Pointwise Addition on Continuous Real Functions forms Group

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Theorem

Let $C$ be the set of all continuous real functions on the set of real numbers $\R$.

Let $f, g \in C$.

Let $f + g$ be the pointwise sum of $f$ and $g$:

$\forall x \in R: \map {\paren {f + g} } x = \map f x + \map g x$


Then $\struct {C, +}$, the algebraic structure on $C$ induced by $+$, forms a group.


Proof

Taking the group axioms in turn:


Group Axiom $\text G 0$: Closure

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

Thus closure is demonstrated.

$\Box$


Group Axiom $\text G 1$: Associativity

Pointwise Addition is Associative.

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element

The constant function $f_0$ defined as:

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

fulfils the role of the Identity:

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

Note that $f_0 \in C$ as the Constant Function is Uniformly Continuous, and hence continuous.

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

From the Multiple Rule for Continuous 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.

$\Box$


All the group axioms are satisfied, hence the result.

$\blacksquare$


Sources