1
$\begingroup$

Take the sequence $f_n(t)=t^n$, $0\le t\le 1$. Then $\{f_n\} \subset \overline{B(0,1)}$, but we have no subsequence of $\{f_n\}$ converging in $C([0,1])$. So the unit ball is not compact in $C([0,1])$?

Could someone elaborate on this? Why do we have no subsequence of $\{f_n\}$ converging in $C([0,1])$? What does such a subsequence look like? Where does the subsequence converge if it does not converge in $C([0,1])$?

$\endgroup$
4
  • 3
    $\begingroup$ What is the metric on $C([0,1])$? $\endgroup$ May 24 '16 at 8:23
  • $\begingroup$ @MatiasHeikkilä It has to be the $\|\cdot\|_{\infty} = \sup_{x \in [0,1]}$ $\endgroup$
    – C.S.
    May 24 '16 at 8:32
  • $\begingroup$ The supremum norm. $\endgroup$
    – csss
    May 24 '16 at 8:32
  • 2
    $\begingroup$ Ah, ok. Then I think you should compare what happens at $t < 1$ and $t=1$. $\endgroup$ May 24 '16 at 8:34
1
$\begingroup$
  • Note $$f(x) = \lim_{x \to \infty} f_{n}(x) = \left\{\begin{array}{cc} 0, & 0\leq x < 1 \\ 1, & x =1 \end{array}\right.$$ Can you see now why $f \not \in C[0,1]$
$\endgroup$

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.