Algebra of Continuous Functions

Pointwise Operations on Continuous Functions

The previous tutorial described how a small family of functions can be dense in a function space. Here we examine a different feature of that space: the functions in $C(K)$ can be added and multiplied, and those operations interact well with the supremum norm. Throughout, $K$ is a nonempty compact subset of $\mathbb{R}$, and $C(K)$ denotes the continuous real-valued functions on $K$, equipped with the supremum norm $\|f\|_\infty=\sup_{x\in K}|f(x)|$.

The operations are defined pointwise. For $f,g\in C(K)$ and $c\in\mathbb{R}$, the functions $f+g$, $cf$, and $fg$ are given by their corresponding operations on values at each $x\in K$. The continuity limit laws ensure that each resulting function is continuous. In particular, $C(K)$ is already known to be a real vector space under pointwise addition and scalar multiplication; pointwise multiplication adds a further algebraic operation.

Definition: A real algebra is a real vector space equipped with a multiplication that is associative and distributive over addition and that satisfies $(cf)g=c(fg)=f(cg)$ for real scalars $c$. It is commutative if $fg=gf$, and unital if it has a multiplicative identity. In $C(K)$, multiplication is pointwise: $(fg)(x)=f(x)g(x)$.

The constant function $\mathbf{1}$, defined by $\mathbf{1}(x)=1$ for every $x\in K$, is the multiplicative identity: $f\mathbf{1}=\mathbf{1}f=f$. The constant zero function is the additive identity and also satisfies $f\mathbf{0}=\mathbf{0}$. Since $K$ is nonempty, $\mathbf{1}$ and $\mathbf{0}$ are different functions. The real-number laws at each point $x$ give associativity, commutativity, and distributivity for the function operations. Thus $C(K)$, with pointwise operations, is a unital commutative real algebra.

The Product and the Supremum Norm

The supremum norm measures the largest size of a function across the whole set. For a product, the pointwise inequality $|f(x)g(x)|\leq\|f\|_\infty\|g\|_\infty$ holds at every $x$. Taking the supremum gives a useful compatibility between multiplication and the norm.

Theorem (Submultiplicativity of the Supremum Norm): If $K$ is nonempty and compact and $f,g\in C(K)$, then $fg\in C(K)$ and $$ \|fg\|_\infty\leq\|f\|_\infty\|g\|_\infty. $$

Proof. The product of continuous real-valued functions is continuous, so $fg\in C(K)$. For each $x\in K$, the definition of the supremum norm gives $|f(x)|\leq\|f\|_\infty$ and $|g(x)|\leq\|g\|_\infty$. Consequently,

|(fg)(x)|=|f(x)|\,|g(x)| \leq\|f\|_\infty\|g\|_\infty.

The right-hand side does not depend on $x$, so it is a uniform bound for $|fg|$ on $K$. Taking the least uniform bound, which is the supremum norm, proves the inequality. $\square$

This estimate is called submultiplicativity. It does not say that the norm of a product must equal the product of the norms. It says that multiplying functions cannot produce a norm larger than that bound. Together with the completeness of $C(K)$ in the supremum norm, established earlier in the course, this makes $C(K)$ a Banach algebra: a complete normed algebra whose norm is submultiplicative.

Worked Example: Adding and Multiplying Two Functions

Let $K=[-1,2]$, $f(x)=x^2-1$, and $g(x)=2x+1$. Both are polynomials and hence continuous on $K$. Their pointwise sum and product are

\begin{aligned} (f+g)(x)&=(x^2-1)+(2x+1)=x^2+2x,\\ (fg)(x)&=(x^2-1)(2x+1)\\ &=2x^3+x^2-2x-1. \end{aligned}

For instance, at $x=-1$, $f(-1)=0$ and $g(-1)=-1$, so $(fg)(-1)=0$. The expanded expression also gives $2(-1)^3+(-1)^2-2(-1)-1=-2+1+2-1=0$. At $x=2$, $f(2)=3$ and $g(2)=5$, so $(fg)(2)=15$; the expanded expression gives $16+4-4-1=15$. These checks illustrate that the product is evaluated pointwise, even when it is subsequently simplified algebraically.

Polynomials of a Function

The algebra operations allow a polynomial to be evaluated at a function. If $p(t)=a_0+a_1t+\cdots+a_nt^n$ is a real polynomial and $f\in C(K)$, define

p(f)=a_0\mathbf{1}+a_1f+\cdots+a_nf^n, \qquad f^0=\mathbf{1},\quad f^{j+1}=f^jf.

Each power $f^j$ is continuous, since it is built using finitely many products, and each finite linear combination of continuous functions is continuous. Therefore $p(f)\in C(K)$. Pointwise, this construction has the simple interpretation $(p(f))(x)=p(f(x))$. It is composition of the scalar polynomial $p$ with $f$, expressed using the operations of the function algebra.

Worked Example: Evaluating a Polynomial at a Function

Take $K=[-1,1]$, $f(x)=x+1$, and $p(t)=t^2-3t+2$. Substituting the function into the polynomial gives

\begin{aligned} p(f)(x) &=(x+1)^2-3(x+1)+2\\ &=x^2+2x+1-3x-3+2\\ &=x^2-x. \end{aligned}

The same result follows by factoring $p(t)=(t-1)(t-2)$: here $f(x)-1=x$ and $f(x)-2=x-1$, so $p(f)(x)=x(x-1)=x^2-x$. For example, at $x=0$, $f(0)=1$ and $p(f)(0)=p(1)=0$, matching $0^2-0=0$. At $x=1$, $f(1)=2$ and $p(f)(1)=p(2)=0$, matching $1^2-1=0$. The resulting function is continuous, as is also evident from its expression as a polynomial in $x$.

When Does a Continuous Function Have an Inverse?

An inverse in this algebra means a multiplicative inverse, not an inverse function in the sense of reversing a map. A function $g\in C(K)$ is a multiplicative inverse of $f$ if $fg=\mathbf{1}$. At any point where $f$ equals zero, this equation is impossible. On a compact set, the converse holds too: if $f$ never vanishes, its values stay a positive distance from zero, so taking reciprocals produces a continuous function.

Definition: A function $f\in C(K)$ is invertible in $C(K)$ if there is a $g\in C(K)$ such that $fg=\mathbf{1}$. Such a function $g$, when it exists, is called the multiplicative inverse of $f$.

Theorem (Characterization of Invertible Functions in $C(K)$): Let $K$ be nonempty and compact, and let $f\in C(K)$. Then $f$ is invertible in $C(K)$ if and only if $f(x)\neq0$ for every $x\in K$. In that case its inverse is the continuous function $1/f$, defined pointwise.

Proof. First suppose $f$ is invertible, so there is a $g\in C(K)$ with $fg=\mathbf{1}$. If $f(x_0)=0$ for some $x_0\in K$, then

$$ (fg)(x_0)=f(x_0)g(x_0)=0, $$

whereas $\mathbf{1}(x_0)=1$. This contradicts $fg=\mathbf{1}$. Thus $f$ has no zeros on $K$.

Conversely, suppose $f(x)\neq0$ for every $x\in K$. The function $|f|$ is continuous on $K$. By the Extreme-Value Theorem, it attains a minimum at some $x_0\in K$. Set $m=|f(x_0)|$. Because $f$ is nonzero at every point, $m>0$, and for every $x\in K$,

|f(x)|\geq m>0.

Define $g(x)=1/f(x)$. The reciprocal function $t\mapsto1/t$ is continuous on $\mathbb{R}\setminus\{0\}$, and $f(K)$ lies in that set, so $g$ is continuous on $K$. Moreover, $|g(x)|\leq1/m$ for every $x\in K$, and $f(x)g(x)=1$ at every $x$. Thus $g\in C(K)$ and $fg=\mathbf{1}$, proving that $f$ is invertible. $\square$

The compactness hypothesis matters in the converse. It supplies a single positive lower bound for $|f|$, not merely nonvanishing at each individual point. That bound also shows that the reciprocal is bounded, as every continuous function on a compact set must be. On a noncompact domain, a continuous nonvanishing function can approach zero and have an unbounded reciprocal.

Worked Example: A Function with a Continuous Multiplicative Inverse

On $K=[0,1]$, let $f(x)=2+x$. Since $2\leq f(x)\leq3$ throughout the interval, $f$ never vanishes. The theorem gives the inverse $g(x)=1/(2+x)$. Direct multiplication verifies the identity:

(fg)(x)=(2+x)\frac{1}{2+x}=1 \qquad (x\in[0,1]).

The reciprocal is continuous, and $1/3\leq g(x)\leq1/2$. In particular, it belongs to $C[0,1]$. In contrast, for $h(x)=x$ on $[-1,1]$, the value $h(0)=0$ rules out any multiplicative inverse: if $hk=\mathbf{1}$, then $(hk)(0)=0$, which cannot equal $1$.

What the Algebra Structure Tells Us

Pointwise multiplication lets us form more than sums and products of a fixed pair of functions. We can build powers, polynomial expressions, and—when a function is nowhere zero on compact $K$—reciprocals. These operations stay within $C(K)$, so the result can be used as an input to further algebraic operations. The submultiplicative estimate controls the size of products, while the invertibility theorem identifies exactly when division by a function is allowed within the space.

A common pitfall is to infer invertibility merely from the fact that a function is nonzero as an element of $C(K)$. A nonzero function need only be nonzero somewhere; invertibility requires it to be nonzero everywhere. For example, $x\mapsto x$ is not the zero function on $[-1,1]$, but it vanishes at $0$ and therefore has no multiplicative inverse in $C[-1,1]$. Nor is a function’s reciprocal automatically continuous and bounded on an arbitrary domain just because it has no zeros there. The compactness argument in the theorem is what guarantees a positive minimum for its absolute value.

Takeaway: Under pointwise operations, $C(K)$ is a unital commutative algebra. Its supremum norm satisfies $\|fg\|_\infty\leq\|f\|_\infty\|g\|_\infty$, and on a nonempty compact set its invertible elements are exactly the continuous functions with no zeros.

Check Your Understanding

Use the definitions, estimates, and proofs in this tutorial to answer the following questions.

What function serves as the multiplicative identity in $C(K)$, and why is it distinct from the zero function when $K$ is nonempty?
Use the pointwise bound for $|f(x)g(x)|$ to explain why $\|fg\|_\infty\leq\|f\|_\infty\|g\|_\infty$.
For $p(t)=t^2+1$ and $f\in C(K)$, explain why $p(f)$ belongs to $C(K)$.
Why does a zero of $f$ prevent $f$ from having a multiplicative inverse?
Where is compactness used to prove that every nowhere-zero function in $C(K)$ is invertible?
Give an example of a continuous, nowhere-zero function on a noncompact interval whose reciprocal is unbounded.