Which of the following are subspaces of $C(\mathbb{R})$ I got a question:  
Let $C(\mathbb{R})$ be the collection of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Then $C(\mathbb{R})$ is a real vector space with pointwise addition and scalar multiplication defined by
$$(f + g)(x) = f(x) + g(x) \text{ and } (rf)(x) = rf(x)$$
for all $f, g \in C(\mathbb{R})$ and all $r, x \in \mathbb{R}$. Which of the following are subspaces of $C(\mathbb{R})$?
I. {$f: f \text{ is twice differentiable and } f''(x) - 2f'(x) + 3f(x) = 0$ for all x}
II. {$g: g \text{ is twice differentiable and } g''(x) = 3g'(x)$ for all x}
III. {$h: h \text{ is twice differentiable and } h''(x) = h(x) + 1$ for all x}  
I believe that all three are correct, since they are all continuous functions on $\mathbb{R}$. However, the answer says that only I and II are correct.  
Could anyone tell me what is wrong with III? Also, I don't see anything meaningful for the two constraints by the pointwise addition and scalar multiplication. Could anyone also tell me why do they matter here and how will they affect the answer to this question if these two constrains are not here?  
Thanks!
---Follow ups from answers:
So is it safe for me to conclude that h fails ALL multiplication tests and addition tests, since the general form of h = $Ae^{-x} + Be^x - 1$, and any attempt to add or multiply to this formula will screw up the constant part -1?
 A: Note that III does not, for example, contain the zero vector/function, so it cannot be a subspace. Remember that subspaces must be non-empty and are closed under addition and scalar multiplication (these latter two conditions both imply that you need the zero vector; sometimes a subspace is defined to contain the zero vector and satisfy the last two conditions). 
You're correct that all three are subsets of continuous functions, but that doesn't mean that they're subspaces. The conditions for I and II guarantee that they're subspaces. Note that III doesn't satisfy the other subspace conditions, either.
A: Consider two functions 
$h_1, h_2 \in C(\Bbb R), \tag 1$
each twice differentiable and satisfying
$h''(x) = h(x) + 1; \tag 2$
then
$h_1''(x) + h_2''(x) = (h_1(x) + 1) + (h_2(x) + 1) = (h_1(x) + h_2(x)) + 2; \tag 3$
it thus appears that $h_1(x) + h_2(x)$ does not satisfy (2); indeed if $h_1(x) + h_2(x)$
did obey (2), some simple algebra would reveal that
$(h_1(x) + h_2(x)) + 2 = h_1''(x) + h_2''(x) + 1 = (h_1(x) + h_2(x)) + 1, \tag 4$
immediately leading to the contradictory statement
$1 = 0. \tag 5$
Thus this class of functions does not form a vector space.
$OE\Delta$.
