1418 CHAPTER 43. INTERPOLATION IN BANACH SPACE
Definition 43.1.11 LetH1 (0,T,X)
denote the functions f ∈ L2 (0,T,X) whose weak derivative f ′ is also in L2 (0,T,X).
Proposition 43.1.12 Let f ∈ H1 (0,T,X). Then f ∈ C0,(1/2) ([0,T ] ,X) and the inclusionmap is continuous.
Proof: First note that
f (t)− f (s) =∫ t
sf ′ (r)dr
and so
∥ f (t)− f (s)∥X ≤∫ t
s
∥∥ f ′ (r)∥∥
X dr ≤ ∥ f∥H1 |t− s|1/2
It follows that
sup0≤s<t≤T
∥ f (t)− f (s)∥|t− s|1/2 ≤ ∥ f∥H1
Also
f (t) = f (0)+∫ t
0f ′ (s)ds
so
∥ f (t)∥ ≤ ∥ f (0)∥+∫ t
0
∣∣ f ′ (s)∣∣ds≤ ∥ f (0)∥+T 1/2 ∥ f∥H1
Now consider ∥ f (0)∥ . Then integrating by parts yields∫ T
0(T − t) f ′ (t)dt = (T − t) f (t) |T0 +
∫ t
0f (t)dt
and so
T ∥ f (0)∥ ≤∫ T
0∥ f (t)∥dt +T
∫ T
0
∥∥ f ′ (t)∥∥dt ≤C (T )∥ f∥H1 .
Hencesup
t∈[0,T ]∥ f (t)∥ ≤C (T )∥ f∥H1
Therefore, this has shown that
∥ f∥C0,(1/2)([0,T ],X) ≡ supt∈[0,T ]
∥ f (t)∥+ sup0≤s<t≤T
∥ f (t)− f (s)∥|t− s|1/2 ≤C (T )∥ f∥H1
You could imagine that other interesting versions of this are available with similar prooffor the case where the function and its weak derivative are in Lp (0,T,X) for p > 1.
With this integration by parts formula, the following interesting lemma is obtained.This lemma shows why it was appropriate to define f as in Definition 43.1.2.