Prove something is in $L^2(\mathbb{R}^+)$

Question

I stumbled with this very unique issue, which seems trivial, but I was not able to prove of disprove the claim. Here it is.

Let $f\in L^2(\mathbb{R}^+)$, such that $f\in H^1(a,\infty)$ for all $a>0$. Hence, $f$ is an $L^2(\mathbb{R}^+)$ function that is nice in the sense that it is absolutely continuous for $x>0$.

Claim: there exists an $\epsilon>0$ such that $$\frac{f(x)}{x^\epsilon}\in L^2(\mathbb{R}^+).$$

In other words, there is always "room" to make $f$ just a little more "bad" at the origin, while staying in $L^2(\mathbb{R}^+)$. That claim would help me in some eigenvalue computation I am currently performing. I believe it is true but I have not found a way to prove or disprove it.

Answer 1

Define $g(x) = \chi_{(0,1/2)}(x) \frac1{\sqrt x |\log x|} $. Then $g \in L^2(0,1)$ but $x^{-\epsilon} g\not\in L^2$ for all $\epsilon>0$.

Now set $f(x) = g(x) - g(1/2)$. It is piecewise smooth and continuous at $x=1/2$, so $f\in H^1((a,\infty))$ for all $a>0$.

Answer 2

To be concrete, consider the following functions. Let $\phi=\chi_{[0,1/\pi]}$ be the indicator function of the interval $[0,1/\pi]$, and define $\psi(t)=\sin(\frac{1}{t})$, which is Borel measurable. Their product $\Psi:=\varphi\cdot \psi$ is an element of $L^2(\mathbb{R}^+)$, and for each $a>0$, $\Psi\in H^1(a,\infty)$ because it is continuous with compact support on $(a,\infty)$. Yet, for each $\epsilon>0$, the function $t\mapsto \frac{\sin(\frac{1}{t})}{t^{\epsilon}}$ is not an element of $L^2(\mathbb{R}^+)$.