I am supposed to prove that an everywhere zero derivative function implies that the original function is constant. Now, I can only think of using the definition of derivative, but I think things get a little more tricky since the denominator is converging to zero as well. Could you guys give any hints please? I cannot use the mean value theorem. My function is from reals to reals: $$ f : (a,b) \rightarrow R $$
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1$\begingroup$ Why exactly is the mean value theorem excluded? You really ought to include that in the body of your post, too, not just the title. $\endgroup$ – rschwieb Jan 20 '14 at 19:50
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$\begingroup$ What is the domain of your function? Can we assume it's an interval in $\mathbb{R}$? $\endgroup$ – Dan Jan 20 '14 at 19:52
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$\begingroup$ Yes, just edited. $\endgroup$ – user191919 Jan 20 '14 at 20:01
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$\begingroup$ Is the fundamental theorem of calculus already available? $\endgroup$ – Daniel Fischer♦ Jan 20 '14 at 20:03
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I am assuming that your domain is an interval of $\mathbb{R}$.
Suppose $f(x) <f(y)$, and let $\phi(t) = f(x+t(y-x))-(f(x)+t (f(y)-f(x))$. Note that $\phi(0)=\phi(1)$ and so $\phi$ has a maximum somewhere with $t \in (0,1)$. At the maximum we have $\phi'(t) = 0$.
However, $\phi'(t) = 0 -(f(y)-f(x)) \neq 0$, which is a contradiction.
answered Jan 20 '14 at 20:04
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$\begingroup$ That's the MVT right? Also, if you drop the condition that $f(x)<f(y)$ then the contradiction goes away and you have a direct proof. $\endgroup$ – WimC Jan 20 '14 at 20:09
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$\begingroup$ @WimC: Well, it depends on your viewpoint. I would view it as a consequence of compactness, as there is no mean value here (sort of $0-0 = 0$, I suppose). $\endgroup$ – copper.hat Jan 20 '14 at 20:12
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Let $\alpha > 0$ be any positive number. Show that for each $x < y$ there are intermediate points $$x=x_1<x_2<\ldots<x_n=y$$ (where $n$ may depend on $x$ and $y$) such that $$|f(x_{k+1})-f(x_k)|<\alpha|x_{k+1}-x_k|$$ for all $k\in\{1,\ldots,n-1\}$.
