18.155 LECTURE 12, 2015
22 OCTOBER
RICHARD MELROSE
Abstract. Notes before and after lecture – if you have questions, ask!
Read:
After lecture
Continued with operators on Hilbert space,
• Bounded operators, B(H).
• Invertible operators, GL(H) = {B ∈ B(H); ∃ E ∈ B(H), EB = BE = Id};
we write E = B −1 of course. An inverse is unique if it exists since F B = Id
implies F = F BE = E using associativity.
•
Proposition 1. GL(H) is open, in fact if B ∈ GL(H) and kAk < 1/kB −1 k
then B − A ∈ GL(H).
Proof by Neumann series:
•
Lemma 1. If kAk < 1 then Id −A ∈ GL(H).
Proof. Consider the Neumann series
∞
X
Ai .
i=1
This converges since it is Cauchy and B(H) is complete. The Cauchy
condition follows from the fact that for N ≥ M
k
N
X
Ai k ≤
i=M
N
X
kAki < kAkN (1 − kAk)−1 .
i=M
The limiting operator E satisfies
X
AE = EA =
Ai = Id −E =⇒ (Id −A)E = E(Id −A) = Id .
i>0
• Proof of Proposition. If B ∈ GL(H) and A ∈ B(H) then
B − A = B(Id −B −1 A)
which has inverse (Id −B −1 A)−1 B −1 since kB −1 Ak ≤ kB −1 kkAk < 1.
1
2
RICHARD MELROSE
• The spectrum and resolvent of a bounded operator A ∈ B(H) are complementary subsets of C :
Res(A) = {z ∈ C; A − z = A − z Id ∈ GL(H)}, Spec(A) = C \ Res(A).
The proposition above shows that if z0 ∈ Res(A), so A − z0 ∈ GL(H) then
if |z − z0 | < 1/k(A − z0 )−1 k it follows that A − z ∈ GL(H). Thus Res(A)
is open. Moveover, if |z| > kAk then A − z = −z(Id − z −1 A) ∈ GL(H) so
Spec(A) ⊂ {|z| ≤ kAk}.
• In general it is quite difficult to find the spectrum of a given operator. It
cannot be empty.
• Let’s consider self-adjoint operators.
Proposition 2. If A ∈ B(H) and A = A∗ then
Spec(A) ⊂ [−kAk, kAk] and either kAk or − kAk ∈ Spec(A).
The first condition is equivalent to Spec(A) ⊂ R, the second says that
one of A ± kAk is not invertible.
• First observe that if A = A∗ and z ∈
/ R then A − z is injective. Indeed,
(1)
Im((A − z)u, u) = − Im zkuk2 .
So if (A − z)u = 0 then u = 0.
• Next ‘recall’ a general fact for bounded operators.
Lemma 2. For any B ∈ B(H),
Ran(B) = (Nul(B ∗ ))⊥ .
Proof. This is really just the defining property of the adjoint
(Bu, v) = (u, B ∗ v) ∀ u, v ∈ H.
Thus if v ⊥ Ran(B) then both sides vanish for all the u and so B ∗ v =
0. Conversely if B ∗ v = 0 then v ⊥ Ran(B). This means Nul(B ∗ ) =
(Ran(B))⊥ . The null space is always closed but the range may not be so
we can only say ((Ran(B))⊥ )⊥ = Ran(B). Taking perps of both sides gives
(2).
• From this it follows for A = A∗ and z ∈
/ R that A − z̄ is also injective and
hence that Ran(A − z) is dense.
Other things from today:1. Unbounded self-adjoint operators
This is a slightly old-fashioned but still usefull concept. Such an operator A is a
linear map
(1)
A : D −→ H
where D ⊂ H is required to be a dense subset. In addition we demand
Symmetry: (Au, v) = (u, Av) ∀ u, v ∈ D.
This of course looks like self-adjointness but we give it a different name because we
require more than this. Namely set
D∗ = {u ∈ H; D 3 v 7−→ (Av, u) ∈ C extends to be continous on H}
L12
3
then we require D∗ = D. If u ∈ D∗ then the continuity assumptions means that
Riesz’ theorem gives us h ∈ H such that (Av, u) = (v, h) for all v ∈ D. Of course if
D∗ = D then symmetry implies that h = Au.
If L ∈ B(H) is bounded, self-adjoint and has dense range D then L−1 is an
unbounded self-adjoint operator and this is almost the general case.
If B is a bounded self-adjoint operator and f is a continous function on (−kAk, kAk)
then f (B) can be shown to be an unbounded self-adjoint operator.
The basic properties of unbounded self-adjoint operators follow from the fact
we can show that A + i is surjective as a map from D to H and has bounded
inverse which is a bounded, normal operator. Then E = (A + i)−1 (A − i)−1 is
1
a positive self-adjoint operator and B = AE 2 is a bounded self-adjoint operator.
The functional calculs for A can be derived from that of B, just use the fact that
1
x(1 + x2 )− 2 is an isomorphism of R to (−1, 1).
Department of Mathematics, Massachusetts Institute of Technology
E-mail address: rbm@math.mit.edu