For $\epsilon > 0$, find $a,b\in \mathbb R^{>0}$ such that $\|\mathcal L f\|^2_2 \ge (1 - \epsilon) \pi \|f\|^2_2$ 
Let $0 < a < b < \infty$, and let $$f(y) = \begin{cases}y^{-1/2} & y\in [a,b]\\ 0 & y\notin [a,b] \end{cases}$$
Show that for each $\epsilon > 0$, we can choose $a$ and $b$ such that $$\|\mathcal L f\|^2_2 \ge (1 - \epsilon) \pi \|f\|^2_2$$
Here, $\mathcal L: L^2(\mathbb R_+)\to L^2(\mathbb R_+)$ such that $$\mathcal L f(x) = \int_0^\infty e^{-xy} f(y)\, dy$$


Context: We have already proved that $\|\mathcal L\| \le \sqrt \pi$. Using the result above, we can get $\|\mathcal L\| \ge \sqrt\pi$, which allows us to conclude $\|\mathcal L\| = \sqrt\pi$ (by taking $\epsilon \to 0$).

My work: For $f$ as defined above, I have shown that $$\|f\|_2^2 = \log \left(\frac b a \right)$$
I am trying to compute/estimate $\|\mathcal Lf \|_2^2$. So far I have $$\|\mathcal L f\|_2^2 =  4\int_0^\infty \left( \int_{\sqrt a}^{\sqrt{b}} e^{-xt^2}\, dt \right)^2 dx$$ on which we want a lower bound (right?) - so I wrote $$\int_{\sqrt a}^{\sqrt b} e^{-xt^2}\, dt = \frac{\sqrt\pi}{\sqrt x} - \int_0^{\sqrt a} e^{-xt^2}\, dt - \int_b^\infty e^{-xt^2}\, dt$$
and I am trying to find upper bounds for $\int_0^{\sqrt a} e^{-xt^2}\, dt$ and $\int_b^\infty e^{-xt^2}\, dt$. Using $\int_\lambda^\infty e^{-x^{2}}dx \le \frac 1{2\lambda}e^{-\lambda^{2}}$ for $\lambda > 0$, I got $$ \int_b^\infty e^{-xt^2}\, dt \le \frac{1}{2x\sqrt b} e^{-bx}$$
I don't know how to bound $\int_0^{\sqrt a} e^{-xt^2}\, dt$, and I also don't know if the bound I have found is useful. My plan is to eventually find a lower bound on $$\frac{\|\mathcal Lf\|_2^2}{\|f\|_2^2}$$ in terms of $a$ and $b$, and then choose $a,b\in \mathbb R^{>0}$ (dependent on $\epsilon$) such that the lower bound is $(1 - \epsilon)\pi$.
I would appreciate any help in completing the solution from here, or any other suggestions on how to go about it. Thank you!

Notation: $\mathbb R_+$ is $[0,\infty)$.
 A: As the OP noticed, for $f(x)=\frac{1}{\sqrt{x}}\mathbb{1}_{(a,b]}(x)$, $0<a<b<\infty$
$$I:=\|\mathcal{L}f\|^2_2=4\int^\infty_0\Big(\int^{\sqrt{b}}_{\sqrt{a}}e^{-xu^2}\,du\Big)^2\,dx$$
By Fubini's theorem
$$ I=4\int^\infty_0\int^{\sqrt{b}}_{\sqrt{b}}\int^{\sqrt{b}}_{\sqrt{a}}e^{-x(y^2+z^2)}\,dz\,dy\,dx=4\int^{\sqrt{b}}_{\sqrt{a}}\int^{\sqrt{b}}_{\sqrt{a}}\frac{dy dz}{y^2+z^2}\\=4\int^{\sqrt{b}}_{\sqrt{a}}\frac{1}{y}\big(\arctan(\sqrt{b}/y)-\arctan(\sqrt{a}/y)\big)\,dy$$
Since $0\leq \arctan(v)\leq \pi/2$ for $v>0$,
$$I\leq \pi\log(b/a)=\pi\|f\|^2_2$$
To complete the problem, it is enough to show that
$$\lim_{n\rightarrow}\frac{4}{\log(b^4)}\int^b_{1/b}\frac{1}{y}\big(\arctan(b/y)-\arctan(1/(by))\big)\,dy=\pi
$$
Let
$$G(b)=4\int^b_{1/b}\frac{1}{y}\big(\arctan(b/y)-\arctan(1/(by))\big)\,dy$$
Monotone convergence implies that $\lim_{b\rightarrow\infty}G(b)=\infty$. We use L'Hospital rule.
$$\begin{align}
G'(b)&=\frac{4}{b}\big(\arctan(1)-\arctan(b^{-2})\big)+\frac{4}{b}\big(\arctan(b^2)-\arctan(1)\big) +\\
&\qquad 4\int^b_{1/b}\frac{1}{y}\Big(\frac{1}{y}\frac{1}{1+\big(\tfrac{b}{y}\big)^2} +\frac{1}{yb^2} \frac{1}{1+\big(\tfrac{1}{by}\big)^2}\Big)\,dy\\
&=\frac{4}{b}\big(\arctan(b^2)-\arctan(b^{-2})\big) +\\
&\qquad 4\int^b_{1/b}\frac{1}{y^2+b^2} + \frac{1}{y^2b^2+1}\,dy\\
&=\frac{4}{b}\big(\arctan(b^2)-\arctan(b^{-2})\big) +\\
&\qquad \frac{4}{b}\Big(\arctan(1)-\arctan(b^{-2})\Big)+ \frac{4}{b}\big(\arctan(b^2)-\arctan(1)\Big)\\
&=\frac{8}{b}\big(\arctan(b^2)-\arctan(b^{-2})\big)
\end{align}
$$
Then
$$
\frac{G(b)}{4\log b}\sim\frac{\frac{8}{b}\big(\arctan(b^2)-\arctan(b^{-2})\big)}{\frac{4}{b}}=2\big(\arctan(b^2)-\arctan(b^{-2})\big)\xrightarrow{b\rightarrow\infty}\pi
$$
Consequently, the conditions of L'Hospital theorem hold and we obtain that
$$\lim_{b\rightarrow\infty}\frac{G(b)}{4\log b}=\pi$$.

Edit: I would like to add to my answer the connection of the Laplace operator considered by the OP and Hilbert's inequality.
$$\begin{align}
\|\mathcal{L}f\|^2_2&=\int^\infty_0\Big|\int^\infty_0 f(y)e^{xy}\,dy\Big|^2\,dx \leq\int^\infty_0\Big(\int^\infty_0|f(y)|e^{-xy}\,dy\Big)^2\,dx\\
&=\int^\infty_0\int^\infty_0\int^\infty_0|f(y)||f(z)|e^{-x(y+z)}\,dzdydx\\
&=\int^\infty_0|f(y)|\Big(\int^\infty_0\frac{|f(z)|}{z+y}\,dz\Big)\,dy
\end{align}$$
The operator $Kf(x)=\int^\infty_0\frac{|f(z)|}{z+x}\,dz$ on $L_p((0,\infty))$ has a kernel $k(x,z)=\frac{1}{z+x}$ that is homogenous of order $-1$. Thus, for any $p\geq1$
$$\begin{align}
\left(\int^\infty_0\Big|\int^\infty_0k(x,z)|f(z)|\,dz\Big|^p\,dx\right)^{1/p}&\leq\left(\int^\infty_0\Big(\int^\infty_0\frac1x|k(1,z/x) f(z)|\,dz\Big)^p\,dx\right)^{1/p}\\
&=\left(\int^\infty_0\Big(\int^\infty_0k(1,u)|f(xu)|\,du\Big)^p\,dx\right)^{1/p}\\
&\leq\int^\infty_0\left(\int^\infty_0|k(1,u)|^p|f(xu)|^p\,dx\right)^{1/p}\,du\\
&=\int^\infty_0\left(\int^\infty_0|f(xu)|^p\,dx\right)^{1/p}|k(1,u)|\,du\\
&=\Big(\int^\infty_0\frac{du}{u^{1/p}(1+u)}\Big)\|f\|_p
\end{align}
$$
The integral $c_p=\int^\infty_0\frac{du}{u^{1/p}(1+u)}$ is finite for $p>1$ and can be easily computed by contour integrals of the Mellin type to obtain $c_p=\frac{\pi}{\sin(\pi/p)}$.
This along with Hölder's inequality yields
$$\|\mathcal{L}f\|_2\leq\Big(\int^\infty_0|f(y)|\Big(\int^\infty_0\frac{|f(z)|}{z+y}\,dz\Big)\,dy\Big)^{1/2}\leq \sqrt{\pi}\|f\|_2$$
