Proof of Completeness Under the Sup Norm

Why Continuity Can Survive Completion

The space of all bounded functions on a nonempty set is complete under the supremum norm. A natural question is whether imposing continuity changes that conclusion. A Cauchy sequence of bounded continuous functions has a uniform limit in the bounded-function space, but to conclude that the continuous functions themselves form a complete space, we must also know that the limit remains continuous.

Earlier in this course, the Uniform Limits of Continuous Functions Are Continuous theorem established precisely this preservation property. Here we combine it with the completeness of bounded function spaces and the theorem on closed subspaces of complete normed spaces. This gives a proof of completeness for bounded continuous functions without needing any compactness assumption on the domain.

Definition: Let $E$ be a nonempty subset of $\mathbb{R}$. Write $C_b(E)$ for the vector space of bounded continuous functions $f:E\to\mathbb{R}$, where continuity is understood relative to $E$. Equip this space with the supremum norm $$ \|f\|_\infty=\sup_{x\in E}|f(x)|. $$

This is a norm because it is the supremum norm on the vector space $B(E)$ of all bounded real-valued functions on $E$, and $C_b(E)$ is a linear subspace of $B(E)$. In particular, sums and scalar multiples of bounded continuous functions are bounded and continuous. The supremum is finite for every $f\in C_b(E)$ because $f$ is bounded.

The Closedness Argument

The key step is to show that a norm limit of functions in $C_b(E)$, when taken in $B(E)$, still belongs to $C_b(E)$. Convergence in the supremum norm is uniform convergence, so the earlier theorem on uniform limits supplies continuity of the limit. Boundedness is already part of being a limit in $B(E)$.

Theorem (Closedness of Bounded Continuous Functions): Let $E$ be a nonempty subset of $\mathbb{R}$. Then $C_b(E)$ is a closed linear subspace of $B(E)$, equipped with the supremum norm.

Proof. We have already noted that $C_b(E)$ is a linear subspace of $B(E)$. To prove that it is closed, let $(f_n)$ be a sequence in $C_b(E)$ that converges in $B(E)$ to some $f\in B(E)$. Thus $$ \|f_n-f\|_\infty\longrightarrow 0. $$ For every $x\in E$, $$ |f_n(x)-f(x)|\leq\|f_n-f\|_\infty. $$ The right-hand side does not depend on $x$ and tends to zero, so $f_n$ converges uniformly to $f$ on $E$. Each $f_n$ is continuous on $E$. By the Uniform Limits of Continuous Functions Are Continuous theorem, $f$ is continuous on $E$. Also $f\in B(E)$ by the choice of the limit, so $f$ is bounded. Therefore $f\in C_b(E)$. Every convergent sequence in $C_b(E)$ with limit in $B(E)$ has its limit in $C_b(E)$, which proves closedness. $\square$

Theorem (Completeness of Bounded Continuous Functions): Let $E$ be a nonempty subset of $\mathbb{R}$. Then $C_b(E)$, with the supremum norm, is complete.

Proof. The space $B(E)$ is complete under the supremum norm by the Completeness of Bounded Functions theorem. The Closed Subspaces of Complete Normed Spaces theorem says that a closed linear subspace of a complete normed space is complete in the inherited norm. By the Closedness of Bounded Continuous Functions theorem, $C_b(E)$ is such a subspace of $B(E)$. Its inherited norm is exactly $\|\cdot\|_\infty$, so $C_b(E)$ is complete. $\square$

The proof has two logically distinct parts. Completeness of $B(E)$ ensures that a Cauchy sequence has a limit among bounded functions. Closedness ensures that this limit still satisfies the extra condition of continuity. Neither part can simply be omitted: completeness of a larger space does not automatically give completeness of every subspace.

Worked Examples

Worked Example: A Uniform Limit on the Whole Real Line

For each positive integer $n$, define $f_n:\mathbb{R}\to\mathbb{R}$ by $$ f_n(x)=\arctan\left(x+\frac1n\right). $$ Each $f_n$ is continuous and bounded, since its values lie between $-\pi/2$ and $\pi/2$. Define $f(x)=\arctan x$, which is also bounded and continuous. The derivative of $\arctan t$ is $1/(1+t^2)$, whose absolute value is at most $1$ for every real $t$. The Mean Value Theorem therefore gives, for every $x\in\mathbb{R}$, $$ |f_n(x)-f(x)| =\left|\arctan\left(x+\frac1n\right)-\arctan x\right| \leq\frac1n. $$ Taking the supremum over $x$ shows that $\|f_n-f\|_\infty\leq1/n$, which tends to zero. Thus the sequence converges in $C_b(\mathbb{R})$. The estimate is uniform even though the domain is unbounded.

Worked Example: A Bounded Continuous Function on a Noncompact Domain

Let $E=(0,1)$ and define $g:E\to\mathbb{R}$ by $$ g(x)=\sin\left(\frac1x\right). $$ The function $x\mapsto 1/x$ is continuous on $(0,1)$, and the sine function is continuous, so $g$ is continuous on $E$. Also $|g(x)|\leq1$ for every $x\in E$, hence $g\in C_b((0,1))$ and $\|g\|_\infty\leq1$. In fact, $g$ takes the value $1$ at $x=1/(\pi/2+2\pi k)$ for all sufficiently large integers $k$, so $\|g\|_\infty=1$.

This function has no continuous extension to $x=0$. For sufficiently large integers $k$, set $$ x_k=\frac{1}{\pi/2+2\pi k}, \qquad y_k=\frac{1}{3\pi/2+2\pi k}. $$ Both sequences lie in $(0,1)$ and tend to zero, while $g(x_k)=1$ and $g(y_k)=-1$. Thus the values do not approach a single limit at the missing endpoint. This does not prevent $g$ from belonging to $C_b((0,1))$: continuity is required only at points of the domain. The completeness theorem applies to this noncompact domain just as it does to a compact one.

Worked Example: A Cauchy Sequence in a Vanishing-at-Zero Subspace

For $n\geq1$, define $h_n:\mathbb{R}\to\mathbb{R}$ by $$ h_n(x)=\left(1-\frac1n\right)\frac{x}{1+|x|}. $$ The map $x\mapsto x/(1+|x|)$ is continuous on $\mathbb{R}$, has absolute value at most $1$, and equals zero at $x=0$. Therefore each $h_n$ is bounded and continuous and satisfies $h_n(0)=0$. Let $$ h(x)=\frac{x}{1+|x|}. $$ Then $h$ has the same properties. For every $x\in\mathbb{R}$, $$ |h_n(x)-h(x)| =\frac1n\frac{|x|}{1+|x|} \leq\frac1n. $$ Moreover, $|x|/(1+|x|)$ approaches $1$ as $|x|$ tends to infinity, so its supremum on $\mathbb{R}$ is $1$. Consequently, $$ \|h_n-h\|_\infty=\frac1n. $$ The sequence converges in the supremum norm to a bounded continuous function that still vanishes at zero.

A Complete Space with a Prescribed Zero Set

The reasoning above also applies when a function is required to vanish on a specified part of the domain. This constraint is useful when functions are subject to boundary or interpolation conditions. The result follows by checking that the constraint is preserved under limits in the supremum norm.

Theorem (Completeness with Prescribed Zero Values): Let $E$ be a nonempty subset of $\mathbb{R}$, and let $A\subseteq E$. Define $$ C_{b,A}(E)=\{f\in C_b(E): f(x)=0\text{ for every }x\in A\}. $$ Then $C_{b,A}(E)$ is complete in the supremum norm.

Proof. The set $C_{b,A}(E)$ is a linear subspace of $C_b(E)$: sums and scalar multiples of functions that vanish on $A$ also vanish on $A$. We show that it is closed in $C_b(E)$. Suppose $f_n\in C_{b,A}(E)$ and $f_n\to f$ in $C_b(E)$. For any $x\in A$, $f_n(x)=0$, and $$ |f(x)|=|f(x)-f_n(x)|\leq\|f-f_n\|_\infty. $$ The right-hand side tends to zero, so $f(x)=0$. This holds for every $x\in A$, hence $f\in C_{b,A}(E)$. Thus $C_{b,A}(E)$ is closed in $C_b(E)$. Since $C_b(E)$ is complete, the Closed Subspaces of Complete Normed Spaces theorem implies that $C_{b,A}(E)$ is complete. $\square$

If $A$ is empty, this theorem reduces to completeness of $C_b(E)$. If $A$ consists of a single point $x_0$, the condition is simply $f(x_0)=0$. More generally, the proof works for any collection of prescribed zero values, including an infinite subset of the domain.

What Completeness Does—and Does Not—Say

Completeness is a statement about Cauchy sequences in the chosen norm: every such sequence must have a limit that belongs to the space. In $C_b(E)$, the supremum norm controls errors uniformly over all points of $E$. A Cauchy sequence therefore has a limit in $B(E)$, and uniform convergence prevents continuity from being lost. This is why the proof works even when $E$ is unbounded or noncompact.

Compactness answers a different question. For example, the theorem does not say that every continuous function on a noncompact set is bounded. The function $x\mapsto x$ on $\mathbb{R}$ is continuous but is not in $C_b(\mathbb{R})$. Nor does completeness say that every bounded continuous function extends continuously to the closure of its domain; the function $\sin(1/x)$ on $(0,1)$ illustrates the failure of such an extension. Completeness concerns limits that are Cauchy in the supremum norm, not extension to new points.

Takeaway: The space of bounded continuous functions on any nonempty subset of $\mathbb{R}$ is complete under the supremum norm because it is closed inside the complete space of bounded functions. Conditions that require functions to vanish on a fixed subset also define complete subspaces.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Why does convergence in the supremum norm imply uniform convergence on $E$?
Which earlier theorem ensures that the limit of a uniformly convergent sequence of continuous functions is continuous?
Why is closedness of $C_b(E)$ in $B(E)$ needed to deduce its completeness?
Why does the function $\sin(1/x)$ belong to $C_b((0,1))$, even though it has no continuous extension to $x=0$?
How does the proof that $C_{b,A}(E)$ is closed use the condition $f_n(x)=0$ for $x\in A$?