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This is a question in Bergman's companion to Rudin's POA.

$f$ is differentiable on $[a,b]$ let g=f' Show that for $x \in (a,b], \ g(x-)\neq \infty \ \text{or} -\infty$

My suspicion is that if the derivative blows up to infinity it must drive the function itself to infinity which will destroy the continuity of the function but I'm having trouble proving this.

Alternative notation:

$f$ is differentiable on $[a,b]$

Show \lim_{t\rightarrow x-}f'(t) \neq \infty \ \text{or} -\infty for any $x \in (a,b]$

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    @Glen: Had guessed so. We happy.2011-08-11

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Fix $x \in (a,b]$. If $f$ is differentiable at $x$, then $ \exists \mathop {\lim }\limits_{y \to x^ - } \frac{{f(x) - f(y)}}{{x - y}} \in \mathbb{R}. $ However, by the mean-value theorem $ \frac{{f(x) - f(y)}}{{x - y}} = g(c), $ for some $c = c(y;x) \in (y,x)$. Hence, if $g(x-) =\pm \infty$, then $ \mathop {\lim }\limits_{y \to x^ - } \frac{{f(x) - f(y)}}{{x - y}} = \mathop {\lim }\limits_{y \to x^ - } g(c(y;x)) = \mathop {\lim }\limits_{y \to x^ - } g(y) = \pm \infty. $

EDIT: In view of the OP's post, it is worth noting that the function $f(x)=\arcsin(x)$, $0 \leq x \leq 1$, is continuous on $[0,1]$ (with $f(1)=\pi/2$) but \lim _{x \to 1^ - } f'(x) = \lim _{x \to 1^ - } \frac{1}{{\sqrt {1 - x^2 } }} = \infty . (Consider also $f(x)=\sqrt{x}$, $0 \leq x \leq 1$, as $x \to 0^+$.)

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    @Olivier: I'm satisfied with my answer.2011-08-11