Let $0<a<d/2$, let $B$ be the unit ball in $\mathbb{R}^{d}$ centered at the origin, and let $f:B \to [0,\infty[$ be a a smooth function such that

(1) $f(x)\geq f(0).$

(2) $\nabla f(x)\neq 0,\quad \forall x\neq 0.$

(3) The Hessian matrix $D^2 f(0)$ is positive definite.

I am trying to prove $f^{-a}$ is integrable on $B$. I am obviously interested in the case $f(0)=0$.

Please keep in mind that I don't know if the conditions (1) through (3) are sufficient for the integrability of $f^{-a}$. But I can't come up with a counterexample.

Since $0$ is a minimum of $f$ we necessarily have $\nabla f(0)=0$. So there exists $\xi \in B$ such that $$f(x)=f(0)+x^T D^2 f(0) x+ \sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi),\quad x \in B.$$

Since $D^2f(0)$ is positive definite, there exists $C>0$ such that $x^T D^2f(0)x\geq C|x|^2$. And, for $x \in B^{\prime}$, a small enough ball centered at $0$, we have $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{10} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\gtrsim |x|^{2}.$$

So we have the integral

$$\int_{B^{\prime}}f^{-a} \lesssim \int_{B^{\prime}}|x|^{-2 a}dx.$$ We are done with this part by the assumption $2a<d$.

I don't know how to get a good enough bound for $f$ away from $0$, if at all possible. Are the conditions (1)-(3) sufficient for $\int_{B\setminus B^{\prime}}f^{-a}$ to be finite ?