42.3. SUBGRADIENTS 1385

Definition 42.3.11 If X is a Banach space define u ∈W 1,p ([0,T ] ;X) if there exists g ∈Lp ([0,T ] ;X) such that

u(t) = u(0)+∫ t

0g(s)ds

When this occurs define u′ (·)≡ g(·) . As usual, p > 1.

The next Lemma is quite interesting for its own sake but it is also used in the nexttheorem.

Lemma 42.3.12 Suppose g ∈ Lp (0,T ;X) . Then as h→ 0,

1h

∫ (·)+h

(·)g(s)dsX[0,T−h] (·)→ g

in Lp ([0,T ] ;X) .

Proof: Let

g̃(u)≡{

g(u) if u ∈ [0,T ]0 if u /∈ [0,T ] , φ h (r)≡

1hX[−h,0] (r) .

Thus g̃ ∈ Lp (R;X) and

g̃∗φ h (t)≡∫R

g̃(t− s)φ h (s)ds.

Then

||g̃∗φ h− g̃||Lp(R;X) ≤(∫

R

(∫R||g̃(t)− g̃(t− s)||X φ h (s)ds

)p

dt)1/p

which by Minkowski’s inequality for integrals is no larger than

≤∫R

φ h (s)(∫

R||g̃(t)− g̃(t− s)||pX dt

)1/p

ds

=1h

∫ 0

−h

(∫R||g̃(t)− g̃(t− s)||pX dt

)1/p

ds <1h

∫ 0

−hεds = ε

whenever h is small enough. This follows from continuity of translation in Lp (R;X) , aconsequence of the regularity of the measure. Thus, g̃∗φ h→ g̃ in Lp (R;X) . Now

g̃∗φ h (t)−1h

∫ t+h

tg(s)dsX[0,T−h] (t)

=

{0 if t ∈ [0,T −h]1h∫ t+h

t g̃(u)du if t /∈ [0,T −h]

and therefore, ∣∣∣∣∣∣∣∣g̃∗φ h (·)−1h

∫ (·)+h

(·)g(s)dsX[0,T−h] (·)

∣∣∣∣∣∣∣∣Lp(R;X)

=