Given $P_n(x)=\frac{n}{1+n^2x^2},$ Prove $f_n(x)=\frac{1}{\pi}\int_{-\infty}^\infty f(x-t)P_n(t)$ converges uniformly 
Let $P_n:\mathbb{R} \rightarrow \mathbb{R}$ be a sequence of functions defined by $P_n(x)=\frac{n}{1+n^2x^2}$.
$(a)$ Prove $\int_{-\infty}^\infty P_n(x) \mathop{dx} = \pi ,$ and that $$\forall \delta>0: \lim_{n\to \infty}(\int_\delta^\infty P_n(x)\mathop{dx})=\lim_{n\to \infty}(\int_{-\infty}^{-\delta} P_n(x)\mathop{dx})=0$$
$(b)$ Let $f:\mathbb{R} \rightarrow \mathbb{C}$ be a continuous $2\pi-$period function, and $f_n(x)=\frac{1}{\pi}\int_{-\infty}^\infty f(x-t)P_n(t).$
Prove $f_n \rightrightarrows f$ in $\mathbb{R}$

I proved $(a),$ and I also found the same question here.
As for $(b),$ it reminds of me of convolution, and I think the limiting function $f$ is $0$, but can't really get further.
Any help appreciated.
 A: This is a general property of such collections of funnctions (approximative unit).
Suppose $(k_n)_{n\geq 1} $ has (i) $\int_\mathbb{R} k_n =1$, (ii) $\int_\mathbb{R} |k_n| < M$, (iii) for any $\delta>0$ $\int_{|x|>\delta} |k_n| \rightarrow 0$, then for any $f\in C(\mathbb{R})$ when the latter is equiiped with the supremum norm; one has $k_n *f \rightrightarrows f$.
The above properties hint that as $n\rightarrow \infty$ the mass of $k_n$ centers around the origin. Then one can understand the convolution against this limit as a convolution against a unity. Note there is no element in $L^p$ such that $\mathbf{1}*f = f$ for all $f$.
Before proving this, if you haven't tried, I know no way except then diving in to the approximations. Something else that can be done is, if the kernel is known (for instance, yours is the Poisson kernel for $L^p(\mathbb{R})$), other strong theorems or approximations may be used, such as monotone convergence theorem / bounded convergence theorem. One might also use the symmetry of that named kernel, and other, perhaps stronger conditions than the third one, which is the one that controls the rate of the convergence.
Or, you can go through the proof for the statement I gave:
Proof: We calculate: $$\begin{aligned} |k_n *f (x)-f(x)|&= \left|\int_\mathbb{R}  k_n(y)f(x-y) -k_n(y)f(x)  \mathrm{d}y \right|\\
&\leq  \int_\mathbb{R} \left|  k_n(y)(f(x-y) -f(x)) \right| \mathrm{d}y 
\\ & = \int_{|y|>\delta} \left|  k_n(y)(f(x-y) -f(x)) \right|\mathrm{d}y + \int_{|y|<\delta} \left|  k_n(y)(f(x-y) -f(x)) \right| \mathrm{d}y
 \end{aligned}$$
now, if $f$ is $2\pi$-periodic continuous function then it is in $C(\mathbb{R})$ and we have $|f(x)-f(y)|\leq 2\|f\|_{L^\infty} <\infty$
so the first integral is bounded by $2\|f\|_{L^\infty} \cdot \int_{|y|>\delta} |k_n(y)|\mathrm{d}y$ which vanishes at the $n\rightarrow \infty$ limit (for any $\delta>0$).
We bound the other integral by:
$$\sup_{|y|<\delta}|f(x-y)-f(x)| \int_{|y|<delta}|k_n(y)|$$
We notice the integral is bounded by $M$ and that the supremum vanishes, due to the uniform continuity of $f$ (as its continuity depends on the way it behaves in the closed $[-\pi,\pi]$ interval, and there it is uniformally continuous). This allows us to pick a $\delta$ small enough, suitable for all $x$.
We finally notice $n,\delta$ were independent of $x$, and so the convergence is uniform.
