Let $\mathbb{C}[0,1]$ be the set of all real-valued functions continuous on
$[0,1] = \lbrace x\mid 0\le x \le 1 \rbrace$. (Included in $\mathbb{C}[0,1]$ are most of the functions that
come up in calculus, such as $\sin x$ and $x^2 — 2$.)
If $f$ and $g$ are in $\mathbb{C}[0,1]$, then
we define $f+ g$ by $(f+ g)(x) =f(x) + g(x)$, which is the usual definition of
addition of functions.
Using any theorems from calculus you wish, show that
$(\mathbb{C}[0,1], +)$ is a group
Need show that the four group properties hold for the given set ($\mathbb{C}[0,1]$), under the operation $+$.
(i) Existence of Identity ($e$) element: Need find any element (let, $c(x)$) in group's set (i.e., a function in $\mathbb{C}[0,1]$) s.t. taking any function $f(x)$; get $f(x)+c(x)= f(x)$.
There seems only one choice for $c(x)$, i.e. a constant function with value $c(x)=0, \forall x \in [0,1]$.
(ii) Inverse ($b$) for any element $a=f(x)$, s.t. $a+b = e\implies b = -f(x)$
To find the opposite $y$ value for function $f(x)$ is possible; but don't know how to prove.
(iii) Closure exists as sum of two continuous functions would yield a continuous function.
(iv) Associativity property exists, as $\forall a$ (let, $f(x)$), $b$ (let, $g(x)$), $c$ (let, $h(x)$) continuous in $[0,1]$. So, $a+(b+c)= (a+b)+c$.
