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I'm trying to generalize Is the derivate on a closed subspace of $C^1[0,1]$ is a continuous linear map? for an exercise in my functional analysis book (Pryce). I want to prove that that the differentiation is continuous even if the function only has a bounded derivative, not necessarily continuous. The problem is that when the derivative isn't continuous the fundamental theorem of calculus doesn't apply.

Suppose that $Z$ is a a closed linear subspace of $\mathscr{C}[0,1]$ s.t. each $f\in Z$ is differentiable on $[0,1]$ with bounded derivative. Show that the mapping $f \mapsto f'$ is a continuous map of $Z$ into $B[0,1]$. Then we can use this to show that the unit ball of $Z$ is compact and Z is thus finite dimensional. The final part is easy.

I've been able to show that if $f_n \rightarrow f \in Z$ uniformly and $f_n'\rightarrow g$ uniformly then $f' = g$ almost everywhere using distribution theory but this seems like overkill for the book from which the exercise is from as it doesn't assume it. I want to use the closed graph theorem as this seemed liked the natural way to prove that differentiation was continuous (and therefore bounded). What am I missing?

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    $\begingroup$ If $f$ is everywhere differentiable, and its derivative is bounded [more generally, if the derivative is locally integrable (in the Lebesgue sense)], then the fundamental theorem of calculus applies, we have $$f(x) = f(x_0) + \int_{x_0}^x f'(t)\,dt.$$ $\endgroup$ – Daniel Fischer Nov 2 '15 at 23:19
  • $\begingroup$ Thank you. So the boundedness of the derivative assures us that the above equality is for all $x \in [0,1]$, not only a.e? $\endgroup$ – MadSax Nov 3 '15 at 9:00
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    $\begingroup$ The boundedness of the derivative implies its Lebesgue integrability, and the integrability (plus the existence of the derivative at all points, a.e. doesn't suffice [but finitely many points at which the one-sided derivatives both exist but are different pose no problem]) gives the equality everywhere (cf. Theorem 7.21 in Rudin's Real and Complex Analysis). $\endgroup$ – Daniel Fischer Nov 3 '15 at 9:17
  • $\begingroup$ Thanks for the reference. This exactly answered my question. $\endgroup$ – MadSax Nov 3 '15 at 9:35
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The problem is that when the derivative isn't continuous the fundamental theorem of calculus doesn't apply.

Rather, an "easy" form of the FTC does not apply. The general form of FTC, due to Lebesgue, applies to absolutely continuous functions. This class includes, in particular, all Lipschitz functions, i.e, those for which there is a constant $L$ such that $|f(x)-f(y)|\le L|x-y|$ for all $x,y$.

An everywhere differentiable function with a bounded derivative is Lipschitz, hence the fundamental theorem of calculus applies to it. This enables an application of the closed graph theorem as you planned: if $f_n\to f$ and $f_n'\to g$ uniformly, then $f$ must be an antiderivative of $g$.

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