If $f_n$ is a sequence of differentiable functions converging to $f$ uniformly on a compact set Suppose $f_n\rightarrow f$ on a compact set in $\mathbb{R}^n$, with $f_n\in C^1$. $f$ is not necessary differentiable. We can easily find a sequence of functions converging to $|f|$, for example.
My question is: does there exist any results which says, for example, the derivative exists at all but finitely many places.
What about if $f_n\in C^2$?
 A: No. Recall that Polynomials are dense in $C(K)$ for every compact $K \subseteq \mathbb R^n$. Now let $f\colon K \to \mathbb R$ a continuous, nowhere differentiable function. There is a sequence of polynomials (hence smooth functions) $f_n$ such that $f_n \to f$ uniformly.
A: No. In particular, the partial sums of the sum defining http://en.wikipedia.org/wiki/Weierstrass_function are smooth and converge uniformly by the M test, but the limit is nowhere differentiable.
A: Define $f(x) = 0 $ for $x \le 0$. Define $f({1 \over n}) = + {1 \over n}$ if $n$ is even and $f({1 \over n}) = - {1 \over n}$ if $n$ is odd, and by straight line interpolation in between. It is easy to see that $f$ is continuous (only $0$ is in question, and $|f(x)| \le |x|$ everywhere). However, $f$ is not differentiable at any ${1 \over n}$.
Now choose $[-1,1]$ (a compact interval) and a sequence of polynomials $p_n$ such that $\max_{|t| \le 1}|f(t)-p_n(t)| \to 0$. Each $p_n$ is smooth, but $f$ is not differentiable at a countable number of points.
