1386 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE
(∫ 0
−h
∣∣∣∣∣∣∣∣1h∫ t+h
tg̃(u)du
∣∣∣∣∣∣∣∣pX
dt)1/p
+
(∫ T
T−h
∣∣∣∣∣∣∣∣1h∫ t+h
tg̃(u)du
∣∣∣∣∣∣∣∣pX
dt)1/p
≤ 1h
(∫ 0
−h
(∫ h
−h||g̃(u)||X du
)p
dt)1/p
+1h
(∫ T
T−h
(∫ T+h
T−h||g̃(u)||X du
)p
dt)1/p
which by Minkowski’s inequality for integrals is no larger than
≤ 1h
∫ h
−h
(∫ 0
−h||g̃(u)||pX dt
)1/p
du+1h
∫ T+h
T−h
(∫ T
T−h||g̃(u)||pX dt
)1/p
du
≤ 1h
∫ h
−hεdu+
1h
∫ T+h
T−hεdu = 4ε
whenever h is small enough because of the fact that ||g̃||pX ∈ L1 (R;X). Since ε is arbitrary,this shows ∥∥∥∥g̃∗φ h (·)−
1h
∫ (·)+h
(·)g(s)dsX[0,T−h] (·)
∥∥∥∥Lp(R;X)
→ 0
and also, it was shown above that
∥g̃∗φ h (·)− g̃∥Lp(R;X)→ 0
It follows that1h
∫ (·)+h
(·)g(s)dsX[0,T−h] (·)→ g̃
in Lp (R;X) and consequently in Lp ([0,T ] ;X) as well. But g̃ = g on [0,T ] .The following theorem is a form of the chain rule in which the derivative is replaced by
the subgradient.
Theorem 42.3.13 Suppose u ∈W 1,p ([0,T ] ;X) ,z ∈ Lp′ ([0,T ] ;X ′), and z(t) ∈ δφ (u(t))a.e t ∈ [0,T ] . Then the function, t→ φ (u(t)) is in L1 (0,T ) and its weak derivative equals⟨z,u′⟩ . In particular,
φ (u(t))−φ (u(0)) =∫ t
0
⟨z(s) ,u′ (s)
⟩ds
Proof: Modify u on a set of measure zero such that δφ (u(t)) ̸= /0 for all t. Next modifyz on a set of measure zero such that for ũ and z̃ the modified functions, z̃(t) ∈ δφ (ũ(t)) forall t. First I claim t→ φ (ũ(t)) is in L1 (0,T ). Pick t0 ∈ [0,T ] and let
z̃(t0) ∈ δφ (ũ(t0)) .
Then for t ∈ [0,T ],⟨z̃(t0) , ũ(t)− ũ(t0)⟩+φ (ũ(t0))≤