Continuous function satisfying axioms

Question

Let $C[0,1]$ be the set of continuous functions from $[0,1]$ to $\mathbb R$. For $g,h∈C[0,1]$ and $a∈\mathbb R$, addition and scalar multiplication are defined respectively as: \begin{align} (g+h)(x)&=g(x)+h(x)\\ (ag)(x)&=ag(x) \end{align} Prove that $C[0,1]$ satisfies the following axioms for being a vector space over $\mathbb R$: a Zero exists, a One exists and the Distributive laws holds.

My issue is that I really don't know what my variables are meant to be in this situation. Would it be the case that saying, for example, $g(x)+0=g(x)$ proves the first axiom or would I need to use something else?

Answer 1

The "variables" (I believe you mean vectors) are continuous functions from $[0,1]\rightarrow \mathbb{R}$. You are correct that the function $h(x) \equiv 0$ is indeed the $0$ vector in this space, and your proof $g(x)+h(x) = g(x)+0=g(x)$ is sufficient. If you want to be more precise you can first say, "Let $g(x)$ be an arbitrary continuous function from $[0,1]$ to $\mathbb{R}$" and also note that $h(x)\equiv 0$ is a continuous function, so it is in the space you are considering.

Answer 2

Yes, but one may doubt that you've understood the proof.

The first axiom is that there is an element $e$ that is the identity under addition. This means for every vector $f$ you would have that $f+e=f$.

To show there's such an vector and it's defined as $e(x)=0$ you must first use the definition of addition given, that is $(f+e)(x) = f(x)+e(x)$. You must also use a definition of equality not given above, but it's defined elsewhere that two functions (with the same domain) are equal if their value coincide at all points, that is since $f(x)+e(x) = f(x) + 0 = f(x)$ for all $x\in\mathbb R$ then $f+e = f$.

Then you have to prove the rest of the axioms for linear spaces in similar way. It doesn't have to be as elaborate - it's just to make sure that the understanding is there too...