3
$\begingroup$

The space of bounded continuous functions from some space to a complete metric space is complete. Now, if we consider the sequence $\lbrace f_n(x)=x^n \mid f_n \colon [0,1]\to [0,1] \rbrace_n^{\infty}$, then $f_n$ is continuous and bounded for all $n$, and converges to a discontinuous function. Since $\lbrace f_n \rbrace$ converges, it is Cauchy. But $\{f_n\}\subset B_C(\mathbb{R})$ (where $B_C (X)$ is the space of bounded continuous functions from $X$ to $X$). How does that comply with the fact that $B_C(\mathbb{R})$ is complete?

Clearly, I'm missing something here. I'd appreciate an explanation. Thanks.

$\endgroup$
4
  • 3
    $\begingroup$ This is what you are missing: Which topology on $B_C(\Bbb R)$ is considered? $\endgroup$ Jul 10, 2018 at 10:42
  • 3
    $\begingroup$ Are you sure that the sequence converges uniformly? $\endgroup$
    – egreg
    Jul 10, 2018 at 10:43
  • $\begingroup$ @HagenvonEitzen Gee... $B_C$ with the sup norm! I completely forgot about that. Thanks! $\endgroup$ Jul 10, 2018 at 10:49
  • $\begingroup$ The first sentence in the question is not generally true. You need that the functions map to a complete metric space. $\endgroup$ Jul 11, 2018 at 2:58

1 Answer 1

3
$\begingroup$

The space of bounded continuous functions is complete with respect to the metric induced by the following {supremum} norm:

$$\|f\|=\sup_{x \in E}|f(x)|$$

where $f: E \to \mathbb{R}$ or $\mathbb{C}$. You can check that this indeed gives us a norm. Now define $d(f,g)=\|f-g\|$ to get a metric on the space of bounded, continuous functions.

When you talk about a metric space, you must know what metric you are working with. Surely, if $f_n \to f$ in the norm given above, $f$ will be continuous. (this is just another way to state the standard theorem that the limit of a uniformly convergent sequence of continuous functions is continuous)

$\endgroup$
1
  • 2
    $\begingroup$ Thanks. Would have voted, but I'm a newbie (in math too) so can't... $\endgroup$ Jul 10, 2018 at 10:51

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.