Prove that for all $x>0$, $f(x) < xf'(x)$

Question

Let $f(x)$ be continuous on $[0,\infty)$ and twice differentiable on $(0,\infty)$. If $f(0)=0$ and $f''(x)>0$ on $(0,\infty)$ prove that for all $x>0$, $f(x) < xf'(x)$.

Hello, can anyone help me with this question? Thank you!

Answer 1

Basically the MVT applies to $f$ on $(0,x)$ to get: for some $c \in (0,x): f(x) - f(0) = f'(c)(x-0) \implies f(x) = xf'(c) < xf'(x)$ since $f'$ is increasing. So $f(x) < xf'(x)$ as claimed.

Answer 2

Let $x > 0$. Since $f(0) = 0$, by the mean value theorem there exists $c\in (0,x)$ such that $\frac{f(x)}{x} = f'(c)$. Since $f''> 0$ on $(0,\infty)$, then $f'$ is strictly increasing in $(0,\infty)$, hence $f'(c) < f'(x)$.