Pointwise Addition on Differentiable Real Functions on Closed Unit Interval forms Group

Theorem
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.

Let $\map {\mathscr D} J$ denote the set of all differentiable real functions from $J$ to $\R$.

Let $\R^J$ denote the set of all mappings from $J$ to $\R$.

Let $\struct {\R^J, +}$ denote the algebraic structure on $\R^J$ induced by addition:
 * $\forall f, g \in \R^J: \map {\paren {f + g} } x = \map f x + \map g x$

Then $\struct {\map {\mathscr D} J, +}$ is a subgroup of $\struct {\R^J, +}$.

Proof
Taking the group axioms in turn:

From the Sum Rule for Differentiable Real Functions, if $f$ and $g$ are differentiable real functions then so is $f + g$.

Thus closure is demonstrated.

Pointwise Addition is Associative.

The constant function $f_0$ defined as:
 * $\forall x \in \R: \map {f_0} x = 0$

fulfils the role of the Identity:


 * $\forall x \in \R: \map {f_0} x + \map f x = 0 + \map f x = \map f x = \map f x + 0 = \map f x = \map {f_0} x$

From Constant Real Function is Differentiable $f_0$ is differentiable.

Hence $f_0 \in \mathscr D$.

From the Multiple Rule for Differentiable Real Functions, if $\map f x$ is continuous then so is $\map g x$ where:
 * $\forall x \in \R: \map g x = -\map f x$.

Then we note that:
 * $\forall x \in \R: \map f x + \paren {-\map f x} = 0 = \paren {-\map f x} + \map f x$

So every element has an inverse.

All the group axioms are satisfied, hence the result.