If $f$ is bounded and twice differentiable in $\mathbb{R}$, show that there exists $\xi\in\mathbb{R}$, s.t. $f''(\xi)=0$.
My idea:
- 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 $f=\frac{1}{1+e^{-x}}$?
- $f$ 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].
Solution 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).
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.
Solution 2:
We don't need Darboux' theorem. Assume that $f''(x)\ne0$ for all $x\in{\mathbb R}$. Let $$\Omega:=\bigl\{(x,y)\in{\mathbb R}^2\>\bigm|\>x<y\bigr\}\ ,$$ and consider the function $g:\>\Omega\to{\mathbb R}$ defined by $$g(x,y):={f'(y)-f'(x)\over y-x}\ .$$ By the MVT for all $(x,y)\in\Omega$ there is a $\tau\in\ ]x,y[\ $ with $$g(x,y)=f''(\tau)\ne0\ .$$ Since $g$ is continuous it follows that the sign of $g$ is constant on $\Omega$, which implies that $f'$ is strictly monotone, say: increasing, on ${\mathbb R}$.
If $f'(b)=:\beta>0$ for some $b\in{\mathbb R}$ then $f'(x)>\beta$ for all $x>b$, and $f$ will be unbounded when $x\to\infty$. Similarly, if $f'(a)=:\alpha<0$ for some $a\in{\mathbb R}$ then $f'(x)<\alpha$ for all $x<a$, and $f$ will be unbounded when $x\to -\infty$. It follows that $f'$ is identically zero – a contradiction.
Solution 3:
Lemma 1: If continuous function $f$ has extreme points, then $\exists x,y,~f(x)=f(y)$.
Proof. Without loss of generality, we assume $f$ has a local maximum point at $t$. According to the definition of local maximum, we have $$\exists \xi>0, \forall x\in\underset{o}{U}(t,\xi), f(x) \leq f(t)$$Let $$M = \max\left\{\min_{x\in(t-\xi, t)}f(x), \min_{x\in(t,t+\xi)}f(x)\right\}$$ By Intermediate Value Theorem, there is function value $(M+f(t))/2$ in both intervals $(t-\xi, t)$ and $(t，t+\xi)$.$\Box$
Lemma 2: If $f$ is continuous in $\mathbb{R}$ and satisfies $\forall x,y, ~f(x)\neq f(y)$, then $f$ is strictly monotone.
Proof. Suppose $f$ is not strictly monotone. Because $\forall x,y, ~f(x)\neq f(y)$, we have $\exists a< b,~f(a)< f(b)$. Let $$\begin{align*}H = \sup_{t>a} \{\forall x\in(a,t), f(x)<f(t)\}\end{align*}\\ L = \inf_{t< a} \{\forall x\in(t,a), f(t)<f(x)\}$$ If $H\to\infty$，$L\to-\infty$, then $f$ is strictly monotone, which contradicts the assumption. So there should be one of them $\nrightarrow \infty$.
Without loss of generality, we assume it is $H$ that $\nrightarrow \infty$, then $\exists \xi, \forall x\in\underset{o}{U}(H,\xi), f(x) < f(H)$ (see the definition of $H$), which indicates that $H$ is a local maximum. According to Lemma 1, $\exists x,y,~f(x)=f(y)$. Contradiction occrus, so $f$ must be strictly monotone.$\Box$
Lemma 3: If $f'$ is strictly monotone and differentiable in $\mathbb{R}$, then $f$ is unbounded.
Proof. $f''>0$ or $f''<0$ since $f'$ is strictly monotone and differentiable . The second-order Taylor series of $f$ is: $$f(x) = f(0) + f'(0)x + f''(\xi)x^2$$ So $f$ is unbounded.$\Box$
Theorem: If $f$ is bounded and twice differentiable in $\mathbb{R}$, then $\exists \xi, f''(\xi)=0$
Proof. $f$ is twice differentiable, so $f'$ is differentiable. Suppose $\forall x,y, ~f'(x)\neq f'(y)$, then $f'$ is strictly monotone (Lemma 2), and $f$ is bounded (Lemma 3), so the assumption doesn't hold, that means $\exists x,y,~f'(x)=f'(y)$.
According to Mean Value Theorem, $\exists \xi, f''(\xi)=0$.$\Box$