Show that the linear operator $(Tf)(x)=\frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy$ satisfies $\|T\|\leq 1$.

Show that the linear operator $T$ given by $(Tf)(x) = \frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy$ is bounded on $L^2(0, \infty)$ with norm $||T|| \leq 1$.

The professor also wrote, $``$In fact, you should observe that $||T|| = 1$, where $$||T|| = \sup\limits_{f\neq 0}\frac{||Tf||_2}{||f||_2}."$$

Upper bound

For any $g\in L^2(0,\infty)$, $$ \begin{align} \int_0^\infty Tf(x)g(x)\,\mathrm{d}x &=\frac1\pi\int_0^\infty\int_0^\infty\frac{f(y)g(x)}{x+y}\,\mathrm{d}y\,\mathrm{d}x\\ &=\frac1\pi\int_0^\infty\int_0^\infty\frac{f(xy)g(x)}{1+y}\,\mathrm{d}y\,\mathrm{d}x\\ &=\frac1\pi\int_0^\infty\int_0^\infty\frac{f(xy)g(x)}{1+y}\,\mathrm{d}x\,\mathrm{d}y\\ &\le\frac1\pi\int_0^\infty\|f\|_2\|g\|_2\frac{y^{-1/2}}{1+y}\mathrm{d}y\\ &=\frac1\pi\|f\|_2\|g\|_2\int_0^\infty\frac2{1+u^2}\,\mathrm{d}u\\ &=\|f\|_2\|g\|_2 \end{align} $$ This says that $\|Tf\|_2\le\|f\|_2$ ; that is, $\|T\|_2\le1$.

Lower bound

Let $$ f(y)=\left\{\begin{array}{} \frac1{\sqrt{y}}&\text{for }1\le y\le\lambda\\ 0&\text{otherwise} \end{array}\right. $$ Then $\|f\|_2=\sqrt{\log(\lambda)}$ . Furthermore, $$ \begin{align} Tf(x) &=\frac1\pi\int_1^\lambda\frac{y^{-1/2}}{x+y}\,\mathrm{d}y\\ &=\frac1{\pi\sqrt{x}}\int_{1/x}^{\lambda/x}\frac{y^{-1/2}}{1+y}\,\mathrm{d}y\\ &=\frac2{\pi\sqrt{x}}\int_{\sqrt{1/x}}^{\sqrt{\lambda/x}}\frac1{1+u^2}\,\mathrm{d}u\\ &=\frac2{\pi\sqrt{x}}\left(\tan^{-1}\left(\sqrt{\lambda/x}\right)-\tan^{-1}\left(\sqrt{1/x}\right)\right)\\ &=\frac2{\pi\sqrt{x}}\tan^{-1}\left(\frac{1-1/\sqrt\lambda}{1+x/\sqrt\lambda}\sqrt{x}\right) \end{align} $$ For any $\alpha\gt0$, $$ \begin{align} \frac{\|Tf\|^2}{\|f\|^2} &=\frac1{\log(\lambda)}\int_0^\infty|Tf(x)|^2\,\mathrm{d}x\\ &\ge\frac4{\pi^2\log(\lambda)}\int_{\lambda^\alpha}^{\lambda^{1-\alpha}}\frac1x\tan^{-1}\left(\frac{1-1/\sqrt{\lambda}}{2}\lambda^{\alpha/2}\right)^2\,\mathrm{d}x\\ &=(1-2\alpha)\frac4{\pi^2}\tan^{-1}\left(\frac{1-1/\sqrt{\lambda}}{2}\lambda^{\alpha/2}\right)^2\\ &\stackrel{\lambda\to\infty}{\to}1-2\alpha \end{align} $$ Since $\alpha\gt0$ was arbitrary, we get that $\|T\|_2\ge1$. Therefore, $\|T\|_2=1$.