Space of bounded continuous functions is complete

Question

I have lecture notes with the claim $(C_b(X), \|\cdot\|_\infty)$, the space of bounded continuous functions with the sup norm is complete.

The lecturer then proved two things, (i) that $f(x) = \lim f_n (x)$ is bounded and (ii) that $\lim f_n \in \mathbb{R}$.

I don't understand why it's not enough that $f$ is bounded. I think the limit of a sequence of continuous functions is continuous and then if $f$ is bounded, it's in $C_b(X)$. So what is this $\lim f_n \in \mathbb{R}$ about? Many thanks for your help.

(i) "$f(x) = \lim f_n(x)$ is bounded" is a terrible way of saying what you want to say - there is a reason why we try to stress the difference between a function and its value at a point. (ii) And what do you mean by "$\lim f_n \in \mathbb{R}$"?! — kahen, Oct 09 '11 at 15:21
I don't understand too: I think first we have to show that for all $x\in X$ the sequence ${f_n(x)}$ has a limit, then put $f(x):=\lim_{n\to\infty}f_n(x)$ and finally check that $f$ is continuous, bounded and that the sequence ${f_n}$ converges to $f$ for the $\sup$ norm. — Davide Giraudo, Oct 09 '11 at 15:22
What you typically prove is that $B(X)$ is a Banach space and that $C_b(X)$ is a closed subspace of $B(X)$. This you do by showing that if $(f_n)$ in $B(X)$ converges uniformly to a function $f: X \to \mathbb{R}$ and the $f_n$ are all continuous at an $x \in X$ then $f$ is continuous at $x$ as well. — kahen, Oct 09 '11 at 15:26
What do you mean by saying "I think the limit of a sequence of continuous functions is continuous," what kind of limit? Never ever say the word limit again without saying what kind of limit you have in mind. It is true that the uniform limit of continuous functions is continuous. However, it is far from true for pointwise limits, or $L^p$-limits. — t.b., Oct 09 '11 at 15:29
@kahen: $B(X)$ (very unfortunately) already denotes already the space of continuous bounded functions here, so your comment does not really make sense :) You probably have $B(X) = \ell^{\infty}(X)$ in mind. — t.b., Oct 09 '11 at 15:30
@t.b.: my mentor (if I may still call you that) just virtually smacked my fingers with a ruler. I won't use the word limit again without saying which limit I'm talking about. : ) — Rudy the Reindeer, Oct 09 '11 at 15:39
Sure, that's fine with me. It's not the first time you do that handwaving with limits, and it's not the first time that I complain :) To come back on topic, @Davide essentially outlined the argument: From the fact that $f_n$ is uniformly Cauchy, you conclude that $f_n(x)$ is Cauchy in $\mathbb{R}$ for each $x$ (since $|f_n(x) - f_m(x)| \leq |f_n - f_m|\infty$), so it has a limit point $f(x)$ by completeness of $\mathbb{R}$. A priori you now know that $f_n \to f$ pointwise, and you want two more things: 1. $f$ is continuous and bounded, 2. $|f_n - f|\infty \to 0$. — t.b., Oct 09 '11 at 15:48
Use the fact that for a fixed $x$ the sequence of real numbers ${f_n}$ is a Cauchy sequence, hence convergent. It defines $f$. Now you have to show that the sequence ${f_n$ converges uniformly to $f$ on $X$. — Davide Giraudo, Oct 09 '11 at 15:49
@t.b. Actually there is only one kind of limit; there are, however, different kinds of convergence to that limit. — ashpool, Aug 17 '19 at 00:00

Rudy the Reindeer · Accepted Answer · 2012-07-08T19:50:26.057

84

Let $(B(X), \|\cdot\|_\infty)$ be the space of bounded real-valued functions with the sup norm. This space is complete.

Proof: We claim that if $f_n$ is a Cauchy sequence in $\|\cdot\|_\infty$ then its pointwise limit is its limit and in $B(X)$, i.e. it's a real-valued bounded function:

Since for fixed $x$, $f_n(x)$ is a Cauchy sequence in $\mathbb R$ and since $\mathbb R$ is complete its limit is in $\mathbb R$ and hence the pointwise limit $f(x) = \lim_{n \to \infty } f_n(x)$ is a real-valued function. It is also bounded: Let $N$ be such that for $n,m \geq N$ we have $\|f_n - f_m\|_\infty < \frac{1}{2}$. Then for all $x$

$$ |f(x)| \leq |f(x) - f_N(x)| + |f_N(x)| \leq \|f - f_N \|_{\infty} + \|f_N \|_{\infty}$$

where $\|f - f_N \|_{\infty} \leq \frac12$ since for $n \geq N$, $ |f_n(x) - f_N(x)| < \frac12$ for all $x$ and hence $|f(x) - f_N(x)| = |\lim_{n \to \infty} f_n(x) - f_N(x)| = \lim_{n \to \infty} |f_n(x) - f_N(x)| \color{\red}{\leq} \frac12$ (not $<$!) for all $x$ and hence $\sup_x |f(x) - f_N(x)| = \|f-f_N\|_\infty \leq \frac12$.

To finish the proof we need to show $f_n$ converges in norm, i.e. $\|f_N - f\|_\infty \xrightarrow{N \to \infty} 0$:

Let $\varepsilon > 0$. Let $N$ be such that for $n,m \geq N$ we have $\|f_n-f_m\|_\infty < \varepsilon$. Then for all $n \geq N$

$$ |f(x) - f_n(x)| = \lim_{m \to \infty} |f_m(x) - f_n(x)| \leq \varepsilon $$

for all $x$ and hence $\|f- f_n\|_\infty \leq \varepsilon$.

edited Jul 08 '12 at 19:50

answered Jul 08 '12 at 07:29

Rudy the Reindeer

48,632

1

Right. That's basically okay now, up to the last paragraph where you'd want $f_n$ instead of $f_N$. – t.b. Jul 08 '12 at 12:15
3

However, you still make this mistake of writing $\lim$ before you justify that the limit exists (twice). I wouldn't start with "Let $f(x) := \lim_{n\to\infty}f_n(x)$." since this doesn't make sense, yet, and is only (somewhat) justified by the next sentence. I'd write "For every $x_0$ the sequence $(f_n(x_0)){n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ because $\lvert f_n(x_0) - f_m(x_0)\rvert \leq \lVert f_n(x_0) - f_m(x_0)\rVert\infty$, so there is $f(x_0) \in \mathbb{R}$ such that $f_n(x_0) \xrightarrow{n \to \infty} f(x_0)$. – t.b. Jul 08 '12 at 12:15
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Worse is the second instance at the end: you write $\lim_{m\to\infty} \lvert f_m(x) - f_N(x)\rvert \leq \lim_{m\to\infty} \lVert f_m - f_N\rVert_\infty$, but you don't know yet that the second limit exists (it would be okay if you wrote $\limsup_{m\to\infty}$ instead of $\lim$.) I'd simply say "For all $x$ we have $\lvert f_m(x) - f_n(x)\rvert \leq \lVert f_m-f_n\rVert_\infty \leq \varepsilon$, so $\lvert f(x) - f_n(x)\rvert = \lim_{m\to\infty} \lvert f_m(x) - f_n(x)\rvert \leq \varepsilon$ and hence $\lVert f - f_n\rVert_\infty \leq \varepsilon$ for all $n \geq N$." – t.b. Jul 08 '12 at 12:16
Finally a minor point: you still repeat (essentially) the same argument twice. Once with $\frac{1}{2}$ and once with $\varepsilon$, so you could combine the two steps into one. – t.b. Jul 08 '12 at 12:17
@t.b. But even if the limit does not exists I can assign it to $f(x)$. The thing $f(x)$ might be undefined for some $x$ but I can give it a name. No? – Rudy the Reindeer Jul 08 '12 at 13:22
So $f$ is an assignment from $\mathbb{R}$ to undefined expressions that will be made sense of later? Come on... :) – t.b. Jul 08 '12 at 13:32
@t.b. Yes. Like variable assignment in programming. You can assign undefined things. – Rudy the Reindeer Jul 08 '12 at 13:33
Be that as it may, I will of course do as I'm told and rewrite it. : ) – Rudy the Reindeer Jul 08 '12 at 13:35
@t.b. I'm getting there. I think now I have what I wanted to write in the first place. Especially the last (duplicate) part since that stays the same in the proof of $(C(K), |\cdot|_\infty)$ is complete. – Rudy the Reindeer Jul 08 '12 at 18:36
@t.b. Though I'm still slightly unsure about the red "<". I get proper less than, don't I? – Rudy the Reindeer Jul 08 '12 at 18:38
No, you don't. You're taking the limit of a Cauchy sequence in $[0,1/2)$. This limit is an element of the closure $[0,1/2]$, so you only get $0 \leq \cdots \mathbin{\color{red}{\leq}} 1/2$ (you can get an easy examples using constant functions $f_n= 1/n$). You still should replace $N$ by $n$ in the last paragraph. The way you phrase it you only show that "for every $\varepsilon \gt 0$ there exists $N$ such that $|f - f_N|_\infty \leq \varepsilon$. What you want to show is that for every $\varepsilon \gt 0$ there exists $N$ such that for all $n \geq N$ you have $|f-f_n| \leq \varepsilon$. – t.b. Jul 08 '12 at 19:23
@t.b. That's what I thought originally then I thought about it for a while confused my self and changed it : / Of course $\leq$. As for the other thing: I see now. It was a flaw in basic logic. – Rudy the Reindeer Jul 08 '12 at 19:41
I'll read it again and check it in a few weeks. Last time you let me get away with a muddled thing. – Rudy the Reindeer Jul 08 '12 at 19:44
Okay, point taken, but I guess I was getting tired of reading the same thing over and over again. Sorry about that... There's a minor typo at the end of the second paragraph of the proof: $|\cdots|$ instead of $|\cdots|_\infty$. Another small cosmetic point: you write $x_0$ at the beginning of the proof, later it's always $x$. – t.b. Jul 08 '12 at 19:47
@t.b. Sure. Sorry for tiring you out. – Rudy the Reindeer Jul 08 '12 at 19:48
Last thing for today: I'd replace $n \gt N$ by $n \geq N$ (twice) -- I guess that's a leftover from trying to show $\color{red}{\lt}$. – t.b. Jul 08 '12 at 19:51
2

@t.b. No that's just a regular typo. – Rudy the Reindeer Jul 08 '12 at 19:52
How are you justifying $|\lim_{n \to \infty} f_n(x) - f_N(x)| = \lim_{n \to \infty}|f_n(x) - f_N(x)|$? – ManUtdBloke Sep 23 '16 at 07:55
2

@eurocoder The norm is continuous. – Rudy the Reindeer Sep 24 '16 at 11:52

Rudy the Reindeer · Answer 2 · 2012-01-08T08:39:45.233

39

To show that $(C_b(X), \| \cdot \|_\infty)$ is complete we first show that there is a pointwise limit function in $\mathbb{R}$ to which $f_n$ converges. For this we note that because $f_n$ is Cauchy with respect to the sup norm, it follows that $f_n(x)$ is a Cauchy sequence in $\mathbb{R}$ for any $x$ in $X$. But $\mathbb{R}$ is complete and hence the limit $\lim_{n \to \infty} f_n (x)$ is in $\mathbb{R}$.

Now let $f(x)$ denote the pointwise limit function of $f_n$. We now want to show that $f$ is bounded, that is, there exists a real constant $K$ such that $\| f \|_\infty < K$. For this we again use that $f_n$ is Cauchy with respect to the sup norm: For every $\varepsilon > 0$ we can find an $N$ such that for $n,m \geq N$ we have that $\| f_n - f_m \|_\infty < \varepsilon$. Using the triangle inequality we have $\| f \|_\infty \leq \| f - f_N \|_\infty + \| f_N \|_\infty$ and because $f_N$ is in $C_b(X)$ we know that there exists an $M$ in $\mathbb{R}$ sucht that $\| f_N \|_\infty \leq M$. We also have $\| f_n - f_N \|_\infty < \varepsilon$ for all $n \geq N$ and hence $\lim_{n \to \infty} \| f_n - f_N \|_\infty = \| f - f_N \|_\infty \leq \varepsilon$. Hence $f$ is bounded.

Now we want to show that $f_n$ converges to $f$ in norm, that is, $\| f - f_n \|_\infty \xrightarrow{n \to \infty} 0$. For this let $\varepsilon > 0$. Then we have that there exists an $N$ such that for $n,m \geq N$, $\| f_n - f_m \|_\infty < \frac{\varepsilon}{2}$, again because $f_n$ is Cauchy.

Using the triangle inequality we get $\| f - f_n \|_\infty \leq \| f - f_N \|_\infty + \| f_N - f_n \|_\infty \leq \varepsilon$. By the same argument as before, that is, because $f_n$ is Cauchy, $\| f_n - f_N \|_\infty < \frac{\varepsilon}{2}$ for all $n \geq N$ and hence $\lim_{n \to \infty} \| f_n - f_N \|_\infty = \| f - f_N \|_\infty \leq \frac{\varepsilon}{2}$.

So $\| f - f_n \|_\infty \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ and as $\varepsilon$ was arbitrary it follows by having $\varepsilon$ tend to $0$ that $\| f - f_n \|_\infty \xrightarrow{n \to \infty} 0$.

Finally, now that we have convergence in norm, we can apply the uniform limit theorem to get that $f$ is continuous and hence $f$ is in $C_b(X)$.

edited Jan 08 '12 at 08:39

answered Dec 20 '11 at 10:52

Rudy the Reindeer

48,632

This looks good! Just a few small things: don't use "any" in place of "every" because "any" is ambiguous. Then, $N$ has two meanings in the second paragraph. It would be a bit more elegant to choose $N$ such that for $n,m \geq N$ we have... Then $|f_N - f|\infty \leq \varepsilon$ follows from the choice of $N$ and then you can simply estimate $|f(x)| \leq \varepsilon + |f_N|\infty$, so $|f|\infty \leq |f_N|\infty + \varepsilon$. Same goes for the second part of the argument: take $N$ such that for $m,n \geq N$... – t.b. Dec 20 '11 at 11:03
I'm still a bit confused about the meaning of $B(X)$: is it the space of bounded functions or the space of bounded continuous functions? You haven't argued that (or mentioned why) $f$ is continuous, so in case the second interpretation applies you should do that. – t.b. Dec 20 '11 at 11:09
@t.b.: Oh, right. In the question it says continuous. I'll do that. Thanks for your comments! – Rudy the Reindeer Dec 20 '11 at 11:37
@t.b. Sorry but I'm not sure I understand your first comment. – Rudy the Reindeer Dec 20 '11 at 12:07
Basically I was only nitpicking over the "for some $N$" in the last sentence of the second paragraph (which says that $N$ has a different meaning from before). If you drop "some" and if you choose $N$ s.t. $n,m \mathrel{\color{red}{\geq}} N$ (as opposed to $\gt$) implies $|f_n - f_m| \leq \varepsilon$ then you can take $m = N$ and you have $|f_n - f_N| \leq \varepsilon$ by definition of $N$. // Similarly, your choice of $N$ in the part following "Now we ..." does not quite imply $|f_n - f_N| \lt \varepsilon/2$ (because you wrote $n,m \mathrel{\color{red}{\gt}} N$). – t.b. Dec 20 '11 at 14:19
Basically, I guess, it boils down to: replace $\gt$ and $\lt$ by $\geq$ and $\leq$ everywhere. One last thing: I'd add the following to "Using the triangle inequality we get $|f - f_n| \leq |f - f_N| + |f_N - f_n| \color{red}{;\leq \varepsilon}$" – t.b. Dec 20 '11 at 14:23
@t.b. Done. I think. Although I think I can leave the $< \varepsilon$ and don't have to replace them with $\leq \varepsilon$. Thank you! – Rudy the Reindeer Dec 20 '11 at 14:29
Very good, now I'm happy. I think. :) Yes you can leave the $\lt \varepsilon$ the way they are. – t.b. Dec 20 '11 at 14:32
@t.b. Hm... I'm not. I think the middle bit (showing convergence in norm) is messed up. There is no need to apply the triangle inequality. If $f_n$ is Cauchy w.r.t. $|\cdot|\infty$ we can show that $|f-f_k|\infty \to 0$ by observing that for $\varepsilon > 0$ there is $N$ such that for $n,m > N$, $|f_m-f_n|\infty < \varepsilon$ and hence in particular, for $k$ large enough, $|f-f_k|\infty = |\lim_{n \to \infty} f_n -f_k|_\infty < \varepsilon$. – Rudy the Reindeer Jul 05 '12 at 09:45
It's the same argument but I think doing a superfluous step in a proof is incorrect. – Rudy the Reindeer Jul 05 '12 at 09:47
I don't follow. How do you pull the limit out of your sleeve at the very end of your comment? You don't know yet that $f$ is the uniform limit of $(f_n)$ which you'd need for that. I think only the order of sentences is a bit off. I'd suggest the following: begin the third from last paragraph with the sentence "By the same argument as before ... for all $n \geq N$" full stop. then "Using the triangle inequality we get ..." and end with "and hence $\lim_{n\to\infty} |f - f_n|_\infty \leq \varepsilon$ (not $\varepsilon / 2$). – t.b. Jul 05 '12 at 22:47
@t.b. I don't know what you mean by "pulling" it "out of my sleeve": $f$ is the pointwise limit of $f_n$. But in this: $$|f - f_n | \leq |f - f_N | + | f_N - f_n|$$ the thing on the LHS is exactly the same as the first thing on the RHS. So if we can argue that the first term on the RHS is less than $\varepsilon$ then we can already argue that the thing on the LHS is, too. – Rudy the Reindeer Jul 06 '12 at 06:05
1

Well, what I'm actually complaining about is the unwarranted use of $\lim$ already at the end of the second paragraph (where the mess starts). What you actually proved there is that $|f|\infty \leq M + \varepsilon$. Then you should argue that for all $x$ and all $m,n \geq N$ you have $|f_n(x) - f_m(x)| \lt \varepsilon$, *so* (letting $n \to \infty$) you have for all $x$ and all $m \geq N$ that $|f(x) - f_m(x)| \leq \varepsilon$ which means that for $m \geq N$ we have $|f-f_m|\infty \leq \varepsilon$ which is indeed what you wanted to show for uniform convergence. – t.b. Jul 06 '12 at 09:28
5

The thing is that $|\cdot|\infty$ is *not* continuous with respect to pointwise convergence so something like $\lim_n |f_n|\infty = |\lim_n f_n|_\infty$ just looks horrible if you only have pointwise convergence of $f_n$ a priori. – t.b. Jul 06 '12 at 09:36
@t.b. But... I don't do $\lim_n |f_n|\infty = |\lim_n f_n|\infty$ anywhere. And I think your penultimate comment is an agreement with my complaint about the triangle inequality. : ) So we agree: we don't want the triangle inequality mentioned in the proof of $|f_n - f| \to 0$. – Rudy the Reindeer Jul 06 '12 at 09:55
I guess I can always delete the answer and rewrite it from scratch. – Rudy the Reindeer Jul 06 '12 at 09:56
and what else is that part "... and hence $\lim_{n\to\infty} |f_n - f_N|\infty = |f - f_N|\infty$." at the end of the second paragraph? Anyway, yes, as soon as you justified that (with $N$ replaced by $m \geq N$) then you're done and need no further application of the triangle inequality. The precise argument is outlined in my previous comments and then you can scratch what follows the second paragraph. I'd also recommend giving the details of the last paragraph (a quick $\varepsilon/3$-argument suffices). – t.b. Jul 06 '12 at 11:41
@t.b. Now I see what you meant : / Yes. Thanks for pointing that out. I'm glad I looked at this again. Well. Sort of. Apart from all the being miserable as a consequence of it. Thank you for your patience. – Rudy the Reindeer Jul 06 '12 at 11:52
@t.b. I have rewritten it ^ – Rudy the Reindeer Jul 08 '12 at 07:30
And I will edit the question to fit the new answer: I dropped the continuity because it distracts from what's relevant. – Rudy the Reindeer Jul 08 '12 at 07:33
@MattN. I read the discussion , and the problem i have is exactly adressed as discussion , but i am still not able to make sense , it sort of looks to me that the argument is going round about ie. in the step u want to show that $f_n$ converges to $f$ , the triangle inequality i am not able to convince myself that $f-f_N \to 0$ . Can you help me out :) . – Theorem Jan 06 '13 at 21:08
@t.b Im reading this as well.. I don't think it is correct. He is not using the pointwise definition of f. But I guess it would be true if we just put in an $\sup_x |f(x) - f_N(x)| = \sup_x |\lim f_n(x) - f_N(x)| \leq \epsilon $. But im not sure – Johan Jan 06 '13 at 22:21
My bad, I see now that this was not the accepted proof. – Johan Jan 06 '13 at 22:24

score 11 · Answer 3 · answered Oct 27 '11 at 20:07

11

Let $(f_n)_{n\geq1}$ be a Cauchy sequence in $B(X)$. Then for every fixed $x\in X$ the sequence $\bigl(f_n(x)\bigr)_{n\geq1}$ is a Cauchy sequence of real numbers, whence convergent to some real number $\xi=:f(x)$. From the definition of the norm in $B(X)$ it follows that the convergence $f_n\to f$ $\ (n\to\infty)$ is actually uniform; therefore the limit function $f$ is continuous on $X$. If $X$ is compact we are done, since then $f\in B(X)$ automatically. Otherwise we argue as follows: There is an $m$ with $\|f_n-f_m\|\leq1$ for all $n\geq m$. Therefore for each $x\in X$ we have $$|f_n(x)|\leq \|f_m\|+1=:C\qquad(n\geq m)\ ,$$ and this implies $|f(x)|\leq C$ for all $x\in X$, whence $f\in B(X)$.

answered Oct 27 '11 at 20:07

Christian Blatter

233,565

1

Doesn't this miss to show that comvergence is also uniform rather than only pointwise? – freishahiri Oct 24 '16 at 08:54
@AlexanderFrei: The sentence "From the definition of the norm $\ldots$ is actually uniform" leaves this as an exercise to the reader. Note that one has $|f_n(x)-f_m(x)|\leq|f_n-f_m|$ for each $x\in X$. – Christian Blatter Oct 24 '16 at 09:25
why did you assume that $X\subset \mathbb{R}$ and $f$ is real valued? The claim holds for any compact set $X$ and space of continuous functions on $X$? – domath Oct 26 '19 at 05:29
@stat_yale: I think my answer refers to an earlier version of the question. See the edit history. – Christian Blatter Oct 26 '19 at 07:59
@ChristianBlatter Just out of curiosity, if I am considering functions in $B(X)$ which vanish at a single fixed point, will the same proof work to prove completeness w.r.t. same norm? – Manoj Kumar Mar 04 '20 at 07:08
@ManojKumar: The proof works for all Cauchy sequences in $B(X)$. If all functions in your sequence happen to vanish at $x_0$ that is some extra information you have about your personal sequence. It implies, of course, that $f(x_0)=0$ as well. – Christian Blatter Mar 04 '20 at 07:57

score 3 · Answer 4 · 2012-07-08T22:19:19.050

3

Several people have already posted a proof of the fact that $(C_b(X),||.||_\infty)$ is complete. In my answer I will show a nice application of this fact.

Let $C_0(\Bbb{R})$ denote the space of all continuous functions from $\Bbb{R}$ to $\Bbb{R}$ such that for any $f(x) \in C_0(\Bbb{R})$, we have $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow -\infty} f(x) = 0$. Then $(C_0(\Bbb{R}),||.||_\infty)$ is complete.

Now I claim that any $f(x)$ in this space is bounded. By the condition concerning the behaviour of $f$ at $\pm \infty$, we immediately deduce that there is $M,N \in \Bbb{R}$ such that for all $x > M$, $|f(x)| < C_1$ for some $C_1$ and for all $x < N$, $|f(x) | < C_2$ for some $C_2$. Now the interval $[N,M]$ is compact and so by the extreme value theorem, there is $C_3$ such that $|f(x)|<C_3$ for all $x \in [N,M]$.

Taking $C = \max\{C_1,C_2,C_3\}$ shows that $f$ is bounded on $\Bbb{R}$ with respect to the Euclidean metric and hence bounded with respect to the sup metric $||.||$. One can show that $C_0(\Bbb{R})$ is a closed subspace of $C_b(X)$ and so by your result, it follows that $C_0(\Bbb{R})$ with the sup norm is complete.

edited Jul 08 '12 at 22:19

answered Jul 08 '12 at 08:28

1

Actually, it would be a bit nicer to say that every $f \in C_0(\mathbb{R})$ extends (uniquely) to a continuous function on the one-point compactification $\mathbb{R} \cup {\infty}$ by setting $f(\infty) = 0$, so $C_0(\mathbb{R})$ is the (maximal) ideal of $C(\mathbb{R} \cup {\infty})$ consisting of the functions vanishing at infinity. Since the evaluation homomorphism $C(K) \to \mathbb{C}$, $f \mapsto f(x)$ is continuous for every $x \in K$, the ideal of functions vanishing at $x$ is closed. – t.b. Jul 08 '12 at 12:22
@t.b. Indeed that is a much nicer way to put it, although I am not familiar with compactifications :D – Jul 08 '12 at 12:37
@t.b. You mean the evaluation homomorphism from $C(K) \rightarrow \Bbb{R}$ is continuous for every $x$, and $\Bbb{R}$ is equipped with the Euclidean topology yes? I gather we want to use the fact that ${0}$ is closed in $\Bbb{R}$ so the preimage of this point which is the maximal ideal is closed in $C_b(X)$. – Jul 08 '12 at 12:40
2

Yes. For continuity simply observe that $|f(x)| \leq |f|_\infty$, so the evaluation homomorphism is $1$-Lipschitz continuous. As for the one-point compactification of a locally compact space $X$, the open neighborhoods of $\infty$ are defined to be the complements of the compact subsets of $X$. The condition on vanishing at infinity in $\mathbb{R}$ just says that setting $f(\infty) = 0$ yields a continuous function on $\mathbb{R} \cup {\infty}$. – t.b. Jul 08 '12 at 12:46
@t.b. Finally some motivation to study functional analysis :D – Jul 08 '12 at 12:50

score 0 · Answer 5 · answered Dec 23 '25 at 16:45

Both answers by @Rudy are correct, but I find their presentation a bit convoluted. I hope to present a proof that clarifies how one can attain uniform convergence from the immediate pointwise limit in this setup.

Outline of the proof

To prove that the space of bounded functions armed with the uniform norm $(B(X), ||\cdot||_u)$ is complete, we must show that every uniformly Cauchy sequence $(f_n)_{n\ge 0}$ converges (in the uniform norm, i.e., uniformly) to a bounded function $f\in B(X)$. We proceed in three steps:

Since the sequence is uniformly Cauchy, the scalar sequence $(f_n(x))_{n\ge 0}$ is Cauchy for every fixed $x\in X$. Therefore, by the completeness of the complex numbers, the functional sequence converges pointwise to some $f$.
The uniform Cauchy property allows us to extend pointwise convergence to uniform convergence via a diagonalization argument.
We use the triangle inequality and uniform convergence to bound $|f|$ uniformly by $||f_N||_u + \varepsilon$, showing that $f$ is in this space of bounded functions.

As a consequence, $f\in B(X)$ is the uniform limit of $(f_n)_{n\ge 0}$.

Details of the proof

By assumption, the sequence $(f_n)$ is uniformly Cauchy, in other words, for every $\varepsilon > 0$ there is some index $N > 0$ such that for all $n, m\ge N$,$ \;\sup_{x\in X} |f_n(x) - f_m(x)| < \varepsilon$. This implies that for any fixed $x \in X$, the sequence of complex numbers $(f_n(x))_{n\ge 0}$ is Cauchy. Since the complex numbers are complete, this scalar sequence must converge. We define the function $f: X \to \mathbb{C}$ by this pointwise limit, that is:

$$f(x) = \lim_{n \to \infty} f_n(x).$$

We now show that $f_n$ converges to $f$ uniformly. Let $\varepsilon > 0$, and pick an $N > 0$ as in the definition of the uniform Cauchy property. Fix an arbitrary $x \in X$. Since $f_m(x) \to f(x)$ as $m \to \infty$, we can find an index $m_x \ge N$, which depends on $x$, such that $f_{m_x}(x)$ is within $\varepsilon/2$ of the limit $f(x)$. We can now use the triangle inequality to perform the diagonalization procedure:

$$|f_n(x) - f(x)| \le \underbrace{|f_n(x) - f_{m_x}(x)|}_{\text{Small by Uniform Cauchy}} + \underbrace{|f_{m_x}(x) - f(x)|}_{\text{Small by Pointwise Limit}}$$

Since both $n$ and $m_x$ are at least $N$, the first term is strictly less than $\varepsilon/2$. By our choice of $m_x$, the second term is also less than $\varepsilon/2$. Thus, $|f_n(x) - f(x)| < \varepsilon$ for all $n \ge N$. Importantly, the choice of index $N$ depended only on $\varepsilon$, not on $x$. This shows that convergence is actually uniform.

Finally, we prove that $f$ is, in fact, in the space $B(X)$. Using the same $N$ and $\varepsilon$ from the previous step, we know that $|f(x) - f_N(x)| < \varepsilon$ for all $x$. Since $f_N$ is a bounded function, we have:

$$|f(x)| \le |f_N(x)| + |f(x) - f_N(x)| \le ||f_N||_u + \varepsilon$$

Since $||f_N||_u + \varepsilon$ is a finite constant independent of $x$, $f$ is bounded, and this concludes the proof.

References

Folland, G. B. (2013). Real analysis: Modern Techniques and Their Applications. John Wiley & Sons.

Tao, T. (2023). Analysis i: Fourth Edition. Springer Nature.

Space of bounded continuous functions is complete

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Outline of the proof

Details of the proof

References