Let (I − C)x_n → y in norm. If {x_n} is bounded, then compactness of C implies that there exists a subsequence x_nk such that C x_nk is norm convergent. So x_nk = (I - C)x_nk + C x_nk is norm convergent, to some x. This gives (I − C)x_nk → (I − C)x = y. The same argument goes through if the distances d(x_n, Ker(I − C)) is bounded.

But d(x_n, Ker(I − C)) must be bounded. Suppose this is not the case. Pass now to the quotient map of (I − C), still denoted by (I − C), on X/Ker(I − C). The quotient norm on X/Ker(I − C) is still denoted by ${\displaystyle \|}$·${\displaystyle \|}$, and {x_n} are now viewed as representatives of their equivalence classes in the quotient space. Take a subsequence {x_nk} such that ${\displaystyle \|}$x_nk${\displaystyle \|}$ > k and define a sequence of unit vectors by z_nk = x_nk/${\displaystyle \|}$x_nk${\displaystyle \|}$. Again we would have (I − C)z_nk → (I − C)z for some z. Since ${\displaystyle \|}$(I − C)z_nk${\displaystyle \|}$ = ${\displaystyle \|}$(I − C)x_nk${\displaystyle \|}$/ ${\displaystyle \|}$x_nk${\displaystyle \|}$ → 0, we have (I − C)z = 0 i.e. z ∈ Ker(I − C). Since we passed to the quotient map, z = 0. This is impossible because z is the norm limit of a sequence of unit vectors. Thus the lemma is proven.

i) Without loss of generality, assume λ = 1. λ ∈ σ(C) not being an eigenvalue means (I − C) is injective but not surjective. By Lemma 2, Y₁ = Ran(I − C) is a closed proper subspace of X. Since (I − C) is injective, Y₂ = (I − C)Y₁ is again a closed proper subspace of Y₁. Define Y_n = Ran(I − C)ⁿ. Consider the decreasing sequence of subspaces

${\displaystyle Y_{1}\supset \cdots \supset Y_{n}\cdots \supset Y_{m}\cdots }$

where all inclusions are proper. By lemma 1, we can choose unit vectors y_n ∈ Y_n such that d(y_n, Y_n+1) > ½. Compactness of C means {C y_n} must contain a norm convergent subsequence. But for n < m

${\displaystyle \left\|Cy_{n}-Cy_{m}\right\|=\left\|(C-I)y_{n}+y_{n}-(C-I)y_{m}-y_{m}\right\|}$

and notice that

${\displaystyle (C-I)y_{n}-(C-I)y_{m}-y_{m}\in Y_{n+1},}$

which implies ${\displaystyle \|}$Cy_n − Cy_m${\displaystyle \|}$ > ½. This is a contradiction, and so λ must be an eigenvalue.

ii) The sequence { Y_n = Ker(λ_i − A)ⁿ} is an increasing sequence of closed subspaces. The theorem claims it stops. Suppose it does not stop, i.e. the inclusion Ker(λ_i − A)ⁿ ⊂ Ker(λ_i − A)ⁿ⁺¹ is proper for all n. By lemma 1, there exists a sequence {y_n}_{n ≥ 2} of unit vectors such that y_n ∈ Y_n and d(y_n, Y_{n − 1}) > ½. As before, compactness of C means {C y_n} must contain a norm convergent subsequence. But for n < m

${\displaystyle \|Cy_{n}-Cy_{m}\|=\|(C-I)y_{n}+y_{n}-(C-I)y_{m}-y_{m}\|}$

and notice that

${\displaystyle (C-I)y_{n}+y_{n}-(C-I)y_{m}\in Y_{m-1},}$

which implies ${\displaystyle \|}$Cy_n − Cy_m${\displaystyle \|}$ > ½. This is a contradiction, and so the sequence { Y_n = Ker(λ_i − A)ⁿ} must terminate at some finite m.

Using the definition of the Kernel, we can show that the unit sphere of Ker(λ_i − C) is compact, so that Ker(λ_i − C) is finite-dimensional. Ker(λ_i − C)ⁿ is finite-dimensional for the same reason.

iii) Suppose there exist infinite (at least countable) distinct eigenvalues {λ_n}, with corresponding eigenvectors {x_n}, such that ${\displaystyle |}$λ_n${\displaystyle |}$ > ε for all n. Define Y_n = span{x₁...x_n}. The sequence {Y_n} is a strictly increasing sequence. Choose unit vectors such that y_n ∈ Yⁿ and d(y_n, Y^{n − 1}) > ½. Then for n < m

${\displaystyle \left\|Cy_{n}-Cy_{m}\right\|=\left\|(C-\lambda _{n})y_{n}+\lambda _{n}y_{n}-(C-\lambda _{m})y_{m}-\lambda _{m}y_{m}\right\|.}$

But

${\displaystyle (C-\lambda _{n})y_{n}+\lambda _{n}y_{n}-(C-\lambda _{m})y_{m}\in Y_{m-1},}$

therefore ${\displaystyle \|}$Cy_n − Cy_m${\displaystyle \|}$ > ε/2, a contradiction.

So we have that there are only finite distinct eigenvalues outside any ball centered at zero. This immediately gives us that zero is the only possible limit point of eigenvalues and there are at most countable distinct eigenvalues (see iv).

iv) This is an immediate consequence of iii). The set of eigenvalues {λ} is the union

${\displaystyle \bigcup _{n}\left\{|\lambda |>{\tfrac {1}{n}}\right\}=\bigcup _{n}S_{n}.}$

Because σ(C) is a bounded set and the eigenvalues can only accumulate at 0, each S_n is finite, which gives the desired result.

v) As in the matrix case, this is a direct application of the holomorphic functional calculus.