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This is part of a broader question that I am about to post.

The point is I would like to know if the following theorem is true and if the demonstration is okay.

Theorem.

Let $y: \mathbb{R} \to \mathbb{R}$ Is $\exists \tau : y'(\tau) = 0$, then $\exists U=U(\tau): y(x) = y(\tau) \ \ \forall x \in U $

($U$ being of course a neighborood of $\tau$)

Demonstration.

We know that $$\lim_{h \to 0} \frac{y(\tau + h) - y(\tau)}{h} = 0$$ That is to say

$$\forall \varepsilon \ \ \exists \delta\ \ \forall\ |h| < \delta: \left|\frac{y(\tau + h) - y(\tau)}{h}\right| < \varepsilon \ $$

Or $$\left|y(\tau + h) - y(\tau)\right| < |h|\varepsilon < \delta \varepsilon$$

Since $\delta\varepsilon$ gets aribitrarily close to $0$ (not sure about that.. I know that if $\varepsilon$ is small, then also $\delta$ is small, but there is a way to prove it? )

then $y(\tau + h) = y(\tau)$ because the left hand side is $\ge 0$ but must be smaller than any number $> 0$. So it has to be $0$

And this has to hold $\forall \ |h| < \delta$, so it is proven that such $U = U(\tau)$ exists.

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The theorem is not correct; Take $y =x^3$ as a counterexample (or $y = \sin x$, or $x^2$, or any number of other functions). In short, have zero derivative at a point does not imply that the function is locally constant. –  user61527 Jan 21 at 19:07
    
You have an incorrect statement: if $|h|<\delta$, how can it be that $|h|\epsilon=\delta\epsilon$? –  MPW Jan 21 at 19:11

1 Answer 1

up vote 2 down vote accepted

As stated in comments, this is quite false: $y = x^2, x^3, \sin x, \cos x$ and so on give lots of counterexamples to the claim. You've concluded correctly that

$$|y(\tau + h) - y(\tau)| < \delta \epsilon \to 0$$

(although you should have $<$, rather than $=$ in one step) as $\epsilon \to 0$. The problem is that this tells you very little: As we make $\epsilon$ small, we have to make $\delta$ small, which forces $h$ to be small; in the limit, all you can conclude is that if $h = 0$, we have

$$|y(\tau + h) - y(\tau)|$$

is smaller than any positive number, so zero. That's a statement that's perfectly true but pretty much useless.

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okay.. so the problem was that it is true that $\delta \varepsilon$ get aribitrarily close to $0$, but so does $h$ so that does not help.. and indeed the theorem is false.. thank you very much! –  Ant Jan 21 at 22:31
    
@Ant Yes, that's exactly the problem. You're very welcome. –  user61527 Jan 22 at 2:38

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