I ran into these 2 questions which made me question my understanding about cosets, and I would like to verify if my understanding is correct.
I understand that given a coset, all elements of the coset would have the same 'output' when ran though the differential equation function. For instance, if I had a coset of $W$ containing $g(x)$, a member of $W$ would look something like this: $f(x)+g(x)$, for some $f(x) \in W$. Then plugging it into the differential equation, I would have:
$$ \frac{d^2(f+g)}{dx^2}-3\frac{d(f+g)}{dx}+2(g)=\frac{d^2g}{dx^2}-3\frac{df+g}{dx}+2g=a(x) $$
So all members of $W+g(x)$ would give me the same $a(x)$, but how do I show that each coset is a solution set of a differential equation? Also, additional question, but why is $g \in C([a,b])$ instead of $C^2([a,b])$? Is there a typo in the question?
For (42), I was tempted to say that a coset containing $W$ contains sequences satisfying $a_{n+3}-2a_n=k$ for some $k \in \mathbb{R}$. However, I realised this is not actually the case, because I could have the coset containing the sequence $(0,1,0,1...0)$ or some other sequence that is not of period 3, and then an element of this coset will break the $a_{n+3}-2a_n=k$ rule. Am I correct in saying this, and if so, what then should an interpretation of cosets in that question be?
For your reference, this is the definition my textbook provided me with cosets:
Let $V$ be a vector space, $W \subseteq V$ a subspace. Then $W+u$ is the coset containing $u$ if $W+u=\{w+u\ |w \in W\}$.
You have to show that for each $u \in C^2[a,b]$, there exists a $g(x) \in C[a,b]$ such that each element in $W+u$ solves the given differential equation. So, $$\frac{d^2(f+u)}{dx^2}-3\frac{d(f+u)}{dx}+2(f+u)=\frac{d^2 u}{dx^2}-3\frac{du}{dx}+2u:=g(x)$$ for all $f\in W$. Which means selecting $g(x)$ as the output of the differential equation for $u$ is enough. Note that $g$ might not be differentiable, but only continuous. This is because we only know that $u$ is twice differentiable.
Similar idea can be applied to the other question.