Spectral Theory in Hilbert Space

CHAPTER 5
Resolvents
We now consider a closed, densely defined operator T in the Hilbert
space H. We define the solvability and deficiency spaces of T at λ by
Sλ = {u ∈ H | (T − λ)v = u for some v ∈ D(T )}
Dλ = {u ∈ D(T ∗ ) | T ∗ u = λu}.
The following basic lemma is valid.
Lemma 5.1. Supposed T is closed and densely defined. Then
(1) D λ = H ⊖ Sλ.
(2) If T is symmetric and Im λ = 0, then Sλ is closed and H =
Sλ ⊕ D λ
(3) If T is selfadjoint and Im λ = 0, then (T − λ)v = u is uniquely
solvable for any u ∈ H ( i.e., Sλ = H), T has no non-real
eigen-values ( i.e., Dλ = {0}), and v ≤ 1
|Im λ| u.
Proof. Any element of the graph of T is of the form (v, λv + u),
where u ∈ Sλ. To see this, simply put u = T v − λv for any v ∈ D(T ).
Now 〈T v, w〉 − 〈v, λw〉 = 〈u + λv, w〉 − 〈v, λw〉 = 〈u, w〉, so it follows
that (w, λw) ∈ GT ∗, i.e., w ∈ Dλ , if and only if w is orthogonal to Sλ.
This proves (1).
If T is symmetric and (v, λv+u) ∈ GT , then 〈λv+u, v〉 = 〈v, λv+u〉,
i.e., Im λv2 = Im〈v, u〉, which is ≤ vu by Cauchy-Schwarz’ inequality.
If Im λ = 0 we obtain v ≤ 1 u, so that v is uniquely
|Im λ|
determined by u; in particular T has no non-real eigen-values. Furthermore,
suppose that u1, u2, . . . is a sequence in Sλ converging to u, and
that (vj, λvj + uj) ∈ GT . Then v1, v2, . . . is also a Cauchy sequence,
since vj − vk ≤ 1
|Im λ| uj − uk. Thus vj tends to some limit v, and
since T is closed we have (v, λv + u) ∈ GT . Hence u ∈ Sλ, so that Sλ
is closed and (2) follows.
Finally, if T is self-adjoint, then T ∗ = T is symmetric so it has no
non-real eigen-values. If Im λ = 0 it follows that Dλ = {0} so that (3)
follows and the proof is complete.
In the rest of this chapter we assume that T is a selfadjoint operator.
We define the resolvent set of T as
ρ(T ) = {λ ∈ C | T − λ has a bounded, everywhere defined inverse} ,
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