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I am working on a problem to prove that the set $\mathcal{C}_b$ of bounded continuous functions is closed in the space of all bounded functions on a set $A \subseteq \mathbb{R}$.

This doesn't seem true to me since the set of functions $$\{ \, f_n(x) = \exp(-nx) \mid n = 0,1,2,\ldots \,\}$$ on $A = [0,1]$ is a subset of $C_b$ (on $A$), but the sequence of functions $f_0, f_1, f_2, \ldots$ converges to $$ f(x) = \begin{cases} 1 & \text{if }x=0\\ 0 & \text{otherwise} \end{cases}$$ which is not continuous but is bounded.

This implies to me that $\mathcal{C}_b$ is not closed on the space of bounded functions on $A$.

What am I missing here?

ted
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    Why do you think it is not true? (In any case, it does not make sense to talk about continuous functions on a set $A$, only on a topological space $A$.) – Eric Wofsey Sep 01 '18 at 21:03
  • Adding to Eric's remark: you must impose a topology so as to make clear what is open or closed, and/or what does "taking a limit" mean. – DonAntonio Sep 01 '18 at 21:26
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    If the toplogy is pointwise convergence, then it is false, even if $A$ is a bounded, closed interval. With different topology, the statement might be true. Add details, please. (Background, source, your thoughts, you know the drill...) – A. Pongrácz Sep 01 '18 at 21:28
  • Hi, I've added clarifications, I hope. Feedback is appreciated. – ted Sep 02 '18 at 18:42
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    It becomes true if you consider the topology induces by the supremum norm (uniform convergence). – Severin Schraven Sep 02 '18 at 18:54
  • @SeverinSchraven - Sorry, not seeing this. Care to elaborate? – ted Sep 02 '18 at 19:05
  • You have to show that the uniform limit of bounded continuous functions is again bounded and continuous. Boundedness is easy (check it) and the continuity part can be found here https://math.stackexchange.com/questions/2164642/proof-of-uniform-limit-of-continuous-functions – Severin Schraven Sep 02 '18 at 19:10
  • @SeverinSchraven - The question asks to prove that $C_b$ is closed on the space of bounded (but not necessarily continuous) functions. And the uniform limit $f(x)$ is not epsilon close to any continuous function under the supremum norm. So the statement I am asked to prove appears false, no? – ted Sep 02 '18 at 19:27
  • Check my answer. I hope it is clear now. – Severin Schraven Sep 02 '18 at 19:57

2 Answers2

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So let us get straight what we actually want to prove.

Let $A\subseteq \mathbb{R}$ be a set. We consider the space of bounded functions on the set $A$ (denote it by $L^\infty(A)$) and endow this space with the supremum norm

$$ \Vert f \Vert_{\sup} := \sup_{x\in A} \vert f(x) \vert. $$

Check that this really is a norm on the space of bounded functions on $A$. Now we consider the following subspace

$$ C_b(A) = \{ f: A \rightarrow \mathbb{R} \ : \ f \text{ is continuous and bounded} \}. $$

We want to prove that this subspace is closed. As we are dealing with normed spaces, this means the following:

For all sequences $(f_n)_{n\geq 1}\subseteq C_b(A)$ which the sequence converge in $L^\infty(A)$ to some function $f$, we actually have $f\in C_b(A)$.

So we have to take a sequence of bounded, continuous functions, which converge uniformly to some function $f$ and we need to show that this function is again bounded and continuous.

I leave boundedness for you to verify and send you to Proof of uniform limit of Continuous Functions to check out the proof that continuity is preserved under uniform convergence.

All in all, yes, the space of continuous, bounded functions is a closed subspace of the space of bounded functions with the supremum norm.

Severin Schraven
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This statement depends on a choice of topology on the space of bounded functions on $A$. If you use the topology of pointwise convergence, then your example is correct and shows that $C_b$ is not closed. However, the usual topology in this context (and presumably the inteded topology in your problem) is not the topology of pointwise convergence but the topology of uniform convergence. That is, we use the topology induced by the norm $\|f\|=\sup\{|f(x)|:x\in A\}$ (so, convergence of a sequence in this topology is uniform convergence of functions).

In particular, in your example, $f_n$ does not converge to $f$ in this topology. In fact, for all $n$, $\|f_n-f\|=1$, since for any $n$, you can choose $x$ very slightly larger than $0$ such that $f_n(x)$ is arbitrarily close to $1$ but $f(x)$ is $0$. In order for $f_n$ to converge to $f$, the distance $\|f_n-f\|$ must converge to $0$.

Eric Wofsey
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  • Thanks all. Now I get it -- it depends on the notion of distance (topology). So I really need to keep the sup norm in mind when I think about sequences of functions. – ted Sep 03 '18 at 00:30