2356 CHAPTER 69. GELFAND TRIPLES

The following is about the Gelfand triple

V = Lp (D)⊆(H1

0)′ ⊆ (Lp (D))′

Lemma 69.1.1 It is possible to consider Lp (D)≡V as a dense subspace of(H1

0)′ ≡H as

follows. For f ∈ Lp (D) and φ ∈ H10 (D) ,

⟨ f ,φ⟩ ≡∫

Df (x)φ (x)dx

One can also consider H ≡(H1

0)′ as a dense subspace of (Lp (D))′ ≡ V ′ as follows. For

−∆φ ∈ H and f ∈ Lp (D) ,

⟨−∆φ , f ⟩ ≡ (−∆φ , f )H ≡ ⟨ f ,φ⟩

−∆ maps H10 (D) to H ≡

(H1

0)′ ⊆V ′.−∆ can be extended to yield a map−∆1 from Lp′ (D)

to V ′.

H10 (D)

−∆→(H1

0)′

Lp′ (D) =V−∆1→ V ′

Proof: First of all, note that by 69.1.1

|⟨ f ,φ⟩| ≤ || f ||Lp ||φ ||Lp′ ≤C || f ||Lp ||φ ||H10

and so it is certainly possible to consider Lp ⊆ H ≡(H1

0)′ as just claimed. Now why

can Lp (D) be considered dense in H ≡(H1

0)′? If it isn’t dense, then there exists ψ ∈

H10 (D) ,ψ ̸= 0 such that

(−∆ψ, f )H = 0

for all f ∈ Lp (D) . However, the above would say that for all f ∈ Lp,

(−∆ψ, f )H ≡ ⟨ f ,ψ⟩ ≡∫

Df ψ = 0

But ψ ∈ Lp′ (D) because H10 (D) embedds continuously into Lp′ (D) and so the above hold-

ing for all f ∈ Lp (D) implies by the usual Riesz representation theorem that ψ = 0 contraryto the way ψ was chosen.

Now consider the next claim. For −∆φ ∈ H ≡(H1

0)′ and f ∈ Lp (D) and from the first

part|⟨−∆φ , f ⟩| ≡ |(−∆φ , f )H | ≡ |⟨ f ,φ⟩| ≤C || f ||Lp ||φ ||H1

0 (D)

Thus −∆φ ∈ H can be considered in (Lp (D))′ . Why should H be dense in (Lp (D))′? If itis not dense, then there exists g∗ ∈ (Lp (D))′ which is not the limit of vectors of H. Then