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Preliminaries: Self-adjoint Operators
Let (H,⟨⋅,⋅⟩) be a Hilbert space with norm ‖f‖2=⟨f,f⟩ and let A be a H-valued densely defined operator on a domain 𝒟(A). We recall the following basic definitions.
- The operator A is said to be closed if xn → x in H and Axn → y in H imply that y=Ax.
- The operator A is said to be symmetric if for f,g ∈ 𝒟(A), ⟨f,Ag⟩ = ⟨Af,g⟩
- The operator A is said to be non-negative symmetric operator if it is symmetric and if for f ∈ 𝒟(A), ⟨f,Af⟩ ≥ 0. It is said to be non-positive, if for f ∈ 𝒟(A), ⟨f,Af⟩ ≤ 0
- The adjoint A* of A is an operator defined on the domain
𝒟(A*) = {f ∈ H : ∃ c(f) ≥ 0, ∀ g ∈ 𝒟(A), |⟨f,Ag⟩| ≤ c(f)‖g‖}
Since for f ∈ 𝒟(A*), the map g → ⟨f,Ag⟩ is bounded on 𝒟(A), it extends thanks to the Hahn-Banach theorem to H. The Riesz representation theorem allows then to define A* by the formula ⟨A*f,g⟩ = ⟨f,Ag⟩ where g ∈ 𝒟(A), f ∈ 𝒟(A*). Since 𝒟(A) is dense, A* is uniquely defined.
- The operator A is said to be self-adjoint if it is symmetric and if 𝒟(A*) = 𝒟(A).
Let us observe that, in general, the adjoint A* is not necessarily densely defined, however it is readily checked that if A is a symmetric operator then, from Cauchy-Schwarz inequality, 𝒟(A) ⊂ 𝒟(A*). Thus, if A is symmetric, then A* is densely defined. The following result is often useful and classical.
Lemma Let A : 𝒟(A) ⊂ H → H be an injective densely defined self-adjoint operator. Let us denote by 𝓡(A) the range of A. The inverse operator A-1 : 𝓡(A) → H is a densely defined self-adjoint operator.
A major result in functional analysis is the spectral theorem.
Theorem (Spectral theorem) Let A be a non-negative self-adjoint operator on H. There is a measure space (Ω,ν), a unitary map U : L2(Ω,ν) → H and a non-negative real-valued measurable function λ on Ω such that
U-1 A U f (x) = λ(x) f(x)
for x ∈ Ω, Uf ∈ 𝒟(A). Moreover, given f ∈ L2(Ω,ν), Uf belongs to 𝒟(A) if only if ∫Ω λ2 f2 dν < +∞.
Definition (Functional calculus) Let A be a non-negative self-adjoint operator on H. Let g : ℝ≥0 → ℝ be a Borel function. With the notations of the spectral theorem, one defines the operator g(A) by the requirement
U-1 g(A) U f (x) = g(λ(x)) f(x)
with 𝒟(g(A)) = {Uf, (g ∘ λ) f ∈ L2(Ω,ν) }.
Exercise Show that if A is a non-negative self-adjoint operator on H and g is a bounded Borel function, then g(A) is a bounded operator on H.
Semigroups and generators
Definition A strongly continuous self-adjoint contraction semigroup is a family of self-adjoint operators (Pt)t≥0 : H → H such that:
- For s, t ≥ 0, Pt ∘ Ps = Ps+t (semigroup property);
- For every f ∈ H, limt → 0 Pt f = f (strong continuity);
- For every f ∈ H and t ≥ 0, ‖Pt f‖ ≤ ‖f‖ (contraction property).
Theorem Let (Pt)t≥0 be a strongly continuous self-adjoint contraction semigroup on H. There exists a self-adjoint, non-positive, and densely defined operator A : 𝒟(A) → H where
𝒟(A) = {f ∈ H : limt → 0 (Ptf – f)/t exists}
such that for f ∈ 𝒟(A),
limt → 0 || (Ptf – f)/t – Af || = 0.
The operator A is called the generator of the semigroup (Pt)t≥0. We also say that A generates (Pt)t≥0. Conversely, if A is a densely defined non-positive self-adjoint operator on H, then it is the generator of the strongly continuous self-adjoint contraction semigroup on H defined as Pt = etA.
Let us consider the following bounded operators on H:
At = (1/t) ∫0t Ps ds
For f ∈ H and h > 0, we have
(1/t)(PtAhf – Ahf) = (1/ht) ∫0h (Ps+tf – Psf) ds
=(1/ht)[ ∫th+t Psf ds – ∫0h Psf ds ]
= (1/ht)[ ∫hh+t Psf ds + ∫th Psf ds – ∫0h Psf ds ]
= (1/ht) ∫0t (Ps+hf – Psf) ds
Therefore, we obtain
limt → 0 (1/t) (PtAhf – Ahf) = (1/h) (Phf – f)
This implies,
{Ahf : x ∈ H, h > 0} ⊂ {f ∈ H : limt → 0 (Ptf – f)/t exists}
Since limh → 0 Ah f = f, we deduce that
{f ∈ H : limt → 0 (Ptf – f)/t exists}
is dense in H. We can then consider
Af := limt → 0 (Ptf – f)/t,
which is of course defined on the domain
𝒟(A) = {f ∈ H : limt → 0 (Ptf – f)/t exists}.
The operator A is closed, indeed if fn → f and Afn → g then, using similar computations as before,
Ahg = 1/h ∫0h Psg ds = limn →+∞ 1/h ∫0h PsAfn ds
= limn →+∞ limt → 0 1/ht ∫0h Ps+t fn – Psfn ds
=limn →+∞ limt → 0 1/ht ∫0t Ps+h fn – Psfn ds
= limn →+∞ 1/h (Phfn – fn) = 1/h(Phf – f)
Taking then the limit as h → 0 yields y = Ax. We now prove that A is a non-positive self-adjoint operator. First, one has for every f ∈ H
⟨Af,f⟩ = limt → 0 ⟨(Ptf – f)/t , x⟩
= limt → 0 (⟨Ptf,f⟩ – ||f||2)/t
=limt → 0 (||Pt/2f||2 – ||f||2)/t ≤ 0
From its definition, it is plain that A is symmetric but proving self-adjointness is a little more involved. Let λ > 0. We will to prove that λId – A is a bijective operator D(A) → H whose inverse is self-adjoint and conclude with a previous lemma.
The formal Laplace transform formula
∫0+∞ e-λt etA dt = (λId – A)-1,
suggests that the operator
Rλ = ∫0+∞ e-λt Pt dt
is the inverse of λId – A. We show this is indeed the case. First, let us observe that Rλ is well-defined as a Riemann integral since t → Pt is continuous and ||Pt|| ≤ 1. We now show that for f ∈ H, Rλx ∈ 𝒟(A). For h > 0,
(Ph – Id)/h Rλ f = ∫0+∞ e-λt (Ph – Id)/h Pt f dt
= ∫0+∞ e-λt (Ph+t – Pt)/h f dt
= eλh∫h+∞ e-λs (Ps – Ps-h)/h f ds
= (eλh/h) (Rλf – ∫0h e-λs Psf ds – ∫h+∞ e-λs Ps-hf ds)
= ((eλh – 1)/h)Rλf – (eλh/h) ∫0h e-λs Psf ds
By letting h → 0, we deduce that Rλf ∈ 𝒟(A) and moreover
ARλf = λRλf – f.
Therefore we proved
(λId – A)Rλ = Id.
Furthermore, it is readily checked that, since A is closed, for f ∈ 𝒟(A),
A Rλf = A ∫0+∞ e-λt Ptf dt = ∫0+∞ e-λt APtf dt = ∫0+∞ e-λt PtAf dt = RλAf.
We therefore conclude
(λId – A)Rλ = Rλ(λId – A) = Id.
Thus,
Rλ = (λId – A)-1,
The operator ∫0+∞ e-λt Pt dt is seen to be self-adjoint (it is symmetric and bounded), thus (λId – A)-1 is also self-adjoint. From the previous lemma, we deduce that λId – A is self-adjoint, from which we conclude that A is self-adjoint (exercise !).
Conversely, let A be a densely defined non-positive self-adjoint operator on H and define Pt = etA. More precisely, from the spectral theorem, there is a measure space (Ω, ν), a unitary map U : L2(Ω,ν) → H and a non-negative real-valued measurable function λ on Ω such that
U-1 A U f (x) = -λ(x) f(x),
for x ∈ Ω, Uf ∈ 𝒟(A). We define then Pt : H → H such that
U-1 Pt U f (x) = e-tλ(x) f(x),
and let as an exercise the proof that (Pt)t≥0 is a strongly continuous self-adjoint contraction semigroup on H with generator A.
Research and Lecture notes 