I just solved an exercises which stated to show $f$ is strictly convex in $(-1, 1)$, knowing that $f: (-1, 1) \to \mathbb{R}$ is: continuous, positive and at least twice differentiable everywhere. Also $f'(x) = 2xf^2(x)$.
Now this was easy since
$$f''(x) = 2f^2(x) + 4xf(x)f'(x) = 2f^2(x)\big(1 + 4f(x)\big)$$
Since $f$ is positive, everything is positive hence the thesis.
- Is it possible to show $f$ is strictly convex by removing the "at least twice differentiability" of the function? Let's just suppose $f$ has all the above properties except for that one, something like (if it makes sense)
$$f(x) \quad \text{is such that} \quad f'(x) = 2xf^2(x) \quad \text{but} \quad f''(x) \quad \text{not guaranteed to exist}$$
So $f$ is a weird function such that $f''$ might not exist (for choice maybe).
Would it be possible to show convexity in this case?