Kenneth Kuttler

1166 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

Proof: From the integral equation 34.2.8,

f (t) = f (s)+∫ t

sf ′ (r)dr

∥ f (t)∥X ≤ ∥ f (s)∥X +

∣∣∣∣∫ t

∥∥ f ′ (r)∥∥

X dr∣∣∣∣

≤ ∥ f (s)∥X +∫ b

∥∥ f ′ (r)∥∥

X dr

and so, integrating both sides with respect to s

(b−a)∥ f (t)∥X ≤ ∥ f∥L1(a,b;X)+(b−a)∥∥ f ′∥∥

L1(a,b;X)

and so

∥ f (t)∥X ≤(

1b−a

+1)(∥ f∥L1(a,b;X)+

∥∥ f ′∥∥

L1(a,b;X)

)Let X be the space of functions f ∈ L1 (a,b;X) such that their weak derivatives f ′ are

also in L1 (a,b;X). Then X is a Banach space with norm given by

∥ f∥X ≡ ∥ f∥L1(a,b;X)+∥∥ f ′∥∥

L1(a,b;X)

This is because the map f → f ′ is a closed map. If fn → f in L1 (a,b;X) and f ′n → ξ inL1 (a,b;X) , then for φ ∈C∞

c (a,b) ,∫ b

aξ φdt = lim

n→∞

∫ b

af ′nφdt = lim

n→∞−∫ b

afnφ′dt =−

∫ b

af φ′dt

showing that ξ = f ′. Thus if you have a Cauchy sequence in X, { fn} , then fn → f inL1 (a,b;X) and f ′n→ ξ in L1 (a,b;X) for some ξ . Hence f ′ = ξ .

Then the above corollary says that pointwise evaluation is continuous as a map fromX to X . This is clearly a linear map. Also the formula obtained shows that in fact, this iscontinuous into C ([a,b] ;X).

∥ f∥C([a,b];X) = supt∈[a,b]

∥ f (t)∥X ≤C(∥ f∥L1(a,b;X)+

∥∥ f ′∥∥

L1(a,b;X)

)=C∥ f∥X .

Now let θ : X→C ([a,b] ;X) be given by θ f (t) ≡ f (t) where f (t) = f (0)+∫ t

0 f ′ (s)ds,f being the continuous representative of f . Then θ is continuous and linear. If θ t f ≡ f (t)so that it is pointwise evaluation at t, then this θ t is also continuous and linear. SupposeX is also reflexive. It follows that if you have a sequence in X { fn} which is convergingweakly to f ∈ X, then you would also have θ t fn = fn (t)→ θ t f ≡ f (t) weakly in X . Ifthis is not so, then since X is reflexive, there is a subsequence, still denoted as fn suchthat fn (t)→ ξ ̸= f (t). However, this says that ( f ,ξ ) is in the weak closure of the graphof θ t . Since this graph is strongly closed and convex, it is also weakly closed and henceξ = θ t f ≡ f (t) , a contradiction. This proves the following nice corollary.

1166 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFFProof: From the integral equation 34.2.8,ro=s0)+ [sarAOlle < Mole +|f Ie OleaAb< i Ole+ [I Olleraand so, integrating both sides with respect to s(b—a)\IF lx SWF llet(anxy) + (0-4) [IF lle aox)and so1IFlle << (GAG+1) (Illacans + lf llayeney)Let X be the space of functions f € L' (a,b;X) such that their weak derivatives f’ arealso in L' (a,b;X). Then X is a Banach space with norm given byIf lle = WA lle (ap:x) + IF Nhetcaox)This is because the map f — f’ is a closed map. If f, — f in L!(a,b;X) and fi > & inL! (a,b;X), then for @ € C2 (a,b),[goa = im, [° seoar= im — [° joo'ar=— f° potenshowing that € = f’. Thus if you have a Cauchy sequence in X, {f,}, then f, > f inL' (a,b;X) and fi, + & in L' (a,b;X) for some &. Hence f’ = &.Then the above corollary says that pointwise evaluation is continuous as a map fromX to X. This is clearly a linear map. Also the formula obtained shows that in fact, this iscontinuous into C ({a,b];X).Illedasixy = Sup If lly $C (Illa aro) + IMF lls(aoexy) = Clflle:te [a,b]Now let 6 : X + C([a,b];X) be given by Of (t) = f(t) where f (t) = f (0) + Jf’ (s)ds,f being the continuous representative of f. Then 6 is continuous and linear. If 0, f = f (t)so that it is pointwise evaluation at t, then this @; is also continuous and linear. SupposeX is also reflexive. It follows that if you have a sequence in X {f,,} which is convergingweakly to f € X, then you would also have 6;f, = fn(t) > 0,f = f(t) weakly in X. Ifthis is not so, then since X is reflexive, there is a subsequence, still denoted as f, suchthat f, (t) > & 4 f(t). However, this says that (f,€) is in the weak closure of the graphof 6;. Since this graph is strongly closed and convex, it is also weakly closed and hence€ = 6,f = f (t), a contradiction. This proves the following nice corollary.