If $w: [0, \infty) \to [0, \infty)$ is upper semicontinuous, increasing, $w(0) = 0$ and $w(r)/r^2 = o(1)$, then $w$ is $C^2$ near $0$.
Why does the above hold?
I understand that, by the definition of derivative, there exists $w'(0) = 0$. However, I cannot proceed further.
Context: Theory of viscosity solutions. More precisely, this is a claim of my teacher when proving that if $(p, X)$ belongs to the superjet of $u$ at $x_0$, then there is a $C^2$ supertangent function $\psi$ such that $D\psi(x_0) = p$ and $D^2\psi(x_0) = X$.
That is not correct. The given conditions do not even imply that $w$ is differentiable near $0$.
As an example, define $f: [0, \infty) \to [0, \infty)$ by $f(0) = 0$, $$ f(x) = \frac{1}{2^n} \quad\text{for }\frac{1}{2^n} \le x < \frac{1}{2^{n-1}}\, , n = 1, 2, 3, \ldots $$ and $f(x) = 1$ for $x \ge 1$. $f$ is increasing and upper semicontinuous, but not differentiable in any interval $[0, \epsilon]$.
Then $w(x) = x^2 f(x)$ is increasing and upper semicontinuous with $w(x)/x^2 = f(x) = o(1)$ for $x \to 0$, but $w$ is not differentiable in a neighbourhood of $0$, and $w''(0)$ does not exist.