The Group of Differentiable Real-Valued Functions

The Group of Differentiable Real-Valued Functions

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

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

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

Let $C^1 [a, b]$ denote the set of differentiable real-valued functions on the interval $[a, b]$ and let $+$ be the operation of function addition defined for $f, g \in C^1 [a, b]$ by:

(1)
\begin{align} \quad (f + g)(x) = f(x) + g(x) \end{align}

Let $f, g, h \in C^1 [a, b]$. We know by Calculus that the sum of two differentiable functions is a differentiable function, so $(f + g) \in C^1 [a, b]$ and hence $C^1 [a, b]$ is closed under $+$.

Furthermore, since $f$, $g$, and $h$ are real-valued functions, then $f(x), g(x), h(x) \in \mathbb{R}$ for all $x \in [a, b]$, and from the associativity of real numbers, we have that:

(2)
\begin{align} \quad (f + [g + h])(x) = f(x) + [g(x) + h(x)] = [f(x) + g(x)] + h(x) = ([f + g] + h)(x) \end{align}

Therefore $+$ is associative.

The identity element is the differentiable function $z(x) = 0$ (which of course is differentiable for any interval $[a, b]$).

For each $f \in C^1 [a, b]$ we know from calculus that for any $k \in \mathbb{R}$ that $kf$ is differentiable on $[a, b]$, that is, $kf \in C^1 [a, b]$. For $k = -1$ we have that $-f \in C^1 [a, b]$ and this function is the inverse of $f$ with respect to $+$ since:

(3)
\begin{align} \quad (f + (-f))(x) = f(x) - f(x) = 0 = z(x) \end{align}
(4)
\begin{align} \quad ((-f) + f)(x) = -f(x) + f(x) = 0 = z(x) \end{align}

Hence $(C^1 [a, b], +)$ is a group. In fact, if we define $C^n [a, b]$ to be the set of $n$-times differentiable real-valued functions on the interval $[a, b]$ then it can be shown that $(C^n[a, b], +)$ is also a group.

Furthermore, we note that the set of continuous real-valued functions on the interval $[a, b]$ denoted $C[a, b]$ and the sets of $n$-differentiable real-valued functions on $[a, b]$, $C^n [a, b]$, are such that:

(5)
\begin{align} \quad C[a, b] \supset C^1 [a, b] \supset C^2 [a, b] ... \supset C^n[a, b] \supset ... \end{align}

Hence we can say that $(C^n [a, b], +)$ is a subgroup of $(C[a, b], +), (C^1[a, b], +), ..., (C^{n-1} [a, b], +)$.

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