3
$\begingroup$

This is a problem in PMA Rudin p.165

Let $C_m=(-\frac{1}{m^2},-\frac{1}{(m+1)^2}), \forall m\in\mathbb{N}$

Define $f_n=\frac{1}{1+n^2x}$ on its natural domain,$\forall n\in \mathbb{N}$.

Define $f(x)=\sum_{n=1}^{\infty} f_n(x)$ where $x\in \mathbb{R}\setminus (\{0\} \cup \{-\frac{1}{n^2}\in \mathbb{R}|n\in\mathbb{N}\})$.

Then $f$ is well-defined.

I have shown that $\sum f_n\rightarrow f$ uniformly on $C_m$ for any $m\in\mathbb{N}$.

Now, i don't know how to prove that $\sum f_n\rightarrow f$ uniformly on $f$'s natural domain. Is there a theorem for this?

  • 1
    Looking at my copy of Rudin, I think you want $f_n(x)={1\over 1+n^2x}$. In this case: consider the values of $f_n(1/n^2)$. This should convince you that $(f_n)$ does not converge uniformly to $0$ on $(0,\infty)$. Then the series can't either. (The series does converge uniformly on intervals of the form $[a,\infty)$, $a>0$ by the $M$-test.)2012-12-19
  • 0
    I meant the series can't be uniformly convergent on $(0,\infty)$ in my previous comment.2012-12-19
  • 0
    @David Oh that's right.. It was a typo2012-12-20
  • 0
    @copper.hat it's edited now Please check it again. Thank you2012-12-20
  • 0
    What you are trying to do won't work. Consider the values of $f_n({-1\over2n^2})$ (then argue as in my previous comment).2012-12-20
  • 0
    @David i got it! Thank you2012-12-20

0 Answers 0