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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.2011-08-11
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    But I might be misreading the question : why do you speak of $f'(x^-)$ when $f$ is differentiable?2011-08-11
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    Could you state your definition of differentiable please?2011-08-11
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    @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. )2011-08-11
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    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, $f'(x^-)=f'(x)=f'(x^+)$...2011-08-11
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    also, does your definition of differentiability allow for infinite values?2011-08-11
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    you can find examples of differentiable functions such that for instance $f'(0)=0$ (for instance) and where there are sequences $t_n<0,t_n\to 0$ and $s_n>0,s_n\to 0$ with $f'(t_n)\to$ whatever you want in $\mathbb{R}\cup\lbrace -\infty,+\infty\rbrace$ and $f'(s_n)\to$ whatever else you want in $\mathbb{R}\cup\lbrace -\infty,+\infty\rbrace$.2011-08-11
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    @Olivier Sometimes $f$ being differentiable is shorthand for $f\in C^1$, and if $f\in C^1$ over a closed interval, then $f'\in C^0$ and is bounded.2011-08-11
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    @ Olivier $f$ is differentiable at $x$ iff $\lim_{t\rightarrow x} \frac {f(t)-f(x)}{t-x}$ exists and is finite2011-08-11
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    one function that exhibits such behaviour is $\mathbb{R}\to\mathbb{R},~x\mapsto x^2\sin(\frac{1}{x^2})$2011-08-11
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    @Olivier Not really, the derivative of that function is bounded on any closed interval not containing zero, and zero is not allowed in the interval because the function is not differentiable at zero. I think you may be confused.2011-08-11
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    @Glenn Wheeler let me state what I forgot, set $f(0):=0$.2011-08-11
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    @Olivier This just bounds the *function itself* at zero. The derivative is still unbounded in any open interval containing zero, and since by hypothesis $f$ is differentiable, zero cannot be in $[a,b]$.2011-08-11
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    @Glenn Wheeler I don't understand your point. The function I defined is differentiable in the usual sense, and indeed the sense of the question (user9352 gave his definition a few comments above). This is one of the most common counter example functions. What do you mean by $0$ can't be in $[a,b]$?2011-08-11
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    @Olivier Let $f$ be your function. $|f'(\epsilon)| > c\epsilon^{-1}$ for some finite $c$.2011-08-11
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    @Glen: Which means that $f$ cannot be $C^1$ but does not prevent $f$ from being *differentiable* (yes, even at $0$).2011-08-11
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    @user9352 you previously wrote in a comment directed at me > "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". In the mean time, you edited your question and it reads > "show that $\lim_{t\to x^-}f'(t)\neq +\infty$ or $-\infty$ for all $x\in(a,b]$ These are different definitions : for the convenience of every one involved, please give the right definitions from the get go.2011-08-11
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    @Glen Wheeler What? Let's define $f(x)=x^2\sin(1/x^2)$ for all non zero $x$ and $f(0)=0$ : $f$ is differentiable out side of $0$ for obvious reasons, and also at $0$, indeed for all non zero $x$, we have that $|x|<\epsilon$ implies $$\left|0-\frac{f(x)-f(0)}{x-0}\right|=|x\sin(1/x^2)|\leq|x|<\epsilon$$ which shows that $f$ is indeed differentiable at $0$ and $f'(0)=0$.2011-08-11
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    @Glen: And you are right to point out that at the same time there exists a sequence $(x_n)$ converging to $0$ such that $f'(x_n)\ge1/x_n$ (but this has nothing to do with the differentiability or non-differentiability of $f$ at $0$).2011-08-11
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    @Glenn Wheeler I have pointed out to you in a comment above that the OP gives the definition of differentiable he's working with. There is no $C^1$ hypothesis, only what the OP wrote down.2011-08-11
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    @Olivier I see that this is my miscommunication. All this time I have been 'explaining' my statement in my first comment. I completely ignored the OP's definition.2011-08-11
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    @Didier I've been thinking of $f\in C^1$.2011-08-11
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    @Glen: Had guessed so. We happy.2011-08-11

1 Answers 1

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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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    I think your last line is incorrect, you need $x$ to move too in the left limit, otherwise this is false.2011-08-11
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    @Olivier: Can you elaborate on your comment (note that $x$ is fixed).2011-08-11
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    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.2011-08-11
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    @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?2011-08-11
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    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 $c(y;x)$.2011-08-11
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    I didn't understand $1)$, which takes care of $f'(x)=\pm\infty$, but the other two points stand.2011-08-11
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    @Olivier: For the first two points, see my previous comment (regarding a proof by contradiction). Concerning your last point, $\mathop {\lim }\limits_{y \to x^ - } f'(y) = \pm \infty $ (which we assume, for a contradiction) implies that also $\mathop {\lim }\limits_{y \to x^ - } f'(c(y;x)) = \pm \infty $.2011-08-11
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    If $\lim \dots =\pm\infty$ means $\lim \dots = +\infty$ or $\lim \dots =-\infty$, then this is not the hypothesis made by the OP. And in the general case, that is if you only make the hypothesis made by the OP : there is a sequence $y_n$ below $x$ such that $\lim_{\infty}f'(y_n)=+\infty$ (for instance), you will not have $\lim_{x^-}f'(c(y,x))=\lim_{x^-} f'(y)$.2011-08-11
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    @Olivier: I'm satisfied with my answer.2011-08-11