Space of bounded continuous functions is complete

I have lecture notes with the claim (Cb(X),, 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.


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.

Source : Link , Question Author : Rudy the Reindeer , Answer Author : Rudy the Reindeer

Leave a Comment