Measure, Integration & Real Analysis, 2021a

Section 10C Compact Operators 317
10.77 closed range
If T is a compact operator on a Hilbert space, then T − αI has closed range for
every α ∈ F with α ̸= 0.
Proof
α ̸= 0.
10.78
Suppose T is a compact operator on a Hilbert space V and α ∈ F is such that
Claim: there exists r > 0 such that
‖ f ‖≤r‖(T − αI) f ‖ for all f ∈ ( null(T − αI) ) ⊥ .
To prove the claim above, suppose it is false. Then for each n ∈ Z + , there exists
f n ∈ ( null(T − αI) ) ⊥ such that
‖ f n ‖ = 1 and ‖(T − αI) f n ‖ < 1 n .
Because T is compact, there exists a subsequence Tf n1 , Tf n2 ,...such that
10.79 lim
k→∞
Tf nk = g
for some g ∈ V. Subtracting the equation
10.80 lim
k→∞
(T − αI) f nk = 0
from 10.79 and then dividing by α shows that
lim
k→∞ f n k
= 1 α g.
The equation above implies ‖g‖ = |α|; hence g ̸= 0. Each f nk ∈ ( null(T − αI) ) ⊥ ;
hence we also conclude that g ∈ ( null(T − αI) ) ⊥ . Applying T − αI to both sides of
the equation above and using 10.80 shows that g ∈ null(T − αI). Thus g is a nonzero
element of both null(T − αI) and its orthogonal complement. This contradiction
completes the proof of the claim in 10.78.
To show that range(T − αI) is closed, suppose h 1 , h 2 ,... is a sequence in
range(T − αI) that converges to some h ∈ V. For each n ∈ Z + , there exists
f n ∈ ( null(T − αI) ) ⊥ such that (T − αI) fn = h n . Because h 1 , h 2 ,...is a Cauchy
sequence, 10.78 shows that f 1 , f 2 ,...is also a Cauchy sequence. Thus there exists
f ∈ V such that lim n→∞ f n = f , which implies h =(T − αI) f ∈ range(T − αI).
Hence range(T − αI) is closed.
Suppose T is a compact operator on a Hilbert space V and f ∈ V and α ∈ F \{0}.
An immediate consequence (often useful when investigating integral equations) of
the result above and 10.13(d) is that the equation
Tg − αg = f
has a solution g ∈ V if and only if 〈 f , h〉 = 0 for every h ∈ V such that T ∗ h = αh.
Measure, Integration & Real Analysis, by Sheldon Axler