Thread: Prove a function is constant.

Problem:

Suppose that $\displaystyle f:R \mapsto R $ is twice differentiable everywhere; $\displaystyle f(x) \leq 0 $ and $\displaystyle f''(x) \geq 0 \ \ \ \ \ \forall x$. Prove that $\displaystyle f:R \mapsto R $ is constant.

I'm trying to prove this thing by showing f'(x) = 0 for all x. Now, I know f' is either monotonically increasing or constant. But I have problem using the f <= 0 for my proof.

Please help, thank you.

Suppose that $\displaystyle f'(x_0)>0$ for some point x_0. Since f"(x)≥0, it follows that f' is an increasing function, and so f'(x)≥f'(x_0) for all x≥x_0. That means that f(x)→+∞ as x→+∞, contradicting the fact that f(x) is never positive.

Similarly, if f'(x) is ever negative then you can show that f(x)→+∞ as x→–∞.