Towards Non-Differentiability?

The simplest counterexample seems to be $f(x) = |x|$ and $f_n(x) = \begin{cases} |x| & \quad & |x| \geq \tfrac{1}{n} \\ \tfrac{n}{2}x^2 + \tfrac{1}{2n} & & |x| < \tfrac{1}{n} \end{cases}$

Then we can directly check that $f_n$ are differentiable and $|f_n - f| \leq \tfrac{1}{n}$ , so $f_n$ converges to $f$ uniformly. However, $f$ is not differentiable at $x=0$ , so we may conclude it is not necessary for $f$ to be differentiable.

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The following is quite a bit more technical, so feel free to skip it:

If we know a little about approximations of the identity of the convolution operation $(f\ast g)(x) = \int_{-\infty}^{\infty} f(x-t)g(t)\,dt$ , we can easily create a large number of nontrivial counterexamples. For instance, if $f$ is a uniformly continuous function which is not differentiable and $\phi$ is any sufficiently smooth, non-negative function which is supported on $[-1,1]$ with $\int_{-1}^1 \phi(x)\,dx = 1$ , then we may define $\phi_n(x) = n\phi(nx)$ $f_n = f \ast \phi_n$ Then it is known that $f_n$ is as smooth as $\phi$ is (so that in the extreme case, if $\phi$ is infinitely differentiable, then every $f_n$ will be infinitely differentiable as well), and it may be shown that $f_n \to f$ uniformly (this uses the uniform continuity of $f$ ).

Therefore, there will be a counterexample to the claim for at least every $f$ which is uniformly continuous and non-differentiable. In particular, since there are examples of uniformly continuous functions which are nowhere differentiable, it is possible for $f_n$ to be differentiable and $f_n \to f$ uniformly even when $f$ is nowhere differentiable!