$L^2$ bound on polynomial

Question

I want to show the following theorem:

Let $\frac{1}{1+|p|} \in L^2(\mathbb{R}^d)$ where $p$ is some polynomial. Then $\forall z \in \mathbb{R} \backslash A$ for $A:=\overline{\{p(x);x \in \mathbb{R}^d\}}$, we have $\frac{1}{p-z}\in L^2(\mathbb{R}^d).$

My ideas so far did not really lead to anything: Of course somehow we have to use the fact that the first function is in $L^2$ to conclude it for the other function. Probably we have to split the domain of integration somehow and use different estimates, i.e. on any compact set $M$ we have $\frac{1}{d(z,A)}$ is an upper bound to $\frac{1}{p-z}$ and square-integrable on $M$. But for the non-compact part $M^C$ we probably have to use $\frac{1}{1+|p|}$ somehow.

Answer 1

It's clear that $p$ has positive degree, hence $p$ tends to infinity at infinity. Choose $R$ so that $$|p(x)|>1+2|z|\quad(|x|>R).$$

The integral of $1/|p-z|^2$ over $|x|\le R$ is finite, simply because you're integrating a continuous function over a compact set. On the other hand, if $|x|>R$ then $$|p(x)-z|\ge|p(x)|-|z|=\frac{|p(x)|+1}2+\frac{|p(x)|-(1+2|z|)}2\ge\frac{|p(x)|+1}2.$$

Answer 2

Let $E_k=\{x\in \mathbb R^d: k\le |p(x)| < k+1\}, k = 0,1,\dots $ Note that because $1/(1+|p|)^2$ is bounded below by $1/(k+2)^2$ on $E_k,$ we have $m(E_k) < \infty$ for each $k.$

Consider the set $\cup_{k<|z|}E_k.$ This is a set of finite measure, hence $1/|p-z|^2,$ being bounded above, is integrable there.

For $k>|z|,$ we have

$$\tag 1\int_{E_k} 1/|p-z|^2 \le 1/(k-|z|)^2m(E_k).$$

On the other hand

$$(1/(k+2)^2)m(E_k) \le \int_{E_k} 1/(1+|p|)^2.$$

Therefore $ \sum_{k>|z|}(1/(k+2)^2)m(E_k) < \infty.$ Now the ratio of $1/(k-|z|)^2$ to $1/(k+2)^2$ tends to $1$ as $k\to \infty.$ Hence

$$\sum_{k>|z|} 1/(k-|z|)^2m(E_k) <\infty.$$

This and $(1)$ show $\sum_{k>|z|} \int_{E_k} 1/|p-z|^2 < \infty,$ hence $1/|p-z|\in L^2$ as desired.