I need to know if this little lemma is true for a result I am working on concerning the spectrum of an operator. I here provide a proof. I am doubting myself because a previous person answering another question doubted it and because my proof seems overly complicated. Any counter-example, error in my proof, or any simplification of my proof would be welcome.
Lemma (the result I am not entirely sure is true):
Let $H^1_0(\mathbb{R})$ be the Hilbert subspace of functions of $H^1(\mathbb{R})$ that are zero at the origin. If $f\in H^1_0(\mathbb{R})$ then $f(\xi)/\xi\in L^2(\mathbb{R})$.
Proof:
By the Sobolev Embedding Theorem, if $s=n/2+\alpha$, with $0<\alpha<1$, then $H^s(\mathbb{R}^n)$ is continuously included into $C^\alpha(\mathbb{R}^n)$. Thus, in our case, with $n=1$ and $\alpha=1/2$, we have that $$ \limsup_{\xi\to 0} \frac{|f(\xi)|}{|\xi|^{1/2}}<\infty.\;\;(1) $$
We write $f$ as $f(\xi)=|\xi|^{1/2}\widetilde{f}(\xi)$. Since $f\in C^{1/2}(\mathbb{R})$, the new function $\widetilde{f}(\xi)$ is continuous, except perhaps at $\xi=0$. $\widetilde{f}(\xi)$ is also bounded because $f$ itself is bounded and because of Condition (1). Furthermore, $\widetilde{f}\in L^2(\mathbb{R})$ since
$$
\int_{-\infty}^{\infty} |{\widetilde{f}} |^2d\xi\leq \int_{-1}^{1} |\widetilde{f}|^2d\xi+\int_{|\xi|>1} |f|^2d\xi<\infty.
$$
In the inequality above, the convergence of the integral on the interval $[-1,1]$ is due to the facts that $\widetilde{f}$ is continuous except, possibly, at $\xi=0$, and because it is bounded.
Furthermore, $\widetilde{f}\in H^1(a,\infty)$ for any $a>0$ since
$$
\widetilde{f}(\xi)=\frac{f}{\xi^{1/2}},\;\;\widetilde{f}'(\xi)=\frac{f'}{\xi^{1/2}}-\frac{f}{2\xi^{3/2}}\in L^2(a,\infty).
$$
We use the fact that $f'\in L^2(\mathbb{R})$ and compute
$$
\int_0^\infty |{f}'(\xi)|^2d\xi=\lim_{a\to 0^+} \int_a^\infty \left|{\xi^{1/2}\widetilde{f}'}+\frac{\widetilde{f}}{2\xi^{1/2}}\right|^2\\
=\lim_{a\to 0^+} \int_a^\infty \left({\xi |\widetilde{f}'|^2}+\frac{|\widetilde{f}|^2}{4\xi}+\frac{1}{2}\left(\widetilde{f}' \overline{\widetilde{f}}+{\widetilde{f}} \overline{\widetilde{f}'}\right)\right)d\xi\\
=\lim_{a\to 0^+} \int_a^\infty \left({\xi |\widetilde{f}'|^2}+\frac{|{f}|^2}{4\xi^2}+\frac{1}{2}(|\widetilde{f}|^2)'\right)d\xi\\
=\lim_{a\to 0^+} \left(\int_a^\infty \left({\xi |\widetilde{f}'|^2}+\frac{|{f}|^2}{4\xi^2}\right)d\xi
-\frac{1}{2}|\widetilde{f}(a)|^2\right)<\infty,\;\;\;(2)
$$
By the Monotone convergence theorem, we have that the limit
$$
\lim_{a\to 0^+} \int_a^\infty \left({\xi |\widetilde{f}'|^2}+\frac{|{f}|^2}{4\xi^2}\right)d\xi,
$$
exists. Since $\widetilde{f}$ is bounded, it follows from the last line of [2] that the limit above is finite. Thus, in particular,
$$
\int_0^\infty \frac{|{f}|^2}{\xi^2}d\xi<\infty.
$$
In the same way, we can show the integral on $(-\infty,0)$ is finite,
thus $f(\xi)/\xi\in L^2(\mathbb{R})$.