1
$\begingroup$

Let $\{f_n\}$ be a sequence of function defined and bounded on $\mathbb R$, and suppose that $\{f_n\} $converges uniformly to $f$ on every finite interval $[a,b]$. Prove or disprove that $$\lim_{n→∞}\sup_{x∈R}⁡f_n (x) =\sup_{x∈R}⁡f(x).$$

I know that $\{f_n\} $converges uniformly to $f$ that mean $\lim_{n\to \infty}\sup |f_n -f| \to0$. But it's not necessary that $\lim_{n→∞}⁡\sup_{x∈R}⁡f_n (x) =\sup_{x∈R}⁡f(x)$.

I don't think this is true, But I can't find counter example either.

$\endgroup$
  • $\begingroup$ Isn't $f_{n}(x) = I(x=n)$ a counter-example? $\endgroup$ – madprob Feb 3 '14 at 21:08
2
$\begingroup$

Let $$ f_n = \left\{ \begin{array}{cc} 1 & \text{ if } x \ge n \\ 0 & \text{ if } x < n \\ \end{array} \right. $$

Given any finite interval $[a,b]$ and any $\epsilon > 0$, just pick $N > b$ and we have that $f_n = 0$ on $[a,b]$ for $n > N$. Thus $f_n \to {\textbf{0}}$ uniformly on $[a,b]$.

$\endgroup$
  • $\begingroup$ I cannot understand why it is downvoted. $\endgroup$ – Sangchul Lee Feb 3 '14 at 21:36
  • $\begingroup$ @sos440 I've receieved 4 or 5 downvotes in the past half hour on some of my recent answers. They are probably hate-downvotes from the same user. $\endgroup$ – 6005 Feb 3 '14 at 21:37
  • $\begingroup$ I cancelled out that suspiciously malicious voting. Hope this will make you feel better. $\endgroup$ – Sangchul Lee Feb 3 '14 at 21:46
  • $\begingroup$ @sos440 thanks :D $\endgroup$ – 6005 Feb 3 '14 at 21:51

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.