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Lemma 4.0.18 Let S ∈ L(H) where H is a Hilbert space. Suppose that Sλ = λI − S with
λ = α + iβ and S = S ∗ . If Sλx ≥ |β|x, with β = 0, then Sλ is invertible.
Proof.
(i) To show that it is injective, let Sλx = 0, then |β|x ≤ 0 which implies that x = 0 for
β = 0. Hence N(Sλ) = {0} and we conclude that Sλ is injective.
(ii) Let us first show that R(Sλ) is closed. If (Sλ(xn)) is a Cauchy sequence so is (xn).
Now since H is complete, xn → x ∈ H. Since Sλ is bounded, then it is continuous. Hence
lim
n→∞ (Sλ(xn)) = (Sλ) lim (xn) = (Sλ)x. Hence (Sλ)x ∈ R(Sλ) and we conclude that R(Sλ) is
n→∞
closed.
Now by direct sum, Theorem 2.3.8, H = R(Sλ) ⊕ R(Sλ) ⊥ . By Theorem 4.0.16 (1), we have that
R(Sλ) ⊥ = N(Sλ ∗ ). We will show that Sλ ∗ is injective and deduce that R(Sλ) ⊥ = N(Sλ ∗ ) = {0}.
Now by Theorem 4.0.16 (1), R(Sλ) ⊥ = N(Sλ ∗ ). For β = 0, N((αI + iβI − S) ∗ ) = N(αI −
iβI − S) = {0} otherwise by Lemma 4.0.17, α − iβ would be an eigenvalue. But α − iβ /∈ R,
so that α − iβ cannot be an eigenvalue. This implies that N(Sλ ∗ ) = R(Sλ) ⊥ = {0}. Hence
H = R(Sλ), showing that Sλ is surjective. We conclude that Sλ is bijective hence invertible.
Theorem 4.0.19 ([Kre78], 7·5-5 (Theorem)) If T is a bounded linear operator on a complex
Banach space, then
ρ(T) = lim T
n→∞ n 1
n . (4.4)
Remark 4.0.20 The proof of this theorem is not trivial and falls outside the scope of this essay.
Corollary 4.0.21 ([Aup91], Corollary 2·3·2) Let H be a Hilbert space and let S be a self-adjoint
operator on H. Then
1. S 2 = S 2 and consequently S = ρ(S).
2. Sp S ⊂ R.
Proof.
1. From Theorem 3.4.4 (5), S ∗ S = S 2 . But S is self-adjoint. Hence S 2 = S 2 and by
induction we have S 2n
= S2n . Now by Theorem 4.0.19, we have that ρ(S) = lim S
n→∞ n 1
n =
S.
2. Let λ = a + iβ with α, β ∈ R and let Sλ = λI − S for all λ ∈ C. Then
Sα + iβI = αI − S + iβI
= (α + iβ)I − S
= λI − S.