Lecture 4. The Lp theory of semigroups and diffusion operators as generators of Dirichlet forms

Contents

The Lp theory of heat semigroups

Our goal, in this section, is to define, for 1 ≤ p ≤ +∞, Pt on Lp := Lp(X,μ). This may be done in a natural way by using the Riesz-Thorin interpolation theorem that we recall below.

Theorem (Riesz-Thorin interpolation theorem)

Let 1 ≤ p0, p1, q0, q1 ≤ ∞, and θ ∈ (0,1). Define 1 ≤ p,q ≤ ∞ by

1/p = (1-θ)/p0 + θ/p1,     1/q = (1-θ)/q0 + θ/q1.

If T is a linear map such that

T : Lp0 → Lq0, ||T||Lp0→Lq0 = M0

T : Lp1 → Lq1, ||T||Lp1→Lq1 = M1,

then, for every f ∈ Lp0 ∩ Lp1,

||Tf||q ≤ M01-θM1θ ||f||p.

Hence T extends uniquely as a bounded map from Lp to Lq with

||T||Lp→Lq ≤ M01-θM1θ.

Remark The statement that T is a linear map such that

T : Lp0 → Lq0, ||T||Lp0→Lq0 = M0

T : Lp1 → Lq1, ||T||Lp1→Lq1 = M1

means that there exists a map T : Lp0 ∩ Lp1 → Lq0 ∩ Lq1 with

supf ∈ Lp0 ∩ Lp1 , ||f||p0 ≤ 1 ||Tf||q0 = M0

and

supf ∈ Lp0 ∩ Lp1 , ||f||p1 ≤ 1 ||Tf||q1 = M1.

In such a case, T can be uniquely extended to bounded linear maps T0 : Lp0 → Lq0 , T1 : Lp1 → Lq1. With a slight abuse of notation, these two maps are both denoted by T in the theorem.

Remark If f ∈ Lp0 ∩ Lp1 and p is defined by

1/p = (1-θ)/p0 + θ/p1,

then by Hölder’s inequality, f ∈ Lp and

||f||p ≤ ||f||p01-θ ||f||p1θ.

We now are in position to state the following theorem:

Theorem Let (Pt)t≥0 be a strongly continuous self-adjoint contraction Markovian semigroup on L2(X,μ). The space L1 ∩ L is invariant under Pt and Pt may be extended from L1 ∩ L to a contraction semigroup (Pt(p))t≥0 on Lp for all 1 ≤ p ≤ ∞: For f ∈ Lp,

||Ptf||Lp ≤ ||f||Lp.

These semigroups are consistent in the sense that for f ∈ Lp ∩ Lq,

Pt(p)f = Pt(q)f.

Proof

If f,g ∈ L1 ∩ L which is a subset of L1 ∩ L, then,

|∫X (Ptf)g dμ| = |∫X f (Ptg) dμ| ≤ ||f||L1 ||Ptg||L ≤ ||f||L1 ||g||L.

This implies

||Ptf||L1 ≤ ||f||L1.

The conclusion follows then from the Riesz-Thorin interpolation theorem.

Exercise Show that if f ∈ Lp and g ∈ Lq with 1/p + 1/q = 1 then,

n f Pt(q) g dμ = ∫n g Pt(p) f dμ.

Exercise 

  1. Show that for each f ∈ L1, the L1-valued map t → Pt(1)f is continuous.
  2. Show that for each f ∈ Lp, 1 < p < 2, the Lp-valued map t → Pt(p)f is continuous.
  3. Finally, by using the reflexivity of Lp, show that for each f ∈ Lp and every p ≥ 1, the Lp-valued map t → Pt(p)f is continuous.

We mention, that in general, the L valued map t → Pt(∞)f is not continuous.

Diffusion operators as generators of Dirichlet forms

Consider a diffusion operator

L = ∑i,j=1n σij(x) ∂2/∂xi∂xj + ∑i=1n bi(x) ∂/∂xi,

where bi and σij are continuous functions on n and for every x ∈ ℝn, the matrix ij(x))1 ≤ i,j ≤ n is a symmetric and non-negative matrix.

Assume that there is Borel measure μ on ℝn which is equivalent to the Lebesgue measure and that symmetrizes L in the sense that for every smooth and compactly supported functions f, g : ℝn → ℝ,

n gLf dμ = ∫n fLg dμ.

For instance, if one can write

Lf = -div(a ∇f),

where a is a smooth field of positive and symmetric matrices, then the Lebesgue measure symetrizes L. From a previous lemma  the quadratic form

ℰ(f,g) = -∫n g Lf dμ, f,g ∈ Cc(ℝn)

is closable. Let ℰ̄ denotes its closure in L2(ℝn,μ).

Proposition The quadratic form ℰ̄ is a Dirichlet form.

Proof

We need to prove that ℰ̄ is Markovian. It is enough to prove that if u ∈ ℱ = 𝒟(ℰ), then |u| ∈ ℱ with ℰ̄(|u|, |u|) ≤ ℰ̄(u,u) and that if u ∈ ℱ with u ≥ 0, then u ∧ 1 ∈ ℱ with ℰ̄(|u|,|u|) ≤ ℰ̄(u,u). We prove the first requirement, the second being established in a similar manner is let as an exercise to the reader.

Let u ∈ Cc(ℝn) and consider

ϕε(x) = (x2 + ε2)^{1/2}, ε > 0.

One can check that ϕε(u) → |u| in L2(ℝn,μ) and that ϕε(u) is a Cauchy sequence for the norm

||f||2 = ||f||2L2(ℝn,μ) + ℰ̄(f,f).

Since ℰ̄ is closed this implies that |u| ∈ ℱ and that ϕε(u) → |u| converges to u in the above norm.

Now, using chain rule we see that for every smooth function u ∈ Cc(ℝn),

ε(u)(x) ≥ u(x)/(u(x)2 + ε2)^{1/2} Lu(x).

Multiplying by ϕε(u) and integrating we get

ℰ(ϕε(u), ϕε(u)) ≤ ℰ(u,u)

Taking the limit ε → 0 yields

ℰ̄(|u|, |u|) ≤ ℰ̄(u,u)

The above inequality extends then to all u ∈ ℱ by using the density of Cc(ℝn) in the || · || norm and the closedness of ℰ.

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