About the Hölder continuity of the power of a function

Let $\Omega$ be a bounded domain and assume that $f: \Omega \subset \mathbb{R}^N \to \mathbb{R}$ is a function in the space $C^{2,\alpha}(\overline{\Omega})$, $0 < \alpha <1$, such that $f(x) > 0$, for all $x \in \Omega$ and $f(x) = 0$, for all $x \in \partial\Omega$. If $a >1$, is it true that $f^a \in C^{2,\alpha}(\overline{\Omega})$ ?

I think when a > 1, we can prove that $f ^a \in C^{1,\alpha}(\overline{\Omega})$. But if $a \leq 2$, I don't know if the above conclusion is true.

For example, we need to show that $$\dfrac{\partial^2 f^a}{\partial x_j\partial x_i}(x) = a(a-1)f^{a-2}(x)\dfrac{\partial^2 f}{\partial x_j\partial x_i}(x) \in C^{0,\alpha}(\overline{\Omega}).$$To do this, it is enough to show that tehre exists $M > 0$ such that $$\dfrac{\left|f^{a-2}(x)\dfrac{\partial^2 f}{\partial x_j\partial x_i}(x) - f^{a-2}(y)\dfrac{\partial^2 f}{\partial x_j\partial x_i}(y)\right|}{|x-y|^{\alpha}} \leq M,$$for all $x,y \in \Omega$, with $x \neq y$. My ideia was the following:

$\left|f^{a-2}(x)\dfrac{\partial^2 f}{\partial x_j\partial x_i}(x) - f^{a-2}(y)\dfrac{\partial^2 f}{\partial x_j\partial x_i}(y)\right| = \left|f^{a-2}(x)\left(\dfrac{\partial^2 f}{\partial x_j\partial x_i}(x) -\dfrac{\partial^2 f}{\partial x_j\partial x_i}(y) \right) - \dfrac{\partial^2 f}{\partial x_j\partial x_i}(y)(f^{a-2}(y) - f^{a-2}(x))\right| \leq |f^{a-2}(x)| \left|\dfrac{\partial^2 f}{\partial x_j\partial x_i}(x) -\dfrac{\partial^2 f}{\partial x_j\partial x_i}(y) \right| + C_2|f^{a-2}(x) - f^{a-2}(y)|$.

Using that $f \in C^{2,\alpha}(\overline{\Omega})$ we have$$\left|\dfrac{\partial^2 f}{\partial x_j\partial x_i}(x) -\dfrac{\partial^2 f}{\partial x_j\partial x_i}(y) \right| \leq C_3|x-y|^\alpha.$$The problem arises when I try to analyze the expressions $f^{a-2}(x) $ and $|f^{a-2}(x) - f^{a-2}(y)|$, because when $x$ and $y$ are very close of $\partial\Omega$, this expressions can become very large. So, I don't know how to solve this problem.