Kenneth Kuttler

1422 CHAPTER 43. INTERPOLATION IN BANACH SPACE

From the first part of this proof which showed that fn → f̂ in Lp (R;V ) and f ′n → f̂ ′ inLp′ (R;V ′) , an application of Holder’s inequality shows the above converges to 0 as n→∞.Therefore, passing to the limit as n→ ∞ in the 43.1.8,∫

∣∣∣ f̂ (t)∣∣∣2H

φ′ (t)dt =−

∫R

2⟨

f̂ ′ (t) , f̂ (t)⟩

φ (t)dt

which shows t→∣∣∣ f̂ (t)∣∣∣2

Hequals a continuous function a.e. and it also has a weak derivative

equal to 2⟨

f̂ ′, f̂⟩

It remains to verify that f̂ is continuous on [0,T ] . Of course f̂ = f on this interval. LetN be large enough that fn (−T ) = 0 for all n > N. Then for m,n > N and t ∈ [−T,2T ]

| fn (t)− fm (t)|2H = 2∫ t

−T

(f ′n (s)− f ′m (s) , fn (s)− fm (s)

)ds

= 2∫ t

−T

⟨f ′n (s)− f ′m (s) , fn (s)− fm (s)

⟩V ′,V ds

≤ 2∫R

∣∣∣∣ f ′n (s)− f ′m (s)∣∣∣∣

V ′ || fn (s)− fm (s)||V ds

≤ 2 || fn− fm||Lp′ (R;V ′) || fn− fm||Lp(R;V )

which shows from the above that { fn} is uniformly Cauchy on [−T,2T ] with values in H.Therefore, there exists g a continuous function defined on [−T,2T ] having values in H suchthat

limn→∞

max{| fn (t)−g(t)|H ; t ∈ [−T,2T ]}= 0.

However, g = f̂ a.e. because fn converges to f in Lp (0,T ;V ) . Therefore, taking a subse-quence, the convergence is a.e. It follows from the fact that V ⊆H = H ′ ⊆V ′ and Theorem43.1.9, there exists f (0) ∈V ′ such that for a.e. t,

f (t) = f (0)+∫ t

0f ′ (s)ds in V ′

Now g = f a.e. and g is continuous with values in H hence continuous with values in V ′andso

g(t) = f (0)+∫ t

0f ′ (s)ds in V ′

for all t. Since g is continuous with values in H it is continuous with values in V ′. Taking thelimit as t ↓ 0 in the above, g(a) = limt→0+ g(t) = f (0) , showing that f (0)∈H. Therefore,for a.e. t,

f (t) = f (0)+∫ t

0f ′ (s)ds in H,

∫ t

0f ′ (s)ds ∈ H.

Note that if f ∈ Lp (0,T ;V ) and f ′ ∈ Lp′ (0,T ;V ′) , then you can consider the initialvalue of f and it will be in H. What if you start with something in H? Is it an initialcondition for a function f ∈ Lp (0,T ;V ) such that f ′ ∈ Lp′ (0,T ;V ′)? This is worth thinking

1422 CHAPTER 43. INTERPOLATION IN BANACH SPACEFrom the first part of this proof which showed that f, > f in L? (R;V) and f, > f’ inL?’ (IR;V’), an application of Holder’s inequality shows the above converges to 0 as n — oo,Therefore, passing to the limit as n > © in the 43.1.8,[|Fol, oa =- [2(Po.Fo)owar2| equals a continuous function a.e. and it also has a weak derivativewhich shows t > Fe)equal to 2 (7).It remains to verify that fis continuous on [0,7]. Of course f= f on this interval. LetN be large enough that f, (—T) =0 for all n > N. Then for m,n > N and t € [—T,2T]|inl) Jn (lee = 2f S(8)— Fils) sFn(5) —Jin(8)) alI)=k| In (8) ~ fin (3) Fi (8) = Fn (8) )yry 82 [Lf (8) — fin) lfn(8) — fn)2||fn — flee! (Rev I|fn — fm\|Lr(e:v)IAIAwhich shows from the above that {f,,} is uniformly Cauchy on [—T,27] with values in H.Therefore, there exists g a continuous function defined on [—T, 27] having values in H suchthatlim max {|f (0) ~()|y it € [-T.27]} =0.However, g = fae. because f, converges to f in L? (0,T;V). Therefore, taking a subse-quence, the convergence is a.e. It follows from the fact that V C H = H’ CV’ and Theorem43.1.9, there exists f (0) € V’ such that for a.e. f,FO) =£0)+ ['F')asinv'Now g= f ae. and g is continuous with values in H hence continuous with values in V’andsog(t) =(0)+ fF (s)asinv'for all t. Since g is continuous with values in H it is continuous with values in V’. Taking thelimit as t | 0 in the above, g (a) = lim,_,0+ g (t) = f (0), showing that f (0) € H. Therefore,for a.e. ft,r= s(0)+ ['F'()asin [racen, |Note that if f € L?(0,7;V) and f’ € L” (0,7;V’), then you can consider the initialvalue of f and it will be in H. What if you start with something in H? Is it an initialcondition for a function f € L? (0,7;V) such that f’ € L? (0,7;V’)? This is worth thinking