Proving a structure is a vector space

Question

I understand that in order to confirm whether or not a given structure is a vector space, I must check to see if it holds to a number of properties that define vector spaces, among the most important of which are that the structure must be closed for vector addition and scalar multiplication.

I will try not to overload this question by proving each individual property, but I would like to know if I'm heading in the right direction.

EDIT I have edited to include proofs for each property.

$V$ = {$f: \mathbb R \to \mathbb R$}, $(f+g)(x) = f(x) + g(x), \forall x \in \mathbb R, \forall f,g \in V,$

$(λ \cdot f) (x) = f(λx), \forall x, λ \in \mathbb R, \forall f \in V.$

I first test:

(1) the set $V$ is closed under vector addition, that is, $v+w \in V$

By definition, $f(x) \in V+ g(x) \in V = (f+g)(x) \in V$ thus the set is closed under addition.

(2) vector addition is commutative, $v+w=w+v$

$f(x) + g(x) = (f+g)(x) = (g+f)(x) = g(x) + f(x)$

(3) vector addition is associative, $(v+w)+u=v+(w+u)$

$(f(x)+g(x))+h(x) = (f+g)(x)+h(x)$

$(f+g)(x)+h(x)=f(x)+(g(x)+h(x))=f(x)+(g+h)(x)=f(x)+g(x)+h(x)$

(4) there is a zero vector $0_v \in V$ such that $v + 0_v = v$ $\forall v \in V$

$f(x) + (0 \cdot g)(x) = f(x) + g(0x) = f(x) + 0 = f(x)$

(5) each $v \in V$ has an additive inverse $w \in V$ such that $w + v = 0_v$

$f(x) + (-1 \cdot f)(x) = f(x) + f(-1x) = f(x) + (-1)f(x) = f(x) - f(x) = 0$

(6) the set V is closed under scalar multiplication, that is, $r \cdot v \in V$

The definition provides this.

(7) addition of scalars distributes over scalar multiplication, $(r+s) \cdot v = r \cdot v + s \cdot v$

$(a+b) \cdot f(x) = ((a+b)\cdot f)(x) = a \cdot f(x) + b \cdot f(x) \neq f(a \cdot x) + f(b \cdot x)$

Property (7) does not hold.

(8) scalar multiplication distributes over vector addition, $r \cdot (v+w) = r \cdot v + r \cdot w$

$(a \cdot f)(x) = f(ax) = a \cdot f(x)$

$f(x) + g(x) = (f+g)(x)$

$a \cdot (f+g)(x) = (a \cdot (f+g))(x) = (f+g)(ax) = f(ax) + g(ax) = a \cdot f(x) + a \cdot g(x)$

(9) ordinary multiplication of scalars associates with scalar multiplication, $(rs) \cdot v = r \cdot (s \cdot v)$

$a(b \cdot f)(x) = a \cdot f(bx) = (a \cdot f)(bx) = f((ab)x) = f(x) \cdot (ab)$

(10) multiplication by the scalar 1 is the identity operation, $1 \cdot v = v$.

$(1 \cdot f)(x) = f(1x) = f(x)$

CONCLUSION

The addition of scalars does not distribute over multiplication (Property (7) does not hold). $V$ is not a vector space.

Answer 1

No, you are not heading in the right direction. The elements of $V$ are the functions from $\mathbb R$ to $\mathbb R$. Where are they? When you try to prove that $V$ is a vector space, you starting talking about matrices with two rows and one column, instead of functions.

Answer 2

Your attempt disregards completely the kind of objects the elements of $V$ are. You can't just write down the proof that $\Bbb R^2$ is an $\Bbb R$-vector space whenever you want to prove that some $V$ is a vector space.