Rudin $7.14$ linked to $7.9$ why does Rudin make boundedness compulsory in uniform convergence?

Question

In Rudin $7.14$, Rudin says

Theorem $7.9$ can be rephrased as follow: a sequence $\{f_n\}$ converge to $f$ with respect to the metric $\mathscr C(X)$ if and only if $f_n \rightarrow f$ uniformly on $X$

Theorem $7.9$ states as follow:

Put $$M_n = \sup_{x\in E} |f_n (x) - f(x)|.$$ Then, $f_n \rightarrow f$ uniformly if and only if $M_n \rightarrow 0$.

$\mathscr C (X)$ is defined as the set of all the complex-valued, continuous and bounded functions with domain $X$.

What I don't understand is why does theorem $7.9$ can be rephrased as above?

I feel that there is an extra information, namely boundedness of the functions $f_n$ and of the function $f$, in "convergence with respect to the metric $\mathscr C (X)$" that wasn't present in theorem $7.9$.

The only attempt I can give at figuring this out is that maybe uniform convergence implies boundedness of the functions $f_n$ and $f$ (but I am absolutely not sure of it).

Edit: thanks for your answer guys! I've had been stuck with this all the afternoon!

Answer 1

The problem is that the usual $\sup$-norm $\lVert - \rVert$ is only defined on the set $\mathscr C(X)$ of continuous bounded functions. This norm induces a metric on $\mathscr C(X)$ via $d(f,g) = \lVert f - g \rVert$. This gives us a concept of convergence in $\mathscr C(X)$ as in any metric space. It has the property that a sequence $(f_n)$ in $\mathscr C(X)$ converges to $f \in \mathscr C(X)$ (with respect to $d$) iff $f_n \to f$ uniformly on $X$.

However, the concept of uniform convergence introduced in Definition 7.7 is not restricted to continuous bounded functions. It can be defined for each sequence of functions $f_n : E \to \mathbb K$ living on an arbitary set $E$. Note that Rudin only considers metric spaces $E$ and focuses on continuous functions (see Theorem 7.12).

Theorem 7.9 is of course correct, but to rephrase it in terms of metric convergence it does not suffice to work with $\mathscr C(X)$. Here are some possiblea approaches.

1. Consider the sets $\mathscr F(E)$ of all functions $f : E \to \mathbb K$ and $\mathscr B(E)$ of all bounded functions $f : E \to \mathbb K$. Here $E$ is any set. $\mathscr F(E)$ is a vector space over $\mathbb K$ and $\mathscr B(E)$ is a linear subspace.
   Clearly $\lVert f \rVert = \sup_{x \in E} \lvert f(x) \rvert$ defines a norm on $\mathscr B(E)$. If $E$ is a metric space, then it contains $\mathscr C(E)$ as a linear subspace.
   For a sequence $(f_n)$ in $\mathscr F(E)$ we have $f_n \to f \in \mathscr F(E)$ uniformly on $E$ iff $f_n - f \in \mathscr B(E)$ for $n \ge n_0$ and $\lim_{n \to \infty} \lVert f_n-f \rVert = 0$.

2. On $\mathscr F(E)$ define a metric $d'$ by $$d'(f,g) = \min (\sup_{x \in E} \lvert f(x) - g(x) \rvert, 1) .$$ It is easy to verify that $d'$ is indeed a metric and that $f_n \to f$ uniformly on $E$ iff $f_n \to f$ with respect to $d'$.
   Moreover, on $\mathscr B(E)$ the metric $d'$ is equivalent to the metric $d$ induced by the $\sup$-norm.

3. Redefine the concept of a metric on a set $E$ by allowing that $d(x,y) = \infty$. Then we can define on $\mathscr F(E)$ $$d''(f,g) = \sup_{x \in E} \lvert f(x) - g(x) \rvert.$$ This is a metric in the generalized sense and we have $f_n \to f$ uniformly on $E$ iff $f_n \to f$ with respect to $d''$.
   Moreover, the metrics $d'$ and $d''$ are equivalent.

Answer 2

Adding boundedness to the assumptions is what makes $\mathscr{C}(X)$ a metric space, since it prevents infinite distances between functions, and even a normed vector space (a complete one at that), but uniform convergence can still of course be defined for unbounded functions.