1
$\begingroup$

I was reading a book on distributions, and the following is classified as a $\delta$-convergent sequence. $$u_n(x)=\frac{n}{\pi(1+n^2x^2)}$$

This was my attempt at understanding why it converges to $\delta(x)$.

$$\displaystyle \lim_{n \to \infty}\left\langle u_n, \phi\right\rangle = \displaystyle \lim_{n \to \infty}\int_{-\infty}^{\infty}\frac{n}{\pi(1+n^2x^2)} \ \phi(x)\ dx$$

Then by the Lebesgue dominated convergence theorem,

$$\displaystyle \lim_{n \to \infty}\left\langle u_n, \phi\right\rangle = \displaystyle \int_{-\infty}^{\infty}\lim_{n \to \infty}\frac{n}{\pi(1+n^2x^2)} \ \phi(x)\ dx$$

$$=\int_{-\infty}^{\infty}\delta(x)\ \phi(x) \ dx\quad\quad (1)$$ $$=\left\langle u, \phi\right\rangle$$ Thus, $$\displaystyle \lim_{n \to \infty}u_n=u=\delta(x)$$

First of all, I'm not sure if my process is correct. Second of all, I'm not certain why I would be able to make the claim in step $(1)$.

Thank you in advance.

$\endgroup$
  • 1
    $\begingroup$ It's not hard to show that $f_n\to\delta$, but your application of DCT is absurd. When you apply a theorem you need to verify the hypotheses - that's clearly impossible here since $\delta$ is not even a function. $\endgroup$ – David C. Ullrich Jun 30 at 23:14
  • $\begingroup$ That's my question. Why does $f_n→\delta$? $\endgroup$ – Salinas Jun 30 at 23:19
2
$\begingroup$

I suppose $u_n$ is same as $f_n$. We have $\int f_n(x)\phi(x)dx=\int \frac 1 {\pi (1+y^{2})} \phi(\frac y n)dy$ (by the substitution $y=nx$). By DCT we get the limit as $\phi(0)$ (because $\frac 1 {\pi (1+y^{2})}$ is integrable). Since $\int \phi d\delta=\delta( \phi)$ the result follows.

$\endgroup$
0
$\begingroup$

Hint: Choose $A>0$ so $|\phi(x)-\phi(0)|<\epsilon$ if $|x|<A$. Then $$\phi(0)-\int\phi(x)f_n(x)=\int(\phi(0)-\phi(x))f_n(x)=\int_{|x|<A}+\int_{|x|>A}.$$

Now $$\left|\int_{|x|<A}\right|<\epsilon\int_{|x|<A}f_n<\epsilon\int f_n=\epsilon,$$while $$\left|\int_{|x|>A}\right|<2||\phi||_\infty\int_{|x|>A}f_n.$$Show that $\int_{|x|>A}f_n\to0$ as $n\to\infty$.

$\endgroup$

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.