How do I prove that the set of differentiable real-valued functions $f$ on the interval $(-4, 4)$, such that $f'(-1) = 3f(2)$ is a subspace of $\Bbb{R}^{(4,-4)}$.
I know I must show that this set contains an additive identity, and is closed under multiplication and addition.
For the additive identity axiom, ( ${0f(x) = 0}$ ) works for this set $\Bbb{R}^{(4,-4)}$.
-> ${f(-1)=3f'(2)}$
-> ${0*f(-1)=0*3f'(2)= 0}$ (by property of equivalence?)
For closure under multiplication & addition, I'm not sure how to approach this. The equation given: $f'(-1) = 3f(2)$ implies closure for the value 2, but how do I prove this for the interval $(-4, 4)$?
I'm very new to linear algebra so please bear with me. From my understanding of the notation, $\Bbb{F}^{s}$ is the set of functions that takes numbers in the an arbitrary set $S$ as an input(which could be $\Bbb{R}$, $\Bbb{N}$, $\Bbb{C}$, etc. ), and outputs numbers belonging to $\Bbb{F}$; correct? So, $\Bbb{R}^{(4,-4)}$ is the set of functions that takes in a number in the interval between (-4) and (4), and outputs a number in $\Bbb{R}$.
I'm also not 100% sure about the phrase "subspace of $\Bbb{R}^{(4,-4)}$". From my understanding, a "subspace" is a subset of a vector-space. Is "subspace" being used here as a more abstract object such that it refers to a subset of anything that has closure of multiplication, addition and the zero vector? Or is it simply implying that $\Bbb{R}^{(4,-4)}$ is a vector space?
I will try to help with some of your confusion.
You have correctly identified what the set $\mathbb{R}*{(4, -4)}$ is. It's a set of functions. That set forms a vector space since you can add those functions, multiply them by scalars (real number) and take their negatives and always get another function of that kind. You don't have to think about multiplying those functions by each other, just by scalars.
Now you are asked to think about a subset of those functions - ones with a particular property. That subset will be a subspace if when you add two such or multiply one by a scalar you get another function in the subset. Can you do that?
(An analogy: in three dimensional with the usual coordinate axes $x$, $y$ and $x$ the $x-y$ plane is a subspace because you don't leave it when you add vectors or multiply them by scalars.)