}$ converge uniformly on $(0,1)$. For any given $c \in(0,1)$, we define functions $g_{c, n}$ by

$$g_{c, n}(x)= \begin{cases}\frac{f_{n}(x)-f_{n}(c)}{x-c} & \text { for } x \neq c, \\ f_{n}^{\prime}(c) & \text { for } x=c .\end{cases}$$

Show that each $g_{c, n}$ is continuous. Using the general principle of uniform convergence (the Cauchy criterion) and the Mean Value Theorem, or otherwise, prove that the functions $g_{c, n}$ converge uniformly to a continuous function $g_{c}$ on $(0,1)$, where

$$g_{c}(x)=\frac{f(x)-f(c)}{x-c} \quad \text { for } x \neq c$$

Deduce that $f$ is differentiable on $(0,1)$.