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I am trying to solve the following problem.

Let $f:(a,b) \to \mathbb{R}$ be a differentiable function with a global isolated minimum at $x_0$. Then is it true that there exist $c < x_0 < d$ such that $f'(x) \le 0$ if $x \in (c,x_0)$ and $f'(x) \ge 0$ if $x \in (x_0, d)$?

I think it is true. I want to prove that for $f(x_0)$ to be a global isolated minimum, there must be an interval $(x_0, d)$ where the function is strictly monotonically increasing and an interval $(c, x_0)$ where the function is strictly monotonically decreasing. This would proof what I want to prove.

Anyway, I have not been able to prove it. Does anyone know if the assertion is true or have any ideas on how to prove it?

Jaramillo
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    If you made the inequalities strict, i.e. $f'(x) < 0$ and $f'(x) > 0$, then yes. Otherwise, any constant function would be a counterexample. – Theo Bendit Nov 08 '23 at 02:24
  • I don't this works, but constructing an example requires some effort. Graphically you can imagine a curve which oscillates between the curves $y=x^2$ and $y=x^4$ for $x\in(-1,1)$ and the global minimum occurs at $x_0=0$. – Paramanand Singh Nov 08 '23 at 06:12
  • @TheoBendit I think you misread the statement. If you ask for strict inequalities, the statement becomes stronger, so it cannot be true when the original statement is false. – Lorenzo Pompili Nov 08 '23 at 17:13
  • Can you clarify what definition of isolated global minimum you're using? – V.S.e.H. Nov 09 '23 at 14:24

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Your claim doesn't hold. Just consider the counterexample given by the function

$$f(x)= \begin{cases} x^2\left(2+\sin \left(\frac 1x\right)\right), & \text{if} \quad x\ne 0, \\ 0, & \text{if} \quad x=0. \end{cases} $$

$f$ is differentiable (it must be shown) and $0=f(0)<f(x), \forall x\in\mathbb{R}\setminus\{0\}.$

Moreover, $\forall x\in\mathbb{R}\setminus\{0\}$ it is $$f'(x)= 2x \sin \left(\frac 1x\right)-\cos \left(\frac 1x \right).$$

If we evaluate $f'(x)$ at points $x=\dfrac 1{n\pi},$ where $n\in\mathbb{Z},$ we get positive or negative values for $f'$ depending if $n$ is even or odd.

mfl
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  • I don't think the minimum in your example is isolated. – V.S.e.H. Nov 08 '23 at 17:13
  • @V.S.e.H. it holds $f(x)>0$ for all $x$, except for $x_0=0$. So, $x_0$ is an isolated, global minimum. – Lorenzo Pompili Nov 08 '23 at 17:17
  • @LorenzoPompili this is not the definition of an isolated local/global minimum. An isolated minimum $x^$ is one, such that there exists an open neighborhood $\mathcal{N}$ at $x^$, such that $x^$ is the only* minimizer within that neighborhood. The function given by this answer clearly has infinitely many local minimizers clustered around $x^*$. – V.S.e.H. Nov 08 '23 at 17:56
  • @V.S.e.H. It seems you are right, I guess it has to be called differently then (although I could argue that the word “global” takes precedence over “isolated”, and in fact $x_0$ is isolated as a point in the set of global minima… X) ). Curiously, I found the exact same example here which is used to show that a strict minimum is not always an isolated minimum with that definition: https://wiki.math.ntnu.no/_media/tma4180/2015v/existence.pdf – Lorenzo Pompili Nov 08 '23 at 18:09
  • @LorenzoPompili Yes. In fact the example given by this answer is almost the same as the one given in "Numerical Optimization" by Nocedal and Wright for a function where its minimum is not an isolated one. – V.S.e.H. Nov 08 '23 at 18:11
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Take a look at this thread Is the derivative always nonnegative in a neighbourhood of a minimum? Using this technique you can build a smooth function with negative derivative in an infinite number of points arbitrarily close to the origin (or what would be $x_0$ in your case). So the answer should be no, but counterexamples are very pathological.

brighton
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