Spectral Theory in Hilbert Space

46 8. COMPACTNESS
weakly convergent sequences are bounded (Theorem 3.9), any subsequence
of {Auj} ∞ 1 has a convergent subsequence. Suppose Aujk → v.
Then for any w ∈ H we have 〈v, w〉 = lim〈Aujk , w〉 = lim〈ujk , A∗w〉 =
〈u, A∗w〉 = 〈Au, w〉, so that v = Au. Hence the only point of accumulation
of {Auj} ∞ 1 is Au, so Auj → Au1 . This completes the proof
of (1). We leave the rest of the proof as an exercise for the reader
(Exercise 8.1).
Theorem 8.3. Suppose T is selfadjoint and its resolvent Rµ is
compact for some µ. Then Rλ is compact for all λ ∈ ρ(T ), and T has
discrete spectrum, i.e., σ(T ) consists of isolated eigenvalues with finite
multiplicity.
Proof. By the resolvent relation Rλ = (I + (λ − µ)Rλ)Rµ where
I is the identity so the first factor to the right is bounded. Hence Rλ
is compact by Theorem 8.2.3.
∆
Now let ∆ be a bounded interval. If u ∈ E∆H then Rλu 2 =
d〈Etu,u〉
|t−λ| 2
≥ Ku 2 where K = inft∈∆|t − λ| −2 > 0 (verify this calcu-
lation!). We have Rλuj → 0 if uj ⇀ 0, so the inequality shows that
any weakly convergent sequence in E∆H is strongly convergent (the
identity operator is compact). This implies that E∆H has finite dimension
(for example since an orthonormal sequence converges weakly
to 0 but is not strongly convergent). In particular eigenspaces are
finite-dimensional. It also follows that any bounded interval can only
contain a finite number of points of increase for Et, because projections
belonging to disjoint intervals have orthogonal ranges (Exercise 8.2).
This completes the proof.
Resolvents for different selfadjoint extensions of a symmetric operator
are closely related. In particular, we have the following theorem.
Theorem 8.4. Suppose a densely defined symmetric operator T0
has a selfadjoint extension with compact resolvent and that dim Dλ <
∞ for some λ ∈ C \ R. Then every selfadjoint extension of T0 has
compact resolvent.
Proof. Let Im λ = 0 and Rλ, ˜ Rλ be resolvents of selfadjoint extensions
of T0. Then A = Rλ − ˜ Rλ has its range in Dλ, since Rλu and
˜Rλu both solve the equation T ∗ 0 v = λv + u. It follows that A is a compact
operator, since if {uj} ∞ 1 is a bounded sequence in H, then {Auj} ∞ 1
is a bounded sequence in a finite-dimensional space. By the Bolzano-
Weierstrass theorem there is therefore a convergent subsequence. If ˜ Rλ
is compact it therefore follows that Rλ = ˜ Rλ + A is compact.
1 If not, there would be a neighborhood O of Au and a subsequence of {Auj} ∞ j=1
that were outside O. But we could then find a convergent subsequence which does
not converge to Au.