Originally Posted by

**emakarov** You are saying this as if these are theorems, but, in fact, these are definitions of addition and multiplications on functions.

You have justified the second equality, but not the first one.

Then it is not just a vector space (but something like an algebra over a field). However, I am not sure talking about vector spaces is needed here; the problem is easy as is.

But you said that it doesn't make sense to multiply vectors (i.e., functions).

By what vector space axiom? This is not multiplication by a scalar; this is application of a function to an argument!

Here is a proof. We need to show that h(f + g) = hf + hg. By definition of (pointwise) equality of functions, this means [h(f + g)](x) = [hf + hg](x) for every x. So fix an arbitrary x.

$\displaystyle \begin{align*}[h(f + g)](x)&=h(x)[(f+g)](x) && \text{by (2)}\\ &=h(x)(f(x)+g(x)) &&\text{by (1)}\\ &=h(x)f(x)+h(x)g(x) &&\text{by distributivity of multiplication on }\mathbb{R}\\ &=[hf](x)+[hg](x)&& \text{by (2)}\\ &=[hf+hg](x)&& \text{by (1)}\end{align*}$