Proving that $\sup f'\left( \left( 0,\infty \right) \right)=0$ under a certain set of conditions.

Question

Let $f$ be a twice differentiable function on $\left( 0,\infty \right)$ s.t. $f''(x)>0$ for all $x\in \left( 0,\infty \right)$. Prove, that if the following conditions are satisfied:

1. $\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=L<\infty $.
2. $\forall x\in \left( 0,\infty \right).f'\left( x \right)<0$.

Then $\sup f'\left( \left( 0,\infty \right) \right)=0$ .

How would you go about proving this?

Answer 1

We do not need the existence of or condition fulfilled by $f''$.

Since $f'(x)<0$, certainly $\sup f'\le 0$. Assume $\sup f'=s<0$. Then $f'(x)\le s$ for all $x$. Consequently, $f(x)\le f(1)+(x-1)s$ for all $x>1$ (MVT). For $x$ big enough, this is $<L-1$, contradicting $\lim_{x\to\infty} f(x)=L$.