Let $V$ be the set of all differentiable real-valued functions defined on $\mathbb R$. Show that $V$ is a vector space under addition and scalar multiplication, defined by
$$(f+g)(t) = f(t) + g(t),\quad (cf)(t) = c[ f(t)],$$ where $f, g \in V$, $c \in F$.
Since addition defined as such is one-to-one, and since $f(t) + g(t)$ is differentiable,
$f+g$ is unique and belongs to $V$. By similar argument, $cf$ is unique and belongs to $V$.
My question has to do with the eight properties of the field $F$. They are so obvious that I have to question if I did it right.
For example, $$f+g = f(t) + g(t) = g(t) + f(t) = g+f.$$
Now, how do I justify that $f(t) + g(t) = g(t) + f(t)$? I do not think that continuity is sufficient, because the proof that
$$\lim_{t \to c} f(t) + g(t) = \lim_{t \to c} g(t) + f(t)$$
already assumes that $f(t) + g(t) = g(t) + f(t)$.
Best Answer
This concept really threw me for a loop when I took linear algebra and afterwards, but what you're using in that step is the fact that the real numbers themselves are commutative, since $f(t)$ and $g(t)$ are real numbers.
You can just quote the fact that addition of real numbers is commutative. You couldn't do it before you evaluated since $f$ is a not a number until you evaluate.
Now, on another note, you can't say that $f+g=f(t)+g(t)$, but you can say that $(f+g)(t)=f(t)+g(t)$.