# Showing the derivative of a differentiable function has a point of continuity

The question goes like this - Let $f:[0,1]\rightarrow \mathbb{R}$ be a differentiable function. Show that $f'(x)$ has a continuity point.

Thanks for the help!

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Can you show that the pointwise limit of continuous functions has a continuity point (in fact many of them)? Then simply note that $f$ must be continuous and $f'(x) = \lim\limits_{n \to \infty} n\cdot(f(x+1/n) - f(x))$. – t.b. Jul 6 '11 at 17:00
differential or differentiable – user9413 Jul 6 '11 at 17:01
What does it mean for a function to be differentiable? – Daniel Freedman Jul 6 '11 at 17:02
@Daniel: Please see en.wikipedia.org/wiki/Differentiable_function – user9413 Jul 6 '11 at 17:03
Could you give a better title? Interesting is a very subjective description. – Asaf Karagila Jul 6 '11 at 17:12

Here are a few thoughts.

1) a theorem of Baire (according to one of my old textbooks): Let $(X,\mathcal{T})$ be a topological space and $f_n$ be a sequence of continuous functions $f_n : X \to \Bbb{R}$ with the property that there exists $f:X \to \Bbb{R},$ $$f(x)=\lim_{n \to \infty} f_n(x),\ \forall x \in X.$$

 A good reference for this (assuming $X$ is metrizable) is Ch. 24.B of Kechris's Classical descriptive set theory (pp. 192ff), given by Theo Buehler in the comments. Many thanks. My source textbook is very old and not known outside my university.

Then the set $D(f)$ of the discontinuity points of the function $f$ is of first Baire category type (it is a countable union of sets $E$ with $\text{int}(\text{cl}(E))=\emptyset$; int is interior, cl is closure).

2) $[0,1]$ with the standard topology is not a space of the first category, because every complete metric space is a Baire space.

3) Let

$$f_n: [0,1] \to \Bbb{R},\ f_n(x)=\begin{cases} \frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}},\ \ x \in \Bigl[0,1-\frac{1}{n}\Bigr]\\ \frac{f(1)-f(1-\frac{1}{n})}{\frac{1}{n}},\ \ x \in \Bigl[1-\frac{1}{n},1\Bigr] \end{cases}$$

The sequence of continuous functions $f_n$ converge pointwise to $f'$ on $[0,1]$. By the first two points, the set of discontinuity points of $f'$ cannot be the whole $[0,1]$.

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you mean every complete metric space in 2). The hard part is to prove 1), of course – t.b. Jul 6 '11 at 17:38
@Beni: Dear Beni, it would be better if you specify your notations. That is please add: $\text{int}$ stands for the interior, and $\text{cl}$ stands for the closure, so that users know what notations you are using. – user9413 Jul 6 '11 at 17:43