Show that $f_n\to f$ uniformly on $\mathbb{R}$ Let $$P_n(x) = \frac{n}{1+n^2x^2}$$.
First, I had to prove that 
$$\int_{-\infty}^\infty P_n(x)\ dx = \pi$$
And that for any $\delta > 0$:
$$\lim_{n\to\infty} \int_\delta^\infty P_n(x)\ dx = \lim_{n\to\infty} \int_{-\infty}^{-\delta} P_n(x)\ dx = 0$$
I've done that easily.
Now I need to prove that for $f:\mathbb{R}\to\mathbb{C}$ which is $2\pi$ periodic and continuous and:
$$f_n(x) = \frac{1}{\pi} \int_{-\infty}^\infty f(x-t)P_n(t)\ dt$$
$f_n\to f$, uniformly on $\mathbb{R}$.
We learned in class about convolution and about Dirichlet/Fejer kernels.
Also, we learned that the trigonometric polynomials, $\{e^{inx}\}_{n\in\mathbb{Z}}$ are a dense set on $C(\mathbb{T})$ and the density is uniform. Meaning, there's a $P_n(x)=\sum c_n e^{inx}$ converges uniformly to $f$ where $f\in C(\mathbb{T})$.
note: $f\in C(\mathbb{T})$ is a continuous and $2\pi$ periodic function (T is for Torus).
 A: To get you started: $$| f_n(x) - f(x)| =\left| (1/\pi) \int_{-\infty}^{\infty} f(x-t) P_{n}(t) \; dt - f(x)\right| = (1/\pi)\left| \int_{-\infty}^{\infty}\left[f(x-t)- f(x)\right] P_n(t) \; dt  \right|$$
Now because $f$ is continuous on $\mathbb{T}$ and $2\pi$-periodic, we can essentially deduce its properties by considering its restriction $f_r$ to $[0,4\pi]$. As a continuous function on a compact interval, $f_r$ is bounded and uniformly continuous. It's not hard to see that these properties carry over to $f$, meaning that $f$ is bounded and uniformly continous on $\mathbb{R}$. This implies that for every $\epsilon > 0$, there is a $\delta 
> 0$ such that 
$$|f(x) - f(x-t)| < \epsilon \quad \forall t \in (-\delta,\delta)\forall x\in \mathbb{R}$$
Now, given an $\epsilon > 0$, we can choose $\delta$ accordingly and then split up the integrals giving
$$|f_n(x) - f(x)| \leq (1/\pi)\left[\int_{-\infty}^{-\delta}C\cdot P_{n}(t) \; dt + \int_{-\delta}^{\delta}\epsilon\cdot P_{n}(t) \; dt + \int_{\delta}^{\infty}C\cdot P_{n}(t) \; dt\right]$$
Because of what you've already shown we know the left and right integral converge to $0$ as $n \to \infty$. But the middle integral can be estimated by $\epsilon$, which concludes the proof.
