Integral of bounded continuous function on $R$

Question

Let $f$ be a bounded continuous function on $R$. Prove that

$$\lim_{n \to \infty} \frac{n}{\pi} \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=f(0)$$

I solved this question as follows, but I ran into a problem:

Solution:

Since $f$ is continuous at $0$, given $\epsilon>0$, there is $\delta >0$ such that $|f(x)-f(0)| \leq \epsilon, \forall x \in [-\delta \hspace{2 mm} \delta]$. Now consider $f_{n}(x)=\frac{nf(x)}{1+n^{2}x^{2}}$,$n \in N$, is a sequence of continuous function that converges to $0$ on $R-[-\delta \hspace{2 mm} \delta]$ when $n \to \infty$. Also, according to guestion, $f$ is bounded (i.e.$|f_{n}(x)|<g(x)$). Therefore according to Dominated Convergence Theorem $$\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} 0 \hspace{2 mm} dt=0$$

Now I have a problem to solve the second part. I want to prove

$\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt = \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt $

which

$\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt =\pi f(0)$

If I can prove that, the rest of question is easy, because I can say:

$\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt+\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ [-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\pi f(0)$

Now how can I prove: $\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt = \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt $

Answer 1

it may help to make the substitution $x=nt$ in the integral, which then becomes $$ \frac1{\pi}\int_{-\infty}^{\infty}\frac{f(\frac{x}{n}) dx}{1+x^2} $$ the sequence of functions $f(\frac{x}{n})$ converges pointwise to the constant function $c(x)=f(0)$ and is bounded, so we may apply the dominated convergence theorem

Answer 2

We have

$$\int_{-\delta}^{\delta}\frac{nf(t)}{1+n^2t^2}\,dt=\int_{-\infty}^{\infty}\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\,du.$$

Since $f$ is bounded, $|f(t)| \leq M$, and it follows that

$$\left|\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\right| \leq \frac{M}{1+u^2}$$

Apply the dominated convergence theorem, noting that

$$\lim_{n\rightarrow \infty}\int_{-\infty}^{\infty}\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\,du=\int_{-\infty}^{\infty}\lim_{n\rightarrow \infty}\left[\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\right]\,du\\=\int_{-\infty}^{\infty}\frac{f(0)}{1+u^2}\,du=\pi f(0)$$