This is an problem from p. 296 of Buck's Advanced Calculus: Let $f$ be continuous on the interval $0\leq x < \infty$ with $|f(x)|\leq M$. Set $$F(u)= \frac{2}{\pi}\int_{0}^\infty \frac{uf(x)}{u^2 + x^2}dx.$$ Show that $\lim\limits_{u\downarrow0}F(u)=f(0)$.
A heavy theme of the preceding chapter has been uniform convergence of improper integrals with parameters, and the operations that uniform convergence justifies. For instance, we might want to express the integrand as itself being an integral in $u$, and then try to switch the order. Or we might try to differentiate with respect to $u$ underneath the integral (and hope that $\int_{0}^\infty \phi_{1}(u,x)dx$ converges uniformly, so we are justified in doing so).
The integral converges uniformly on $u\in [0, L]$ for $L>0$, but I don't think it converges uniformly on $u\in \mathbb{R}$ (perhaps I am wrong).
At first I thought of integrating $F(u)du$ on $[0,L]$, and switching the order (which is justified by uniform convergence), and then seeing what we can come up with there. If indeed it is true that $\lim\limits_{u\downarrow0}F(u)=f(0)$, then $F$ should be integrable on $[0,L]$.
$$\int_0^L F(u)du = \int_0^L\frac{2}{\pi}\int_0^\infty\frac{uf(x)}{u^2+x^2}dx\,du=\frac{2}{\pi}\int_0^\infty f(x)\int_{0}^L \frac{u}{u^2+x^2}du \, dx$$ $$=\frac{1}{\pi}\int_0^\infty f(x) \ln\left(\frac{L^2+x^2}{x^2}\right) \, dx,$$ but I'm not sure that's getting me anywhere.
Any ideas?
Given $u>0$, letting $x=ut$ in the integral expression of $F(u)$, then $$F(u)=\frac{2}{\pi}\int_0^\infty\frac{f(ut)}{1+t^2}dt.$$ Since $$\int_0^\infty\frac{1}{1+t^2}dt=\frac{\pi}{2},$$ it follows that $$F(u)-f(0)=\frac{2}{\pi}\int_0^\infty\frac{f(ut)-f(0)}{1+t^2}dt.$$ Then for any $a>0$, $$|F(u)-f(0)|\le \frac{2}{\pi}\left(\int_0^a\frac{|f(ut)-f(0)|}{1+t^2}dt+\int_a^\infty\frac{|f(ut)-f(0)|}{1+t^2}dt\right)\le \frac{2}{\pi}\left(\int_0^a\frac{|f(ut)-f(0)|}{1+t^2}dt+2M\int_a^\infty\frac{1}{1+t^2}dt\right).$$ Letting $u\to 0^+$, we have $$\limsup_{u\to 0^+}|F(u)-f(0)|\le \frac{4M}{\pi}\int_a^\infty\frac{1}{1+t^2}dt.$$ Letting $a\to\infty$, the conclusion follows.