Section 1.5 The Space \(C_\bdd(X)\)
For a given set \(S\text{,}\) recall the space \(B(S)\) of bounded functions \(f \maps S \to \R\) with the supremum norm. If we equip \(S\) with a metric (giving rise to a metric space), we have an important subspace comprising all the continuous functions. We have already seen it in the special case where \(S\) is a bounded, closed interval. We will use a slightly different notation here, however, because the notation \(C^0(X)\) is traditionally used for something else (which coincides with the following notion when \(X\) is a closed, bounded interval).
Definition 1.48.
Let \((X,d)\) be a metric space. Then \(C_\bdd(X)\) is the normed subspace of \(B(X)\) comprising all continuous, bounded functions \(f \maps X \to \R\text{.}\)
Remark 1.49.
As \(C_\bdd(X)\) is a subset of \(B(X)\text{,}\) according to Convention 1.16 we will, unless stated otherwise, always equip it (and any of its subsets) with the supremum norm.
Theorem 1.50.
For any metric space \((X,d)\text{,}\) the space \(C_\bdd(X)\) is a Banach space.
The key to the proof of this statement is the following.
Theorem 1.51. Uniform limit theorem.
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces, and let \(\seq
fn\) be a sequence of continuous maps \(X \to Y\text{.}\) If \(\seq fn\) converges uniformly to a map \(f \maps X \to Y\) in the sense that
\begin{equation*}
\lim_{n \to \infty} \sup_{x \in X} d_Y(f_n(x),f(x)) = 0,
\end{equation*}
then \(f\) is also continuous.
Proof.
This is similar to the proof of the corresponding statement for functions on an interval. You are asked to give the details in Exercise 1.5.1.
Proof of Theorem 1.50.
We know that \(C_\bdd(X)\) is a normed space, so it suffices to verify that it is complete. Moreover, we know that \(B(X)\) is complete by Theorem 1.40, thus by Theorem 1.41, it suffices to show that \(C_\bdd(X)\) is closed in \(B(X)\text{.}\)
We prove closedness with the help of Theorem 1.35. Let \((f_n)_{n \in \N}\) be a sequence in \(C_\bdd(X)\) and suppose that we have \(f_0 \in B(X)\) with \(f_0 = \lim_{n \to \infty} f_n\) (in \(B(X)\)). Then \(f_0\) is continuous by the Uniform limit theorem. Hence \(f_0 \in C_\bdd(X)\text{,}\) and it follows that \(C_\bdd(X)\) is closed.
Exercises Exercises
1. Uniform limit of continuous maps.
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces. Suppose that \((f_n)_{n \in \N}\) is a sequence of continuous maps \(f_n \maps X \to Y\) that converges uniformly to \(f \maps X \to Y\) in the sense that
\begin{equation*}
\lim_{n \to \infty} \sup_{x \in X} d_Y(f_n(x),f(x)) = 0.
\end{equation*}
Show that \(f\) is continuous.
Hint 1.
You have seen a proof of this when \(X=Y=\R\text{;}\) the proof here is very similar. The usual way to do it is a so-called ‘\(\varepsilon/3\)’ argument.
Hint 2.
Note that when choosing \(\delta\) in the definition of continuity for the functions \(f_n\text{,}\) the value you get will depend not only on \(\varepsilon\) (and the base point \(x_0\)) but also on the index \(n \in \N\text{.}\) Worryingly, we might even have \(\delta \to 0\) as \(n \to \infty\text{!}\) A way around this is to first choose \(N \in \N\) appropriately, and then appeal to the continuity of single function \(f_N\text{.}\)
Solution.
Let \(x_0 \in X\) and fix \(\varepsilon \gt 0\text{.}\) Choose \(N \in \N\) such that \(d_Y(f_n(x), f(x)) \lt \varepsilon/3\) for all \(n \ge N\) and all \(x \in X\text{.}\) Furthermore, choose \(\delta \gt 0\) such that
\begin{equation*}
d_Y(f_N(x), f_N(x_0)) \lt \frac{\varepsilon}{3}
\end{equation*}
for all \(x \in X\) with \(d_X(x, x_0) \lt \delta\text{.}\) Then if \(d_X(x,
x_0) \lt \delta\text{,}\) we find that
\begin{align*}
d_Y(f(x), f(x_0))
\amp\le d_Y(f(x), f_N(x)) + d_Y(f_N(x), f_N(x_0)) + d_Y(f_N(x_0), f(x_0)) \\
\amp\lt \varepsilon\text{.}
\end{align*}
Therefore, the map \(f\) is continuous.
Comment.
In the above solution we first fix \(n=N\) appropriately and then choose \(\delta \gt 0\) based on the continuity of \(f_N\text{.}\) In previous years, many students tried to choose \(\delta\) based on the continuity of some general \(f_n\text{.}\) This works, up to a point, but the value \(\delta=\delta_n\) that you get will in general depend \(n\text{.}\) This is a bit worrying, since ultimately we want a single value of \(\delta \gt 0\) that works for the limiting function \(f\text{,}\) and if we are very unlucky then perhaps \(\delta_n \to 0\) as \(n \to \infty\text{.}\)
