Let $g=g(\xi)$ be a bounded continuous function on $\Bbb R$. For $x \in \Bbb R, y>0$, define $f(x,y)$ by the improper integral $$ f(x,y)=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y}{(x-\xi )^{2} + y^{2}} g(\xi ) d\xi . $$
- Prove that $f(x,y)$ converges.
Write $f(x,y)=I_1+I_2$ where $I_1=\frac{1}{\pi} \int_{-\infty}^{0} \frac{y}{(x-\xi )^{2} + y^{2}} g(\xi ) d\xi, I_2=\frac{1}{\pi} \int_{0}^{\infty} \frac{y}{(x-\xi )^{2} + y^{2}} g(\xi ) d\xi$. It suffices to prove that $I_1,I_2$ both converge. Let $M>0$ be the number such that $|g(\xi)| \le M$. By Direct comparison test, it's sufficient to prove that $J_1,J_2$ both converge where $J_1=\frac{M}{\pi} \int_{-\infty}^{0} \frac{y}{(x-\xi )^{2} + y^{2}} d\xi, J_2=\frac{M}{\pi} \int_{0}^{\infty} \frac{y}{(x-\xi )^{2} + y^{2}} d\xi$. We can compute directly $$J_1=\frac{M}{\pi}(\frac{\pi}{2}-\arctan\frac{x}{y}),J_2=\frac{M}{\pi}(\frac{\pi}{2}+\arctan\frac{x}{y}), J_1+J_2=M.$$ Thus, $f(x,y)$ (absolutely) converges. Does the proof looks fine?
- Prove that if $y \downarrow 0$, then $f(x,y)$ converges to $g(x)$.
I was completely lost at this point. I don't know where to get started.
Any help would be much appreciated.
Your proof for 1. is fine.
For 2., note that $f(x,y)=1$ when $g\left(\xi\right)=1$ for any real number $\xi$. Therefore, using the substitution $s=x- \xi$, we have to show that $$ \lim_{y\downarrow 0} \int_{- \infty}^{+ \infty}\frac y{s^2+y^2}\left(g\left(x-s\right)-g\left(x\right)\right)\mathrm ds=0.$$ Using the substitution $s=ty$, the problem becomes $$ \lim_{y\downarrow 0} \int_{- \infty}^{+ \infty}\frac 1{t^2+1}\left(g\left(x-ty\right)-g\left(x\right)\right)\mathrm dt=0.$$ This can be done in the following way: fix $\epsilon$ and find $R$ (independent of $y$) such that the integrals $\int_R^{+\infty}$ and $\int_{-\infty}^{-R}$ have a contribution which is smaller than $\epsilon$. Then use uniform continuity of $g$ on $[-R,R]$.