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Let $(E,d_\infty)$ the metric space of continuous functions defined on $ [0,1] $, with $$ d_\infty(f,g)=\sup\{ |f(x) - g(x)| : x\in [0,1] \}. $$ For all $ n\in \mathbb{N} $ let $$ F_n = \{ f: \exists x_0\in [0,1-1/n] \forall x\in [x_0,1]\left(|f(x)-f(x_0)|\leq n(x-x_0)\right) \}. $$ Let $D$ the set of continuous functions which have a finite derivate on the right for at least one point of $[0, 1[$. I need to show that $$ D = \bigcup_{n\in\mathbb{N}} F_n. $$ This is part of problem 38 of section 8 of Royden's Real Analysis book.

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    Hint: the point with finite right derivative is $x_0$.2011-05-22
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    This looks wrong to me. Are you sure you want $x_{0} \in [0,1/n]$ and not $x_{0} \in [0,1-1/n]$? Then you should be able to apply the hint given by @Yuval.2011-05-22
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    Yes, that is the point @Yuval. And You are right @Theo, thank you.2011-05-22
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    You should probably draw a picture. You know that the graph lies below some line on the interval $[x_0, x_0 + \delta]$. Now you also know that the function attains its maximum $M$ somewhere. If the line intersects the vertical line at $x_{0} + \delta$ at a higher point than $M$ then you're done. If not, simply make the line steeper to achieve that.2011-05-22
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    Whoa, wait a minute! I only commented on the very first comment with a formula. I didn't see the others (I had a coffee in the meantime and left this page open) and I haven't read them. It seems to me that it should be *much* simpler.2011-05-22
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    The other inclusion should follow from the same idea.2011-05-22
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    Are both inclusions simpler?2011-05-22
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    I think so. Did you understand what I was trying to say in the comment "You should probably draw a picture"?2011-05-22
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    Thank you by your help @Theo.2011-05-23
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    If you've found the answer, you should add it here, for future reference and other users.2012-10-27

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