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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.

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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.

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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.

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