I need to figure out which of the given real-valued continuous functions defined (-inf, inf) with conditions define a subspace.
I know how to do that with vectors, but not with functions. I know I need to show:
a) non empty (contains zero function)
b) holds under addition
c) holds under scalar multiplication
Here are some of my functions:
$x(1)+x(2)=2$,
$x(1)=0$,
$x$ is periodic with period $2\pi$.
Disclaimer: While calling a function "$x$" isn't forbidden in any way, when dealing with functions, I'll avoid using $x$ or $y$. Your misconception in the comment is the exact reason why. So for the rest of the answer, functions will be noted $f$ or $g$. In each question, I'll also note $A$ the subset defined by your equation.
You say you know how to do it with vector, and even give the conditions you have to check. It seems that you're stressing to much about your objects beeing functions, but you'll treat these exactly like you'll do with vectors.
So, let's treat your functions 1 by 1.
1/ $f(1)+f(2)=2$
This question as already been answered in the comment by mfl: the function $f: x \rightarrow 0$, AKA the $0$ function doesn't verify the equation, so it doesn't belong to the subset. If the 0 function isn't is the subset, it can't be a subspace.
2/$f(2)=0$
Once again, the comments solved it:
a/ The $0$ function trivially belongs to the subset
b/ It holds under addition, in fact:
$\forall f,g \in A, (f+g)(2) = f(2)+g(2)= 0+0 = 0 $
c/ It holds under scalar multiplication:
$\forall \lambda \in \mathbb R,\forall f \in A, (\lambda * f)(2)= \lambda * f(2)= \lambda * 0 = 0$
Therefore,your subset $A$ is a subspace
3/ f is periodic of period $2\pi$
This one is a bit more tricky (not much), because you don't have a defined value for f(x). All you know is that:
$\forall k \in \mathbb Z, \forall x \in \mathbb R, f(x+2k\pi) = f(x)$
By using this, you can prove that $A$ is once again a subspace