1
$\begingroup$

I was attempting this qualifying problem. $\{f_n\}_{n=1}^{\infty}$ is a sequence of real valued functions on $\mathbb{R}$. If $ f_n $ converges to $f$ uniformly then $f$ must be continuous.

I am little wondering about this problem . I know uniform limit of sequence of continuous functions must be continuous.

If it has to do something with 'real valued functions on $\mathbb{R}$', then I might be missing that trick , otherwise I do not see the reason that the statement is true. But I found this problem as one the past qualifying exam on the website of the university.

Anyone has some idea about this?

Thanks

$\endgroup$
3
  • $\begingroup$ Maybe every $f_n$ has to be continuous. $\endgroup$ Jul 3 '15 at 2:34
  • $\begingroup$ If that was the case , there would not be any problem. I thought the same thing at the beginning. $\endgroup$
    – mp100
    Jul 3 '15 at 2:36
  • $\begingroup$ On the other hand, if it is not the case, there are trivial counter-examples. $\endgroup$ Jul 3 '15 at 2:37
2
$\begingroup$

Take any discontinuous function $f$ and define the constant sequence $f_{n}=f$ for all $n\in\mathbb{N}$. Then $(f_{n})_{n=1}^{\infty}$ converges uniformly to $f$, but $f$ is not continuous. So the claim must be assuming that each $f_{n}$ is continuous.

$\endgroup$
0
$\begingroup$

that's wrong. Take $$ f_n(x) = \begin{cases} 1, & x = 0, \\ 0, & x \ne 0. \end{cases} $$ Then, $f_n$ converges uniformly to $f_1$, which is not continuous.

$\endgroup$
4
  • $\begingroup$ I did not get your explanation. $\endgroup$
    – mp100
    Jul 3 '15 at 2:36
  • $\begingroup$ OP is saying the $f_i$ must be continuous $\endgroup$
    – user217285
    Jul 3 '15 at 2:40
  • $\begingroup$ @Nitin Hasn't he $\endgroup$
    – user251257
    Jul 3 '15 at 2:41
  • $\begingroup$ My comment was directed at @mp100 $\endgroup$
    – user217285
    Jul 3 '15 at 2:41

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.