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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 $f'(x) = \lim\limits_{n \to \infty} n\cdot(f(x+1/n) - f(x))$.2011-07-06
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    differential or differentiable2011-07-06
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    What does it mean for a function to be differentiable?2011-07-06
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    @Daniel: Please see http://en.wikipedia.org/wiki/Differentiable_function2011-07-06
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    Could you give a better title? Interesting is a very subjective description.2011-07-06
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    @Theo,mixedmath: Sorry, I completely misunderstood the question.2011-07-06

1 Answers 1

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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 $f:X \to \Bbb{R},$ $$ f(x)=\lim_{n \to \infty} f_n(x),\ \forall x \in X.$$

[edit] 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 course2011-07-06
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    See also this thread: http://ask.metafilter.com/98118/A-nowhere-continuous-derivative2011-07-06
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    @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.2011-07-06
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    @Theo: Sorry. I forgot to mention the completeness in 2)2011-07-06
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    @Chandru: I noticed that on the site "int" and "cl" are used for denoting the interior and closure of a set. I use different notations, but complied to the usage I've seen. I will mention the definitions in following posts.2011-07-06
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    @Beni: I was just suggesting you to do it. Anyway, I have no regrets :)2011-07-06
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    @Beni: in order to have a reference: the statement you're referring in 1) to can be easily extracted from Ch. 24.B of Kechris's *[Classical descriptive set theory](http://books.google.com/books?id=pPv9KCEkklsC)* (pp. 192ff).2011-07-06
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    @Theo: on a Kechris riff today, eh?2011-07-06
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    @Willie: Yeah... But come on, it's only the third time I mention it (once on MO, twice here) :) One of my not-so-secret secret weapons...2011-07-06
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    Thank you very much for your answer! I knew that Bier's theorems must be used but couldn't figure out how.2011-07-06