If $f'$ exists on an open interval, and there is some point $c$ where $f'(c) > 0$, then there exists a d-neighborhood $\{x \in \mathbb{R} : |x - c| < d\} = V_d(c)$ around c in which $f'(x) > 0$ for all $x \in V_d(c).$

1. How to presage this is false?

2. Where did this counterexample feyly emanate from?

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f is differentiable everywhere, including $x = 0$.
To prove $f'(x)$ takes on negative values..., compute $f'(x) = \dfrac{1}{2} + 2x\sin\dfrac{1}{x} - \cos \dfrac{1}{x}$.

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3. How does $x(k)$ 'enter'?

4. Why will $g'(x_k) \in (-d, d)$ for k large enough?

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    What does the shortcut *S.A.* in the title of your question mean? – Martin Sleziak Feb 27 '14 at 11:52
  • BTW derivative of part of your function is calculated here: http://math.stackexchange.com/questions/393602/differentiability-of-fx-x2-sin-frac1x-and-f and http://math.stackexchange.com/questions/232672/show-that-the-function-gx-x2-sin-frac1x-g0-0-is-everywhere – Martin Sleziak Feb 27 '14 at 11:56

1 Answers1

  1. It isn't immediately clear whether this is false or not. We can safely leave the art of divination to that small fraction of Hogwarts' staff and such people.
  2. It is part of a known set of counterexamples in analysis/topology. They are (loosely speaking) collected under the umbrella term of Topologist's sine curve. At the heart is the function $h(x)=\sin(1/x)$ that has no limit (finite or infinite) as $x\to0$. Then the slightly modified function, $r(x)=xh(x), r(0)=0,$ does have a limit at $x=0$, and is continuous there. But it has no derivative there, because the difference quotient $(r(0+h)-r(0))/h$ oscillates violently between $\pm1$ as $h\to0$. Continuing along the same lines, the function $s(x)=x^2\sin(1/x),s(0)=0,$ behaves better still, and is differentiable at $x=0$. For the purposes of this counterexample, an extra term $x/2$ is needed. As the violently oscillating term $\in[-1,1]$ it is essential that the additional term has slope $1/2$, so that oscillations force sign changes.
  3. By the definition of the limit of a sequence $x_k\in(-\delta,\delta)$ for large enough $k$. I assume that `enter' is used here to indicate that for small $k$ this is not necessarily the case, but as $k$ grows this is bound to happen - irrespective of how small $\delta>0$ is.
  4. `This point' refers to $x_k$. It does not refer to $g(x_k)$ as you seem to think.
Jyrki Lahtonen
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