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).

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.

• Thank you @user159517. I'm trying to think how to utilize the $f(x-t)-f(x)$ expression but can't think of anything good. Could you help me proceed please? I guess I should rely on what I already found about the integral of $P_n(x)$ but the $f(x-t)-f(x)$ expression is "interfering" – Elimination Jan 23 '16 at 14:31
• I edited my answer to include a little more information. If you still dont see what I'm getting at, I can make a full answer out of it. – user159517 Jan 23 '16 at 14:40
• @user159517 I'm sorry but how did you move the $f(x)$ into the integral? On the first line the function is not being integrated, but then on the second line it is? – TomGrubb Jan 23 '16 at 14:50
• @bburGsamohT note that $f(x)$ does not depend on the integration variable $t$. So because of $\int_{-\infty}^{\infty} P_n(t) dt = \pi$, we have $(1/\pi)\int_{-\infty}^{\infty}f(x) P_n(t) dt = f(x)$ – user159517 Jan 23 '16 at 14:54
• @Elimination no, it's the same $\delta$. The left and right integral converge to $0$ no matter what $\delta$ we choose, so we can take it to be the $\delta$ from the uniform continuity. – user159517 Jan 23 '16 at 20:01