# Sequence of $C^1[0,1]$ functions $(f_n) \to f$ but $f \notin C^1[0,1]$

Question:

Give an example of a sequence of continuously differentiable function $(f_n)$ on $[0,1]$ so that $f_n \to f$ uniformly, but $f$ is not differentiable at all points of $[0,1]$.

My Thoughts:

Would a Fourier Series be a correct answer to this question? Take the Triangle Wave for instance. Wikipedia gives me the following equation:

Here $\omega$ is the angular frequency. Instead of $\infty$ in the sum, could each of my $f_n$ be $\sum_{k=0}^n$. In the limit, this sum of continuously differentiable functions converges to a function that is not differentiable at its cusps.

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Close but not quite. The triangular wave is not differentiable only at the "corners". –  timur Aug 23 '12 at 19:29
@timur I think that is what I mean by "cusps". For example, $|x|$ has a cusp at $x = 0$. –  KingOliver Aug 23 '12 at 19:30
Ok. I think I misunderstood the question. I thought you want $f$ to be nowhere differentiable. –  timur Aug 23 '12 at 19:32
Was just about to submit the answer $f_n:x \mapsto x^n$, but then noted the pretty important "uniformly" in the statement of the question. Anyway i'm not an analist at all, but i recall that if the convergence is uniform, the limit inherits the differentiability properties, if they are shared by all $f_n$. Maybe an analist here can give a (reference to) a proof or a counterexample.. –  Joachim Aug 23 '12 at 19:37
No, the limit doesn't inherit diffentiability properties from the sequence. The triangle wave given in the question is a counterexemple : all partial sums are $C^1$, they converge uniformly to the limit, but the limit is not differentiable. In order to get results, you have to put assumptions on the convergence of the derivatives. –  Ahriman Aug 23 '12 at 19:41

A simpler alternative to the triangular wave is $f_n(x)=\sqrt{(x-\frac12)^2+\frac1n}$ and $f(x)=|x-\frac12|$.