Prove that the limit as $y$ goes to $0$ of $V(x,y)=F(x)$ if $V(x,y)=\frac{1}{\pi}\int^{\infty}_{-\infty}\frac{yF(\lambda)}{y^2+(\lambda-x)^2}d\lambda$

Question

Question: If $F(x)$ is bounded and continuous for $-\infty<x<\infty$ and

$$V(x,y)=\frac{1}{\pi}\int^{\infty}_{-\infty}\frac{yF(\lambda)}{y^2+(\lambda-x)^2}d\lambda$$

Prove that the limit as $y$ goes to $0$ of $V(x,y)=F(x)$

I'm not too sure how to begin. I found this in the section about the uniform convergence of improper integrals and it's properties. Basically, what it says is that if an integral converges uniformly, I can differentiate the inside with respect to another variable. However, in this case, differentiating w.r.t to $x$ doesn't seem to get me any closer to solving the integral.

Hints and solutions appreciated. I've been stuck on this fo a while now.

Answer 1

I assume that the limit is taken as $y$ goes down to $0$ from the positive side. Let $t=\frac{\lambda-x}{y}$. The integral becomes $$V(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{F(x+yt)}{1+t^2}dt.$$ Since $F$ is bounded, there exists $M>0$ such that $|F(z)|\leq M$ at all $z\in \Bbb{R}$.

Fix $x\in\Bbb{R}$ and $\epsilon>0$. There exists $L(\epsilon)>0$ such that $$\int_{L(\epsilon)}^\infty\frac{1}{1+t^2}dt<\frac{\pi\epsilon}{8M}.\tag{1}$$ For example, you may take $L(\epsilon)=\frac{8M}{\pi\epsilon}$.

Recall that $F$ is continuous. Pick $\delta>0$ so small that $\big\vert F(z)-F(x)\big\vert<\frac{\epsilon}{2}$ for all $z\in\Bbb{R}$ such that $|z-x|<L(\epsilon)\delta$. Thus, for $0<y<\delta$, $$\big\vert V(x,y)-F(x)\big\vert\leq \frac{1}{\pi}\int_{-L(\epsilon)}^{L(\epsilon)}\frac{\big|F(x+yt)-F(x)\big|}{1+t^2}dt+\frac{1}{\pi}\int_{L(\epsilon)}^\infty\,\frac{2M}{1+t^2}dt+\frac{1}{\pi}\int_{-\infty}^{-L(\epsilon)}\frac{2M}{1+t^2}dt.$$ That is, $$\big\vert V(x,y)-F(x)\big\vert\leq \frac{1}{\pi}\int_{-L(\epsilon)}^{L(\epsilon)}\frac{\frac{\epsilon}{2}}{1+t^2}dt+\frac{4M}{\pi}\int_{L(\epsilon)}^\infty\,\frac{1}{1+t^2}dt.$$ From (1), we have $$\big\vert V(x,y)-F(x)\big\vert< \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\frac{\epsilon}{2}}{1+t^2}dt+\frac{4M}{\pi}\left(\frac{\pi\epsilon}{8M}\right)=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ Thus, $$\lim_{y\searrow0}V(x,y)=F(x).$$