The Group of Continuous Real-Valued Functions

The Group of Continuous Real-Valued Functions

Recall from the Groups page that a group is a set $G$ paired with a binary operation $\cdot : G \times G \to G$ where:

  • 1) For all $a, b, c \in G$ we have that $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ (Associativity of $\cdot$).
  • 2) There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (The existence of an identity for $\cdot$).
  • 3) For all $a \in G$ there exists a $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ (The existence of inverses for each element in $G$).

We will now look at the group of continuous real-valued functions on a fixed interval $[a, b]$.

Consider the set of continuous functions $f$ on the interval $[a, b]$ where $a \leq b$, which we denote by $C[a, b]$. Let $+$ be the operation of function addition where for all $f, g \in [0, 1]$ we have that:

\begin{align} \quad (f + g)(x) = f(x) + g(x) \end{align}

To see if $(C[a, b], +)$ is a group, we must verify that all three points in the definition. From calculus we know that if $f$ and $g$ are continuous on $[a, b]$ then the sum $(f + g)$ is continuous on $[a, b]$ so $C[a, b]$ is closed under function addition.

To show that the associativity property holds for $+$, let $f, g, h \in C[a, b]$. Then:

\begin{align} \quad (f + (g + h))(x) = f(x) + (g + h)(x) = f(x) + (g(x) + h(x)) = (f(x) + g(x)) + h(x) = (f + g)(x) + h(x) = ((f + g) + h)(x) \end{align}

Therefore function addition is commutative. Furthermore, if we let $e(x) = 0$ (which is continuous on all of $\mathbb{R}$ and so is continuous on any interval $[a, b]$) we have that:

\begin{align} \quad (f + e)(x) = f(x) + e(x) = f(x) + 0 = f(x) \quad \mathrm{and} \quad (e + f)(x) = e(x) + f(x) = 0 + f(x) = f(x) \end{align}

Therefore $e(x) = 0$ is our identity element. Lastly, recall from calculus that if $f$ is continuous on $[a, b]$ then for any $k \in \mathbb{R}$ we have that $kf$ is continuous on $[a, b]$. If $k = -1$ then $-f$ is continuous on $[a, b]$. So for all $f \in C[a, b]$ we have that $-f \in C[a, b]$ and:

\begin{align} \quad (f + (-f))(x) = (0)(x) = 0 = e(x) \end{align}

Therefore for each $f \in C[a, b]$ there exists an (additive) inverse, $-f \in C[a, b]$. Thus $(C[a, b], +)$ is a group.

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