Linear Algebra- Subspace proof involving operator Let $\mathbb{R}^\mathbb{R}$ be the real vector space of all functions $f:\mathbb{R}→\mathbb{R}$ and let $∆:\mathbb{R}^\mathbb{R} →\mathbb{R}^\mathbb{R}$ be the linear operator defined by  $$∆[f](x) := f(x + 1) − f(x).$$
(a) As usual, $∆^2 := ∆ ◦ ∆$ denotes the composition of $∆$ with itself. Prove that the subset $W$, consisting of all functions $f ∈ \mathbb{R}^\mathbb{R}$ such that $(∆^2)[f] + (7∆)[f] + 3f = 0$, is a linear subspace.
(b) Is the endomorphism $∆ ∈ \mathrm{End}(\mathbb{R}^\mathbb{R})$ injective? Explain.
I am confused by the use of $∆$ and I don't understand how squaring $∆$ actually affects the equation the linear operator is defined by. While I understand how the subspace test works, I am wondering if someone can explain how they would go about proving it for this subspace.

I'll leave (b) for now, and instead clarify the meaning of $\Delta^2$. 
Specifically, let's consider a simple example, $f(x) = x^2$. Then say $g = \Delta f$, so $g(x) = f(x+1) - f(x) = 2x+1$. But what about $\Delta^2$? Well, $$\Delta^2 f= (\Delta \circ \Delta) f = \Delta (\Delta f) = \Delta g,$$
and $\Delta g = g(x+1) - g(x) = 2$, so $\Delta^2 f(x) = 2$.
Now, onto the subspaces. First observe that $\Delta$ is a linear operator, so $\Delta^2$ is also a linear operator. From here, you know $\Delta^2[f+g] = \Delta^2 f + \Delta^2 g$, so you should be able to continue the calculations to see $W$ is a linear subspace. 
(Alternatively, you could observe that $\Delta^2 + 7 \Delta + 3$ is a linear operator also, and $W$ is its kernel, so $W$ is a linear subspace.)
