Space of bounded continuous functions is complete

Solution 1:

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

Solution 2:

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 $C_b(X)$.

Solution 3:

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 $f\in B(X)$.

Solution 4:

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.