I see what you're trying to do, but is this enough? All you've shown is that the spectrum of the limit is just $1$, which I don't think uniquely characterizes the identity. Or does it?

Honest question: Must a self-adjoint operator with $1$ as the only point in its spectrum be the identity?

@Potato I think so, and that this is a consequence of the spectral theorem. If not, perhaps we additionally need to use the fact that the limit $A$ is such that $A-I$ is compact; I'm rusty on the details of the spectral theorem for operators.

@Potato we could also have used the explicit diagonalization of $T$ and ostensibly applied the same analysis

Right, that works. I'm wondering if there's a way to see it without appealing to the diagonalization, though. (Because if you're going to diagonalize, why not just do that in the first place and be more explicit?)

@Potato well for one, maybe I just like being mysterious with my hints. However, as I indicated, I can't answer off the top of my head whether or not the compactness of $T$ is necessary, and if my approach allows for some generalization of the desired result.

@Potato: For selfadjoints $A=A^*$ one has $r(A)=\|A\|$. So $\sigma(A)=(0)$ implies $\|A\|=0$ and $A=0$. (Note this argument only works in C-algebras, i.e. Hilbert spaces!)*

@Freeze_S yep, that will do it. Thanks!

@Potato: For arbitrary elements that is wrong. Some nontrivial nilpotents have zero only in their spectrum. (So the assertion is wrong in Banach algebras, i.e. Banach spaces.)

@Omnomnomnom: You're welcome! :)

@Omnomnomnom: Besides, was your argument Banach fixed point?

@Freeze_S That's where I was heading, but showing that $\phi$ is contractive over all of $K$ is tricky and apparently unnecessary. I suppose we can restrict our analysis to the set of operators that are "simultaneously diagonalizable with $T$" or something of the like.

@Omnomnomnom hmm, I'm afraid I'm not following your train of thought, although the concepts you're referring to are right on track with my toolset for analysis. It sounds like the goal is to determine that the spectrum of the limit is just $\{1\}$? I can see how we'd be done once we get there, but I don't yet see how $f$ and $\phi$ help us get there.

@user218389 the sequence is defined by $A_n = \phi(A_{n-1})$. Note also that for any $\lambda \in [1/2,3/2]$, if we define $$ a_n = f(a_{n-1}), \quad a_0 = \lambda $$ then we have $a_n \to 1$.

@user218389 have you read Potato's answer? It's incomplete, but it provides a useful intuition here. Also, because $T$ is compact, the "matrix" analogy is particularly apt.

@Omnomnomnom ok, so are you saying that for any eigenvalue $\lambda$, $f(\lambda) \rightarrow 1$, and that gives us that $f(\sigma(A_n)) \rightarrow \{1\}$ which gives $\sigma(A_n) \rightarrow \{1\}$?

And one more question: what type of convergence does this argument give us? It shouldn't be uniform, because each $A_n$ is compact, and compact operators can't converge uniformly to a non-compact operator like $I$. But I've never seen an argument for convergence based on convergence of the spectrum, so I'm not sure how to analyze the type of convergence we have here.

Your first comment doesn't quite make sense as written, and I don't know what you mean. The $A_n$ are not compact, but they are equal to a compact operator plus a multiple of the identity.

See my latest edit.

That last edit cleared it up for me. Thanks so much for your help!