[Math] If $f$ is bounded and twice differentiable in $\mathbb{R}$, show that there exists $\xi\in\mathbb{R}$, s.t. $f”(\xi)=0$.

derivatives

My idea:

  1. If $f$ has maximum and minimum, then $f'=0$ at these two points, and the conclusion is further derived using Mean Value Theorem. But what if $f$ has no maximum/minimum, like $f=\frac{1}{1+e^{-x}}$?
  2. $f$ is twice differentiable in $\mathbb{R}$, so $f'$ is continuous in $\mathbb{R}$. If $\forall x,y,~f'(x)\neq f'(y)$, then $f'$ is a monotonic function (how to prove this?).
    Then use second-order Taylor approximation to show $f$ is unbounded (inspired by Simon S), which contradicts the assumption. So $\exists x,y,~f'(x)=f'(y)$ and $\exists \xi,~ f''(\xi)=0$ [Mean Value Theorem].

Best Answer

  1. if a function $f$ is differentiable on an interval and the derivative is always different from zero then the function is either increasing or decreasing (the derivative has constant sign, by the Darboux property of derivatives).

  2. Suppose a function $f$ has second derivative always different from zero. Apply 1. to the first derivative to find that the given function is either strictly concave or strictly convex. Take a point where the derivative is not zero, the tangent line is unbounded by above and below. So the function is unbounded because it lies either above or below the tangent line.

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