# Does a bounded function $f$ on $\mathbb R$ has zero for its second derivative?

Ask: Does a bounded function $$f$$ on $$\mathbb R$$, which is continuous and twice differentiable on $$\mathbb R$$, has zero for its second derivative, i.e. there exists a point $$x_0 \in \mathbb R$$ such that $$f''(x_0)=0$$ My idea is considering the contrapositive for the statement above. Let say $$f''(x_0)>0$$ , or equivalently the function $$f$$ is strictly convex. (Another case: Consider the function $$-f$$, which is strictly concave and having $$-f''(x_0)<0$$.) Then the function $$f$$ supposes to be strictly increasing, so $$f$$ is not bounded on $$\mathbb R$$.

Then it follows the conclusion is true.

I am doubted whether my proof is correct or not, and am interested in finding a relevant example, or a counterexample for the conclusion.

## 2 Answers

There exists functions $$f$$ that are strictly increasing but bounded, take $$f = \arctan(x)$$. So that implication in the last statement of your proof is not right.

Your idea is correct. However you need to prove why a convex map can’t be bounded.

An argument is that such a map is always above its tangent. Then consider the value of $$f^\prime(x)$$. I can’t be always vanishing.

If $$f^\prime(x_0)>0$$ then $$\lim\limits_{x \to \infty} f(x) =\infty$$. And if $$f^\prime(x_0)<0$$ then $$\lim\limits_{x \to -\infty} f(x) =\infty$$. Due again to the fact that a convex map lies above its tangents.