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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 $f_n\in C^2$?

Lost1
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3 Answers3

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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.

martini
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  • Why is the convergence obviously uniform? – Lost1 Jul 14 '14 at 11:46
  • That is by the Stone–Weierstrass Theorem: Suppose $X$ is a compact Hausdorff space and $A$ is a subalgebra of $C(X, \mathbb R)$ which contains a non-zero constant function. Then $A$ is dense in $C(X, \mathbb R)$ if and only if it separates points. And the polynomials seperate points. – martini Jul 14 '14 at 11:48
  • Oh that was not what I meant? I want confirmation that the topology you equip the $C(K)$ with is the topology of uniform convergence, but it seems that must be the case – Lost1 Jul 14 '14 at 11:54
  • Yes, uniform convergence. – martini Jul 14 '14 at 12:00
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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.

Ian
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  • I was then thinking whether it holds everywhere but a set of measure 0. You shattered my hope... – Lost1 Jul 14 '14 at 11:47
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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.

copper.hat
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