Let $f \in L^1(\mathbb{R})$ and $p$ a polynomial of deg $m$. Does $f / p \in L^1$ imply $x^m f / p \in L^1$ or vice versa?

Question

Let $f \in L^1(\mathbb{R})$ be an integrable function on the real line.

Let $ p = x^m + a_1 x^{m-1} + \cdots + a_m \in \mathbb{R}[x] $ be a real polynomial of degree $m$.

Consider the function $h$ defined (almost everywhere) by

$$ h(x) = \frac{x^m f(x)}{p(x)} $$

We want to study the integrability of $h$. Near infinity, $h$ behaves like $f$, so $h$ is integrable on $\{|x| > R\}$ for some large $R > 0$. But near zeros of $p$, $h$ might blow up to infinity, which could potentially screw up the integrability.

To investigate how the zeros of $p$ may affect the integrability of $h$, we can also consider the integrability of the function $f(x)/p(x)$, without the $x^m$ in the numerator.

Question:

Consider the following two statements:

1. $ h \in L^1 \implies \frac{f}{p} \in L^1 $
2. $ \frac{f}{p} \in L^1 \implies h \in L^1 $

Is either one of the statements true? If both statements are false, could you give some counterexamples?

Answer 1

1. If both $f$ and $f/p$. Are integrable then $h(x)=x^m\frac{f(x)}{p(x)}$ is also integrable. The reason is that $$\lim_{x\rightarrow\pm\infty}\Big|\frac{x^m}{p(x)}\Big|$$ exists, and it is finite. Hence, for some constant $c$ and for all $x$ large enough, say $|x|>A$ for some $A>0$, $$ \Big|\frac{x^m}{p(x)}\Big|\leq c$$ Consequently $$|h|\leq A^m\Big|\frac{f}{p}\Big|\mathbf{1}(|x|\leq A) + c|f|\mathbf{1}(|x|>A)$$ The conclusion then follows by dominated convergence.

Answer 2

1 is false, by example take $f(x) = e^{-x^2}$ and $p(x)=x$. Then $h(x) = f(x) \in L^1$, $\dfrac{f}{p} = \dfrac{e^{-x^2}}{x} \notin L^1.$