43.4. SOME IMPLICIT INCLUSIONS 1441

and so

2⟨Tu,u⟩= 2∫ T

0

⟨−Bu′,u

⟩dt = ⟨Bu(0) ,u(0)⟩ ≥ 0

Next, why is T ∗ = L? Let u ∈ D(L) . Then for v ∈ D(T ) ,

⟨T v,u⟩ =∫ T

0

⟨−Bv′,u

⟩dt =

∫ T

0

⟨−Bu,v′

⟩dt = ⟨−Bu,v⟩ |T0 +

∫ T

0

⟨(Bu)′ ,v

⟩ds

= ⟨Lu,v⟩

Hence u ∈ D(T ∗) and T ∗u = Lu since D(T ) is dense. Thus D(L)⊆ D(T ∗) and on D(L) ,these two are equal. Next suppose u ∈ D(T ∗) . Then for all v ∈ D(T ) ,⟨T v,u⟩ ≤C∥v∥V .Thus, by density and the Riesz representation theorem, there exists a unique g∗ ∈ V ′ suchthat

⟨T v,u⟩=∫ T

0⟨g∗,v⟩dt =

∫ T

0

⟨−Bv′,u

⟩dt =−

∫ T

0

⟨Bu,v′

⟩dt

In particular, it follows from the definition of weak V ′ valued distributions that g∗ = (Bu)′ .Simply specialize to letting v(t)= vφ (t) where φ ∈C∞

c (0,T ). Thus in particular (Bu)′ ∈V ′

and the above reduces to

⟨T v,u⟩=∫ T

0

⟨(Bu)′ ,v

⟩dt

⟨T v,u⟩ =∫ T

0

⟨−Bv′,u

⟩dt =

∫ T

0

⟨−Bu,v′

⟩dt = ⟨−Bu,v⟩ |T0 +

∫ T

0

⟨(Bu)′ ,v

⟩dt

= ⟨(Bu)(0) ,v(0)⟩+∫ T

0

⟨(Bu)′ ,v

⟩dt = ⟨(Bu)(0) ,v(0)⟩+ ⟨T v,u⟩

Thus also (Bu)(0) = 0. Hence D(T ∗)⊆ D(L) and this shows that L = T ∗ as claimed.Why is L monotone? From the material on weak derivatives,

Bu(t) =∫ t

0(Bu)′ (s)ds

and now use Theorem 43.3.2 to obtain

0≤ ⟨Bu(t) ,u(t)⟩= 2∫ t

0

⟨(Bu)′ ,u

⟩ds.

In particular, this holds for t = T and so ⟨Lu,u⟩ ≥ 0.Why is L∗ monotone? This follows from Proposition 25.8.2 and the fact that L = T ∗

shown above.G (L∗) = (τG (L))⊥

Consider (τS)⊥. To say that (x,y∗) ∈ (τS)⊥ is to say that if (a,b∗) ∈ S, then

⟨(x,y∗) ,(−b∗,a)⟩= 0

or in other words, ⟨x,b∗⟩ = ⟨y∗,a⟩ . To say that (x,y∗) ∈ τ

(S⊥)

is to say that (x,y∗) =(−c,d∗) where

⟨(d∗,c) ,(a,b∗)⟩= 0