54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1715

For u ∈V, define ∇u to be that element of L2 (Ω;Cn,a(x)dmn) , the space of vector valuedL2 functions taken with respect to the measure a(x)dmn which satisfies

|∇u−∇un|L2(Ω;Cn,a(x)dmn)→ 0.

Denote this space by W for simplicity of notation.

Observation 54.3.12 V is a Hilbert space with inner product given by

(u,v)1 ≡∫

Ω

(auv+a(x)∇u ·∇v

)dx

Everything is obvious except completeness. Suppose then that {un} is a Cauchy se-quence in V. Then there exists a unique u ∈ L2 (Ω) such that |un−u|L2(Ω)→ 0. Now let

|wn−un|L2(Ω)+ |∇wn−∇un|W < 1/2n

It follows {∇wn} is also a Cauchy sequence in W while {wn} is a Cauchy sequence inL2 (Ω) converging to u. Thus the thing to which ∇wn converges in W is the definition of∇u and u ∈V. Thus

||un−u||1 ≤ ||un−wn||1 + ||wn−u||1

<12n + ||wn−u||1

and the last term converges to 0. Hence V is complete as claimed.Then it is clear V is a Hilbert space. The next observation is a simple one involving the

Riesz map.

Definition 54.3.13 Let V be a Hilbert space and let V ′ be the space of continuous conju-gate linear functions defined on V . Then define R : V →V ′ by

Rx(y)≡ (x,y) .

This is called the Riesz map.

Lemma 54.3.14 The Riesz map is one to one and onto and linear.

Proof: It is obvious it is one to one and linear. The only challenge is to show it is onto.Let z∗ ∈V ′. If z∗ (V ) = {0} , then letting z = 0, it follows Rz = z∗. If z∗ (V ) ̸= 0, then

ker(z∗)≡ {x ∈V : z∗ (x) = 0}

is a closed subspace. It is closed because z∗ is continuous and it is just z∗−1 (0) . Sinceker(z∗) is not everything in V there exists

w ∈ ker(z∗)⊥ ≡ {x : (x,y) = 0 for all y ∈ ker(z∗)}

and w ̸= 0. Then

z∗(

z∗ (x)w− z∗ (w)x)= z∗ (x)z∗ (w)− z∗ (w)z∗ (x) = 0