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Fix $\alpha \in (0,1)$ and $\psi\in C^{\infty}_{c}(\mathbb{R}\to \mathbb{R})$. For a smooth function $\phi\geq 0$ define the integral $$J_{\alpha}(\phi):=\int \frac{\psi}{\phi^{\alpha}}.$$ If $|\phi^{\prime}|$ is bounded below away from zero, then $J_{\alpha}$ exists for all $\alpha<1$. Indeed, one can simply integrate by parts to get $$J_{\alpha}=-\frac{1}{1-\alpha}\int \phi^{1-\alpha}\left(\frac{\psi}{\phi^{\prime}}\right)^{\prime}.$$

My question is about the case where $\phi^{\prime}$ changes sign finitely many times in the support of $\psi$. Using a smooth partition of unity, one may assume $\phi^{\prime}(x)=0$ at exactly one point $x_{0}$ in the support of $\psi$.

For simplicity, let us further assume that $|\phi^{\prime\prime}|$ is bounded below away from zero near $x_{0}$.

Obviously, we necessarily have $\alpha<1/2$. Just take $\phi$ to be a quadratic function.

I figured this out when $\phi^{\prime\prime}>C>0$: Write $$\phi(x)=\phi(x_{0})+(x-x_{0})^{2}\int_{0}^{1}(1-t) \phi^{\prime\prime}(x_{0}+t(x-x_{0}))dt\geq \frac{C}{2}(x-x_{0})^{2}$$ In this case we have $$|J_{\alpha}(\phi)|\leq \frac{2}{C}\int \frac{|\psi|}{|x-x_{0}|^{2\alpha}}$$ which exists for all $\alpha<1/2$.

What is left is the case $\phi^{\prime\prime}<c<0$.

Summary : Fix $\alpha \in (0,1/2)$ and $\psi\in C^{\infty}_{c}(\mathbb{R}\to \mathbb{R})$. Let $\phi\geq 0$ be a smooth function with a unique stationary point $x_{0}$ in the support of $\psi$. Assume that $\phi^{\prime\prime}<c<0$. Show that $J_{\alpha}(\phi)=\int \frac{\psi}{\phi^{\alpha}}$ is finite ?

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  • $\begingroup$ It seems you basically answered your own question already. If $|\phi''|\ge c>0$, then $\phi(x)\gtrsim (x-x_0)^2$ by a Taylor expansion, so all is well for $\alpha< 1/2$. $\endgroup$ Jun 25, 2022 at 16:15
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    $\begingroup$ Also, either $\phi>0$, so $\phi\ge \delta>0$ on the support of $\psi$, or $\phi(x_0)=0$, but then $\phi$ has a minimum at $x_0$ and $\phi''(x_0)\ge 0$. $\endgroup$ Jun 25, 2022 at 16:23
  • $\begingroup$ Thank you Christian. I don't know how to show $\phi(x)\gtrsim (x-x_{0})^2$ when $\phi^{\prime\prime}<c<0$. This is the last missing piece indeed. Could you make it rigorous ? $\endgroup$
    – Medo
    Jun 25, 2022 at 17:00
  • $\begingroup$ In fact, $\phi''(x_0)<0$ is impossible when $\phi(x_0)=0$, see my second comment. If $\phi(x_0)>0$, then of course there are no problems since $\phi\ge\delta >0$ on the support of $\psi$. $\endgroup$ Jun 25, 2022 at 17:07
  • $\begingroup$ When $\phi(x_{0})>0$ and $\phi^{\prime\prime}<0$, one can look at a small enough neighborhood of $x_{0}$. But in this case $\phi(x)$ is merely bounded below by $\phi(x_{0})$. This finishes the argument. The statement $\phi(x)\gtrsim (x-x_{0})^2$ is not true however when $\phi^{\prime\prime}<0$ as shown by the counterexample $\phi(x)=1-x^2$ with $x_{0}=0$. Thank you for the useful comments. $\endgroup$
    – Medo
    Jun 25, 2022 at 17:24

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$\newcommand\al\alpha\newcommand\R{\mathbb R}$Let us prove the following generalization of your desired statement:

Let $\psi\colon\R\to\R$ be a continuous function with compact support $S$. Let $\phi\colon\R\to\R$ be a nonnegative twice differentiable function with the following property: for any $x_0\in S$ such that $\phi'(x_0)=0$ we have $\phi''(x_0)<0$. Then $J_\al(\phi):=\int\frac{\psi}{\phi^\al}$ is finite for any real $\al>0$.

The proof is based on a comment by Christian Remling. Indeed, take any $x_0\in S$ such that $\phi(x_0)=0$. Since $\phi$ is nonnegative, it follows that $\phi'(x_0)=0$ and hence $\phi''(x_0)<0$, which implies that $\phi<0$ in some punctured neighborhood of $x_0$. This contradicts the condition that $\phi$ is nonnegative.

So, there is no $x_0\in S$ such that $\phi(x_0)=0$. So, $\phi>0$ on $S$. Since $\phi$ is continuous and $S$ is compact, we have $\phi\ge b$ on $S$ for some real $b>0$. Thus, for any real $\al>0$ $$|J_\al(\phi)|\le\int_S\frac{|\psi|}{b^\al}<\infty,$$ since $\psi$ is continuous and $S$ is compact.

We conclude that $J_\al(\phi)$ is finite for any real $\al>0$, as claimed. $\quad\Box$

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