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I was wondering if there exists a sequence $(f_n)$ of nowhere differentiable functions $(f_n) \rightarrow f$ uniformly, BUT $f$ everywhere differentiable. I have a hunch this violates a theorem of differentiability, but I can't put my tongue on which one nor on how to prove it. Does such a counter example exist?

Wow McWow
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Let $g$ be your favorite bounded nowhere differentiable function and define

$$f_n(x) = \frac{g(x)}{n}$$

  • So I'm assuming we can say $f_n \rightarrow f$ uniformly iff g(x) is bounded because $f_n(x)$ < M, so |M/n - 0| is independent of $x$ for any epsilon?

    In other words, this will ONLY work if $g(x)$ is bounded?

     – Wow McWow Mar 31 '16 at 01:23
  • This particular example only works because $g$ is bounded, yes. For unbounded examples, take $h(x) + g(x) / n$ with $h$ an everywhere differentiable function. –  Mar 31 '16 at 01:25