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).


1 Answer 1


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.

  • $\begingroup$ 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" $\endgroup$ Jan 23, 2016 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, 2016 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, 2016 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, 2016 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, 2016 at 20:01

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