Is $h^{+}(x) = \max\{h(x), 0\}$, where $h: \mathbb{R}^n \to \mathbb{R}$ and $h \in C^1$ continuously differentiable?

Question

In this article (DOI: 10.2307/2319406 ), the author define a function as

$$h^{+}(x) = \max\{h(x), 0\},$$ where $h: \mathbb{R}^n \to \mathbb{R}$ and $h \in C^1 (G), G\subseteq \mathbb{R}^n$.

Is it the case that $h^{+}$ is always continuously differentiable ?

My gut says that it should not necessarily be because at the points where $h(x) = 0$, $h^+$ might be not continuously differentiable because of that steep change fro the constant value $0$ to $h(x) > 0$.

Edit:

In case you cannot access to the article: This is the section that the claim is made.

Answer 1

The author considers the derivative of $[h_r^+]^2$ only, not that of $h_r^+$. In contrast to $h_r^+$, $[h_r^+]^2$ is continuously differentiable. Incidentally, at points where $h_r^+$ is non-zero, he can use the chain rule to differentiate $[h_r^+]^2$ because there $h_r^+$ behaves as $h$.

Answer 2

Consider $h(x) = \sin(x)$. Then $h^{+}(x)$ is not differentiable at $x = \pi$ (look at limits).