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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 $f(0)=0$.

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 $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$

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 $2a<d$.

What about the easier integral $$\int_{B\setminus B^{\prime}}f^{-a}$$

Edit: I think the answer to my question is simple. This is also inspired by the helpful comments of @Fedor Petrov and @Bazin

Since $D^{2}f(0)$ is positive definite then $f$ is strictly convex in some neighborhood of the origin, say $B(\delta_{0})\subseteq B^{\prime}$. We claim that if $|x|>\delta_{0}$ then $f(x)>\frac{1}{4} C\delta_{0}^2$. (Recall that we have shown $|y|\geq \frac{1}{2}C|y|^2$ for all $y\in B^{\prime}$). Otherwise $f(x)\leq f(y)$ for some $y \in B(\delta_{0})$. This can not occur since $f$ has no critical points outside $B(\delta_{0})$, by the condition $(2)$.

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 $f(0)=0$.

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 $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$

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 $2a<d$.

What about the easier integral $$\int_{B\setminus B^{\prime}}f^{-a}$$

Edit: I think the answer to my question is simple. This is also inspired by the helpful comments of @Fedor Petrov and @Bazin

Since $D^{2}f(0)$ is positive definite then $f$ is strictly convex in some neighborhood of the origin, say $B(\delta_{0})\subseteq B^{\prime}$. We claim that if $|x|>\delta_{0}$ then $f(x)>\frac{1}{4} C\delta_{0}^2$. (Recall that we have shown $f(y)\geq \frac{1}{2}C|y|^2$ for all $y\in B^{\prime}$). Otherwise $f(x)\leq f(y)$ for some $y \in B(\delta_{0})$. This can not occur since $f$ has no critical points outside $B(\delta_{0})$, by the condition $(2)$.

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 $f(0)=0$.

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 $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$

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 ?

Edit: I think the answer to my question is affirmative and simple:

Let $\delta>0$ be the radius of $B^{\prime}$. If $|x|>\delta$ then $f(x)>\frac{1}{2} C|x|^2$. Otherwise $f(x)\leq f(y)$ for some $y \in B^{\prime}$. This does not occur unless $f$ has a critical point, which contradicts condition $(2)$.

I think more is true: The function $f$ is convex. It is strictly convex near $0$ by assumption $(3)$, and the assumption $(2)$ (that $\nabla f\neq 0$ at any point other than $0$) guarantees that $f$ has no critical points other than $0$. Makes sense ?

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 $f(0)=0$.

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 $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$

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 $2a<d$.

What about the easier integral $$\int_{B\setminus B^{\prime}}f^{-a}$$

Edit: I think the answer to my question is simple. This is also inspired by the helpful comments of @Fedor Petrov and @Bazin

Since $D^{2}f(0)$ is positive definite then $f$ is strictly convex in some neighborhood of the origin, say $B(\delta_{0})\subseteq B^{\prime}$. We claim that if $|x|>\delta_{0}$ then $f(x)>\frac{1}{4} C\delta_{0}^2$. (Recall that we have shown $|y|\geq \frac{1}{2}C|y|^2$ for all $y\in B^{\prime}$). Otherwise $f(x)\leq f(y)$ for some $y \in B(\delta_{0})$. This can not occur since $f$ has no critical points outside $B(\delta_{0})$, by the condition $(2)$.

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 $f(0)=0$.

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 $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$

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 ?

Edit: I think the answer to my question is affirmative and simple:

Let $\delta>0$ be the radius of $B^{\prime}$. If $|x|>\delta$ then $f(x)>\frac{1}{2} C|x|^2$. Otherwise $f(x)\leq f(y)$ for some $y \in B^{\prime}$. This does not occur unless $f$ has a stationary point, which contradicts condition $(2)$.

I think more is true: The function $f$ is convex. It is strictly convex near $0$ by assumption $(3)$, and the assumption $(2)$ (that $\nabla f\neq 0$ at any point other than $0$) guarantees that $f$ has no critical points other than $0$. Makes sense ?

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 $f(0)=0$.

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 $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$

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 ?

Edit: I think the answer to my question is affirmative and simple:

Let $\delta>0$ be the radius of $B^{\prime}$. If $|x|>\delta$ then $f(x)>\frac{1}{2} C|x|^2$. Otherwise $f(x)\leq f(y)$ for some $y \in B^{\prime}$. This does not occur unless $f$ has a critical point, which contradicts condition $(2)$.

I think more is true: The function $f$ is convex. It is strictly convex near $0$ by assumption $(3)$, and the assumption $(2)$ (that $\nabla f\neq 0$ at any point other than $0$) guarantees that $f$ has no critical points other than $0$. Makes sense ?