C(K) as a Normed Space

From Compact Domains to Function Spaces

For a continuous function on a nonempty compact set, the Extreme-Value Theorem from the previous tutorial guarantees boundedness. This means that the supremum norm, already introduced for bounded functions, can be used for every continuous function on such a domain. The continuous functions also remain continuous under pointwise addition and scalar multiplication, so they form a vector space equipped with a natural norm.

Throughout this tutorial, \(K\subseteq\mathbb{R}\) is a fixed, nonempty compact set. Continuity on \(K\) is relative to \(K\). The nonempty hypothesis matters for the usual supremum norm: it ensures that the supremum is taken over a nonempty domain and that constant functions provide useful test functions. The empty set has only one real-valued function on it, but we will not use the standard supremum-norm construction for that case.

The definition of \(C(K)\) identifies the set of functions; the normed-space structure requires two further checks. First, the pointwise vector operations must keep us inside \(C(K)\). Second, the norm must be finite for every member of \(C(K)\). Compactness supplies the second fact through boundedness in the Extreme-Value Theorem.

The Normed-Space Structure

Proof. The zero function is continuous, so \(C(K)\) is nonempty. If \(f,g\in C(K)\), then \(f+g\) is continuous; if \(c\in\mathbb{R}\), then \(cf\) is continuous. These are the usual continuity laws for sums and scalar multiples. The pointwise operations inherit the vector-space identities from the real numbers: for example, \((f+g)(x)=f(x)+g(x)\) and \((c f)(x)=c f(x)\) for every \(x\in K\). Thus \(C(K)\) is a real vector space.

By the Extreme-Value Theorem, \(f\) is bounded on \(K\). Hence \(\sup_{x\in K}|f(x)|\) is finite, so the supremum norm is defined on every element of \(C(K)\). In the earlier tutorial “The Supremum Norm,” it was proved that this formula is a norm on \(B(E)\) for any nonempty set \(E\). Since \(C(K)\subseteq B(K)\), the same norm properties hold on \(C(K)\). Therefore \(C(K)\), with this norm, is a normed space. \(\square\)

This proof uses an important structural shortcut. Once \(C(K)\) is known to be a vector subspace of \(B(K)\), the norm axioms need not be checked again from the beginning: the supremum norm on \(B(K)\) restricts to a norm on the subspace. Compactness is what puts every continuous function into \(B(K)\).

The Supremum Is a Maximum on \(C(K)\)

For an arbitrary bounded function, its supremum need not be attained at any point. For a continuous function on compact \(K\), the situation is stronger. The function \(x\mapsto |f(x)|\) is continuous, so the Extreme-Value Theorem applies directly to its values.

Proof. The absolute-value function on \(\mathbb{R}\) is continuous, so the composition \(x\mapsto |f(x)|\) is continuous on \(K\). By the Extreme-Value Theorem, it attains its maximum at some \(x_0\in K\). Since the maximum of \(|f|\) is also its supremum, $$ \sup_{x\in K}|f(x)|=|f(x_0)|. $$ The left side is \(\|f\|_\infty\) by definition. \(\square\)

Attainment is useful when a norm calculation needs a lower bound: rather than searching for points whose function values approach the supremum, one can identify a point where the absolute value is largest. It also shows that the closed unit ball in \(C(K)\) can be described pointwise as the continuous functions satisfying \(|f(x)|\leq1\) for every \(x\in K\).

Worked Examples

Point Evaluation Is Controlled by the Norm

A norm measures a function globally, but it also controls every individual function value. Fix \(a\in K\), and consider the map that reads off the value at \(a\). This map is linear, and its output cannot be larger in absolute value than the supremum norm of the input.

Proof. For \(f,g\in C(K)\) and \(c\in\mathbb{R}\), $$ \delta_a(f+g)=(f+g)(a)=f(a)+g(a)=\delta_a(f)+\delta_a(g) $$ and $$ \delta_a(cf)=(cf)(a)=c f(a)=c\delta_a(f). $$ Thus \(\delta_a\) is linear. For every \(f\in C(K)\), the definition of the supremum gives $$ |\delta_a(f)|=|f(a)|\leq\sup_{x\in K}|f(x)|=\|f\|_\infty. $$ So \(C=1\) is a valid bound. To see that no smaller nonnegative bound works, take the constant function \(\mathbf{1}(x)=1\). Then \(|\delta_a(\mathbf{1})|=1\) and \(\|\mathbf{1}\|_\infty=1\). Any valid \(C\) must satisfy \(1\leq C\cdot1\), hence \(C\geq1\). Therefore the smallest bound is \(1\). \(\square\)

In particular, if \(f_n,f\in C(K)\) and \(\|f_n-f\|_\infty\to0\), then for every fixed \(a\in K\), $$ |f_n(a)-f(a)|\leq\|f_n-f\|_\infty\longrightarrow0. $$ So convergence in the norm implies convergence of function values at every point. The supremum norm gives the stronger, uniform control: the same bound works for all \(a\in K\).

Norm Distance and Uniform Distance

For \(f,g\in C(K)\), their difference is continuous and therefore belongs to \(C(K)\). The norm distance between them is $$ \|f-g\|_\infty=\sup_{x\in K}|f(x)-g(x)|. $$ Thus the norm records the largest possible discrepancy across the whole domain. By the definition of uniform convergence, a sequence \((f_n)\) in \(C(K)\) converges uniformly to \(f\in C(K)\) exactly when \(\|f_n-f\|_\infty\to0\). This is the function-space form of the relationship between uniform convergence and the supremum norm established earlier in the course.

This identification is useful because it lets us use the language of normed spaces for uniform approximation. For example, to show that a continuous function can be approximated uniformly by another function, one estimates a single norm, rather than handling each point of \(K\) separately. Conversely, convergence at each individual point alone does not provide such a uniform estimate.

Being a normed space records the vector operations and the geometry supplied by the norm. Those facts alone do not establish that every Cauchy sequence in \(C(K)\) has a limit in \(C(K)\). Completeness requires a separate argument, using both the behavior of uniform Cauchy sequences and the preservation of continuity under uniform limits.