(e) An operatorS ∈ B(H)commutes with everyL∈A if and only ifS commutes with every projection E(ω).
Now, we have a spectral theorem found bounded normal operators.
Theorem 3.19. [18, Theorem 12.23 on p.324] If L ∈ B(H) and L is normal, there exists a unique resolution of the identity E on the Borel subsets of σ(L)
which satisfies:
L=
Z
σ(L)
λdE(λ)
Furthermore, every projection E(ω) commutes with every S ∈ B(H), which com- mutes with L. This E is also called the spectral decomposition of L.
Remark 3.20. Iff is a bounded Borel function onσ(L), givenEis the spectrum decomposition of L, then the operator:
Φ(f) =
Z
σ(L) f dE
is well-defined. In particular, if f ∈C(σ(L)), then f → f(L) is an isomorphism onC(σ(L)) satisfying: kf(T)xk2 = Z σ(L) |f|2dE x,x
3.5
Spectrum Theorem for Unbounded Normal
Operators
In most cases, operators are not bounded. Let H be a Hilbert space. An un- bounded operator is a pair D(L), L, where D(L) ⊂ H and L : D(L) → H is linear. The graph G(L) of L is defined as the set
{(x, y)⊂H×H|Lx=y,∀x∈ D(L)}
We say thatL is a closed operator if its graph is a closed subspace ofH×H. A operatorS is said to be an extension ofLif G(L)⊂ G(S) (That isD(L)⊂D(S) and Lx=Sxfor x∈D(L)).
In this situation, D(L∗) 6= D(L) in general. Consider a continuous linear func- tional
x→
Lx, y
By Hahn-Banach theorem, it can be extended to anyx∈H. If y∈D(L∗), there exists a L∗y ∈H satisfying:
Lx, y =x, L∗y, x∈D(L)
Therefore, L∗y is uniquely determined if and only L is densely defined (That is
D(L) is dense inH).
Definition 3.21. An operator L inH is said to be symmetric if: (Lx, y) = (x, Ly)
where x, y ∈D(L).
It follows that L ⊂ L∗. Further, L is said to be self-adjoint if L = L∗. A symmetric operator L is said to be maximally symmetric if L has no proper symmetric extension. (That is L⊂S and S is symmetric, then L=S)
Theorem 3.22. Let L be symmetric operator on H (not necessarily densely de- fined), the following statements hold:
(a) kLx+ixk2 =kxk2+kLxk2, ∀x∈D(L)
(b) L is a closed operator if and only of R(L+iI) is closed. (c) L+iI is one-to-one.
(d) If R(L+iI) =H, then L is maximally symmetric (e) Statements 1-4 are still true if we replace i by −i. Proof. Since L is a symmetric operator, then:
kLx+ixk2 =kxk2+kLxk2+ (ix, Lx) + (Lx, ix)
=kxk2+kLxk2−(Lx, ix) + (Lx, ix) (By symmetricity) =kxk2+kLxk2
It proves (a). For (b), the result in (a) shows there is an one-to-one correspondence between the range ofL+iIandD(L). Hence, (c) is also followed. For (d), assume that there exists a proper extension S. Since the range of L+iI is the whole spaceH, thenS+iI is not one-to-one. By the statement (c), Sis not symmetric. It finished the proof of (d). Finally, all proofs above work when we replace i by
3.5. SPECTRUM THEOREM FOR UNBOUNDED NORMAL OPERATORS47 Since the above theorem shows that
kLx+ixk2 =kxk2+kLxk2 =kLx−ixk2
then there exists an isometry U such that:
U(Lx+ix) = Lx−ix for x∈D(L) with
D(U) =R(L+iI), R(U) =R(L−iI)
Since (L+iI)−1 is a map from D(U) to D(L), thus U can be written as: U = (L+iI)(L−iI)−1
The map L → U is called the Cayley transform of the symmetric operator L. Several important statements about Cayley transform are stated below.
Theorem 3.23. Suppose U is the Cayley transform of a symmetric operator L
in H. Then, the following statements are true: (a) U is closed if and only if L is closed.
(b) R(I −U) = D(L), I−U is one-to one and T can be reconstructed from U
by
L=i(I+U)(I−U)−1
(c) U is unitary if and only if T is self-adjoint.
Proof. I will mainly prove statement (b), which will be used in constructing the spectral theorem for unbounded operators. Others can be found in [18, Theorem 13.19 on p.358]. By Theorem 3.22, L+iI is one-to-one and thus there exists one to one correspondence between D(L) and R(L+iI). Denote z =Lx+ix∈
R(L+iI) and U z = Lx−ix for x ∈ D(L). Subtracting and summing them lead to (I −U)z = 2ix and (I +U)z = 2Lx. Thus, I −U is one-to-one and
R(I−U) = D(L). It follows that (I−U)−1 maps fromD(L) onto H and so,
Theorem 3.18 holds for everyf ∈L∞(E) and Φ(f)∈ B(H). Given the same resolution of the identity E : M → B(H), it can extend to any unbounded measurable functions. The following lemma insures that such a construction is valid.
Lemma 3.24. [18, Lemma 13.23 on p.361] Let f : Ω→C be measurable. Let
Df ={x∈H : Z
Ω
|f|2dE
x,x <∞}
Then Df is a dense subset of H. Further, if x, y ∈H, then Z Ω |f|d|Ex,y| ≤ kyk nZ Ω |f|2dE x,x o12
Theorem 3.25. [18, Theorem 13.24 on p.362] LetE be a resolution of the iden- tity on a set Ω.
(a) To every measurable f : Ω→Ccorresponds a densely defined closed operator
Φ(f) in H, with domain D(Φ(f)) =D(f), which is characterised by
Φ(f)x, y = Z Ω f dEx,y and satisfies kΦ(f)xk2 = Z Ω |f|2dEx,x
(b) If and g are measurable, then
Φ(f)Φ(g) = Φ(f g) and D(Φ(f)Φ(g)) = Dg∩Df g
Hence, Φ(f)Φ(g) = Φ(f g) if and only if Df g ⊂Dg
(c) For every measurable f : Ω→C
Φ(f∗) = Φ(f)
and
Φ(f)Φ(f)∗ = Φ(|f|2) = Φ(f)∗ Φ(f)
3.5. SPECTRUM THEOREM FOR UNBOUNDED NORMAL OPERATORS49 Theorem 3.26. [18, Theorem 13.27 on p.366] Suppose E is a resolution of the identity on Ω. Let f : Ω→C be a measurable function, and:
ωα ={p∈Ω :f(p) =α} α ∈C
(a) If α is in the essential range of f and E(ωα) 6= 0, then α is in the point
spectrum of Φ(f).
(b) If α is in the essential range off but E(ωα) = 0, then α is in the continuous
spectrum of Φ(f).
(c) σ(Φ(f)) is the essential range of f.
Finally, we reach our main theorem.
Theorem 3.27. Let H be a Hilbert space. Let L be a self-adjoint operator in H. There exists a unique resolution E of the identity such that:
Lx, y =
Z
σL
tdEx,y(t) (∀x∈D(L).∀y∈H)
In this case, we also call E the spectral decomposition.
Proof. LetU be the Cayley transform of a self-adjoint operator L. Let ¯E be the spectral decomposition of U. By Theorem 3.23, U is unitary (so it is normal) and I−U is one-to-one. Therefore, 1 is not an eigenvalue of L. Referring to the page 328 Rudin’s book [18], then ¯E({1}) = 0. Since the unitary operator U is bounded and normal, then applying Theorem 3.19
U x, y=
Z
Ω
λE¯x,y(λ)
where Ω is a unit ball without point 1 and Ex,y in unque. Define a measurable
function
f(λ) = i(1 +λ)
1−λ (λ∈Ω)
and thus by Theorem 3.25, there exists a densely defined closed operator:
Φ(f)x, y =
Z
Ω
f dEx,y
Since f(λ)(1−λ) = i(1 +λ), then by (b) of Theorem 3.25 Φ(f)(I −U) = i(I+U)
It impliesR(I−U)⊂D(Φ(f)). In addition, by Theorem 3.23, we can reconstruct L from U such that:
L(I−U) = i(I+U)
Therefore, D(L) = R(I −U) ⊂ D(Φ(f)). Since f is a one-to-one map from a unit circle to the real line, then Φ(f) is a self-adjoint extension of L. Every self-adjoint operator is maximally symmetric, then Φ(f) =L. By (c) of Theorem 3.26, σ(L) is the essential range of f. Hence, define E(f(ω)) := ¯E(ω) for every Borel set ω ⊂Ω, we have
Lx, y= Z Ω f dE¯x,y = Z σ(L) tdEx,y(t), (∀x∈D(L),∀y∈H)
The uniqueness of E comes from the uniqueness of ¯E.