Solving for the spectrum and eigenvectors of the "shift operator(?)" $T$ in $P_3(\mathbb{R})$?

Question

This question is inspired from accidentally misreading this question in my tutorial exercise in my linear algebra course because I forgot the 3 in $P_3(\mathbb{R})$

(The actual question can be easily solved by considering the matrix of $T$ and then the eigenvector is found to be the constant polynomials with eigenvalues 1,1,1,1)

Attempt at solving the (misreaded) question (Let's call that Question $5a^*$)

Since I cannot solve for the spectrum and eigenvectors of $T$ in question $5a^*$ using characteristic polynomial because the matrix of $T$ will have countably infinite size, I then go back to the fundamental definition of eigenvectors

$$T(v)=\lambda v,\lambda\in\mathbb{F}$$

For $\mathbb{F}=\mathbb{R}$, plugging in the information given in the question gives the following recurrence relation

$$p(t+1)=\lambda p(t)$$

We know that if $t=0$, then $p$ will be a constant polynomial.Using this will obtained

$$p(1)=\lambda p(0)=\lambda c,c\in\mathbb{R}$$ $$p(2)=\lambda p(1)=\lambda^2 c$$ $$\vdots$$ $$p(t)=\lambda p(1)=\lambda^{t-2} c,t\in \mathbb{Z}$$

So the constant functions are eignevectors of $T$ as expected by intuition

But then

Q1. How to take account of noninteger cases e.g. $p(0.2)$ (thus possibly showing that there are no other eigenvectors)?

Q2a. How to find the other eigenvectors, if any?

*Q3. How to find the eigenvalues $\lambda$?

Extended question (Question $5a^{**}$)

What if $T$ is in $C^{\infty}(\mathbb{R})$, then the recurrence relation reads:

$$f(t+1)=\lambda f(t)$$

Q4a. Is this the simplest we can get, i.e. the eigenvectors are any periodic functions with period 1?

Q4b. How to find its spectrum?

Answer 1

I can answer question 5a*. I'm not sure about 5a** because I don't know any functional analysis.

Consider a polynomial $p(t) = a_nt^n + a_{n-1} t^{n-1} + \ldots + a_0 \in P(\mathbb R)$ with $a_n \neq 0$. Then $Tp(t) = p(t+1) = a_n(t+1)^n + \ldots + a_0$. Using the binomial theorem, we can expand this and find the first two coefficients. We then have $$Tp(t) = a_nt^n + (na_n + a_{n-1})t^{n-1} + \ldots$$ Of course, two polynomials are equal exactly when their coefficients are the same, so the condition $Tp(t) = \lambda p(t)$ can only be satisfied (for nonzero $p(t)$) if $\lambda = 1$. Hence $1$ is the only eigenvalue of $T$.

Moreover, we will have $Tp(t) = 1 p(t) = p(t)$ only if $na_n + a_{n-1} = a_{n-1}$, i.e. $na_n = 0$. Since $a_n \neq 0$, we conclude that $n=0$, i.e. $p(t)$ must be constant. Thus the constant polynomials are indeed the only eigenvectors of $T$.