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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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My suspicion is that this is wrong. Are you assuming only $f$ to have a derivative everywhere in $[a,b]$? Then the conclusion is wrong. –  Olivier Bégassat Aug 11 '11 at 14:46
    
But I might be misreading the question : why do you speak of $f'(x^-)$ when $f$ is differentiable? –  Olivier Bégassat Aug 11 '11 at 14:48
    
Could you state your definition of differentiable please? –  Olivier Bégassat Aug 11 '11 at 14:50
    
@Olivier Yes the only assumption is that $f$ is differentiable on the closed interval. We say $g(x-)$ the idea is that a sequence of points say $t_n$ that approach $x$ from below are such that the derivative of f at these points blows up to infinity. (Bergman includes a note that f'(x-) would obviously be finite because f is differentiable. ) –  user9352 Aug 11 '11 at 14:51
    
please state your definition of differentiability and why there would be a minus sign in the derivative, which usually means derivative on the left of $x$, but, if $f$ differentialbe, $f'(x^-)=f'(x)=f'(x^+)$... –  Olivier Bégassat Aug 11 '11 at 14:55
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1 Answer

up vote 1 down vote accepted

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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I think your last line is incorrect, you need $x$ to move too in the left limit, otherwise this is false. –  Olivier Bégassat Aug 11 '11 at 15:14
    
@Olivier: Can you elaborate on your comment (note that $x$ is fixed). –  Shai Covo Aug 11 '11 at 15:21
    
There are several things wrong. First, the left limit is equal to $f'(x)$ since $f$ is differentialble. Thus it can't be equal to $\pm\infty$. Also $\lim f'(y)$ doesn't have to exist. –  Olivier Bégassat Aug 11 '11 at 15:26
    
@Olivier: 1) This is a proof by contradicion; 2) We suppose, for a contradiction, that $\mathop {\lim }\limits_{y \to x^ - } f'(y): = \mathop {\lim }\limits_{y \to x^ - } g(y) = \pm \infty $. Can you find something wrong now? –  Shai Covo Aug 11 '11 at 15:31
    
Finally, $\lim_{y\to x^-} f'(c(y;x))\neq \lim_{y\to x^-} f'(y)$ simply because $c(y;x)$ has no incentive to visit all of $(x-\epsilon,x)$, what I mean is that there can be great gaps in the values taken by $c(y;x)$. –  Olivier Bégassat Aug 11 '11 at 15:35
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