# Pointwise Addition on Continuous Real-Valued Functions forms Group

## 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 $g$:

$\forall x \in R: \left({f + g}\right) \left({x}\right) = f \left({x}\right) + g \left({x}\right)$

Then $\left({C, +}\right)$, the algebraic structure on $C$ induced by $+$, forms a group.

## Proof

Taking the group axioms in turn:

### G0: Closure

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

Thus closure is demonstrated.

$\Box$

### G1: Associativity

$\Box$

### G2: Identity

The constant function $f_0$ defined as:

$\forall x \in \R: f_0 \left({x}\right) = 0$

fulfils the role of the Identity:

$\forall x \in \R: f_0 \left({x}\right) + f \left({x}\right) = 0 + f \left({x}\right) = f \left({x}\right) = f \left({x}\right) + 0 = f \left({x}\right) = f_0 \left({x}\right)$

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

$\Box$

### G3: Inverses

From the Multiple Rule for Continuous Functions, if $f \left({x}\right)$ is continuous then so is $g \left({x}\right)$ where:

$\forall x \in \R: g \left({x}\right) = - f \left({x}\right)$.

Then we note that:

$\forall x \in \R: f \left({x}\right) + \left({- f \left({x}\right)}\right) = 0 = \left({- f \left({x}\right)}\right) + f \left({x}\right)$

So every element has an inverse.

$\Box$

All the group axioms are satisfied, hence the result.

$\blacksquare$