8
$\begingroup$

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).

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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" $\endgroup$ – Elimination Jan 23 '16 at 14:31
  • $\begingroup$ 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. $\endgroup$ – user159517 Jan 23 '16 at 14:40
  • $\begingroup$ @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? $\endgroup$ – TomGrubb Jan 23 '16 at 14:50
  • 1
    $\begingroup$ @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 $(1/\pi)\int_{-\infty}^{\infty}f(x) P_n(t) dt = f(x)$ $\endgroup$ – user159517 Jan 23 '16 at 14:54
  • 2
    $\begingroup$ @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. $\endgroup$ – user159517 Jan 23 '16 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.