PROBLEM SET 9, 18.155
DUE MIDNIGHT FRIDAY 18 NOVEMBER, 2011
Mostly about unbounded self-adjoint operators which are linear maps
A : D −→ H from a linear subspace D ⊂ H into H such that
(1) D ⊂ H is dense
(2) A is symmetric:- hAu, vi = hu, Avi for all u, v ∈ D.
(3) If v ∈ H and D 3 u 7−→ hAu, vi extends to a continuous linear
functional on H then v ∈ D.
The last condition is important of course – it actually says that the
‘maximal’ domain of the ‘adjoint’ is D. Note that the domain D is part
of the data for ‘an unbounded self-adjoint’ operator. The discussion
below effectively extends the spectral theorem from to the case of unbounded self-adjoint operators. After doing the problmes you might
consider the problem of showing that it f : R −→ R is a continous
function then f (A) is a well-defined unbounded self-adjoint operator
(defining the domain is part of the question). This is not required
for the problem set. Note that one important example of such an unbounded self-adjoint operator is the Laplacian on any (geodesically)
complete manifold.
(1) Show that an unbounded self-adjoint operator is closed in the
sense used in the last problem set.
(2) Derive the Closed Graph Theorem from the Open Mapping
Theorem: That is, show that if B : H −→ H is a linear map
with a closed graph (on a Hilbert space but it works between
Banach spaces) then B is bounded. Hint: The graph ΓB is
a closed subset of H ⊕ H so is itself a Hilbert space with the
projections onto the two factors πi : ΓB −→ H both bounded.
The first projection π1 is a bijection and B = π2 ◦ (π1 )−1 .
(3) If A is unbounded self-adjoint and t ∈ R \ {0} show that A +
it Id : D −→ H is a bijection with graph closed in H ⊕ H
and hence has a bounded inverse – meaning a bounded map
B : H −→ H with range contained in D such that (A + it)B =
IdH , B(A + it) = IdD .
(4) Show that (A ± i)−1 : H −→ D are adjoints of each other.
(5) Show that U = (A − i)(A + i)−1 is (defined and is) a unitary
operator on H and that if λ ∈ R then A − λ is a bijection from
D to H if and only if (λ − i)(λ + i)−1 is not in the spectrum
1
2
PROBLEM SET 9, 18.155 DUE MIDNIGHT FRIDAY 18 NOVEMBER, 2011
of U. Hint: The map R 3 λ 7−→ s = (λ − i)(λ + i)−1 is a
bijection to the complement of 1 in the unit circle and has
inverse λ = i(1 + s)(1 − s)−1 . With this relationship between s
and λ you can check that (A−λ)−1 = (1−s)−1 (A+i)−1 (U −s)−1 .