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.

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    $\begingroup$ (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 "$\lim f_n \in \mathbb{R}$"?! $\endgroup$ – kahen Oct 9 '11 at 15:21
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    $\begingroup$ 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. $\endgroup$ – Davide Giraudo Oct 9 '11 at 15:22
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    $\begingroup$ 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. $\endgroup$ – t.b. Oct 9 '11 at 15:29
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    $\begingroup$ @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. $\endgroup$ – t.b. Oct 9 '11 at 15:30
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    $\begingroup$ @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. : ) $\endgroup$ – Rudy the Reindeer Oct 9 '11 at 15:39

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 $\|f- f_n\|_\infty \leq \varepsilon$.

  • $\begingroup$ Right. That's basically okay now, up to the last paragraph where you'd want $f_n$ instead of $f_N$. $\endgroup$ – t.b. Jul 8 '12 at 12:15
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    $\begingroup$ 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 $f_n(x_0) \xrightarrow{n \to \infty} f(x_0)$. $\endgroup$ – t.b. Jul 8 '12 at 12:15
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    $\begingroup$ 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 $n \geq N$." $\endgroup$ – t.b. Jul 8 '12 at 12:16
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    $\begingroup$ @t.b. No that's just a regular typo. $\endgroup$ – Rudy the Reindeer Jul 8 '12 at 19:52
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    $\begingroup$ @eurocoder The norm is continuous. $\endgroup$ – Rudy the Reindeer Sep 24 '16 at 11:52

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 $C_b(X)$.

  • $\begingroup$ 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 $m,n \geq N$... $\endgroup$ – t.b. Dec 20 '11 at 11:03
  • $\begingroup$ 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. $\endgroup$ – t.b. Dec 20 '11 at 11:09
  • $\begingroup$ @t.b.: Oh, right. In the question it says continuous. I'll do that. Thanks for your comments! $\endgroup$ – Rudy the Reindeer Dec 20 '11 at 11:37
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    $\begingroup$ 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. $\endgroup$ – t.b. Jul 6 '12 at 9:28
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    $\begingroup$ 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. $\endgroup$ – t.b. Jul 6 '12 at 9:36

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 $f\in B(X)$.

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    $\begingroup$ Doesn't this miss to show that comvergence is also uniform rather than only pointwise? $\endgroup$ – C-Star-W-Star Oct 24 '16 at 8:54
  • $\begingroup$ @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 $x\in X$. $\endgroup$ – Christian Blatter Oct 24 '16 at 9:25
  • $\begingroup$ 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 $X$? $\endgroup$ – domath Oct 26 '19 at 5:29
  • $\begingroup$ @stat_yale: I think my answer refers to an earlier version of the question. See the edit history. $\endgroup$ – Christian Blatter Oct 26 '19 at 7:59
  • $\begingroup$ @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? $\endgroup$ – Manoj Kumar Mar 4 '20 at 7:08

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.

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    $\begingroup$ 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. $\endgroup$ – t.b. Jul 8 '12 at 12:22
  • $\begingroup$ @t.b. Indeed that is a much nicer way to put it, although I am not familiar with compactifications :D $\endgroup$ – user38268 Jul 8 '12 at 12:37
  • $\begingroup$ @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 $C_b(X)$. $\endgroup$ – user38268 Jul 8 '12 at 12:40
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    $\begingroup$ 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 $\mathbb{R} \cup \{\infty\}$. $\endgroup$ – t.b. Jul 8 '12 at 12:46
  • $\begingroup$ @t.b. Finally some motivation to study functional analysis :D $\endgroup$ – user38268 Jul 8 '12 at 12:50

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