43.1. SOME STANDARD TECHNIQUES IN EVOLUTION EQUATIONS 1421

provided n is large enough. This follows from continuity of translation in Lp with Lebesguemeasure. Since ε > 0 is arbitrary, it follows fn → f̂ in Lp (R;V ) . Similarly, fn → f inL2 (R;H). This follows because p ≥ 2 and the norm in V and norm in H are related by|x|H ≤C ||x||V for some constant, C. Now

f̂ (t) =

Ψ(t) f (t) if t ∈ [0,T ] ,Ψ(t) f (2T − t) if t ∈ [T,2T ] ,Ψ(t) f (−t) if t ∈ [0,T ] ,0 if t /∈ [−T,2T ] .

An easy modification of the argument of Lemma 43.1.13 yields

f̂ ′ (t) =

Ψ′ (t) f (t)+Ψ(y) f ′ (t) if t ∈ [0,T ] ,Ψ′ (t) f (2T − t)−Ψ(t) f ′ (2T − t) if t ∈ [T,2T ] ,Ψ′ (t) f (−t)−Ψ(t) f ′ (−t) if t ∈ [−T,0] ,0 if t /∈ [−T,2T ] .

.

Recall

fn (t) =∫ 1/n

−1/nf̂ (t− s)φ n (s)ds =

∫R

f̂ (t− s)φ n (s)ds

=∫R

f̂ (s)φ n (t− s)ds.

Therefore,

f ′n (t) =∫R

f̂ (s)φ′n (t− s)ds =

∫ 2T+ 1n

−T− 1n

f̂ (s)φ′n (t− s)ds

=∫ 2T+ 1

n

−T− 1n

f̂ ′ (s)φ n (t− s)ds =∫R

f̂ ′ (s)φ n (t− s)ds

=∫R

f̂ ′ (t− s)φ n (s)ds =∫ 1/n

−1/nf̂ ′ (t− s)φ n (s)ds

and it follows from the first line above that f ′n is continuous with values in V for all t ∈ R.Also note that both f ′n and fn equal zero if t /∈ [−T,2T ] whenever n is large enough. Exactlysimilar reasoning to the above shows that f ′n→ f̂ ′ in Lp′ (R;V ′) .

Now let φ ∈C∞c (0,T ) .∫

R| fn (t)|2H φ

′ (t)dt =∫R( fn (t) , fn (t))H φ

′ (t)dt (43.1.7)

=−∫R

2(

f ′n (t) , fn (t))

φ (t)dt = −∫R

2⟨

f ′n (t) , fn (t)⟩

φ (t)dt

Now ∣∣∣∣∫R ⟨ f ′n (t) , fn (t)⟩

φ (t)dt−∫R

⟨f ′ (t) , f (t)

⟩φ (t)dt

∣∣∣∣≤

∫R

(∣∣⟨ f ′n (t)− f ′ (t) , fn (t)⟩∣∣+ ∣∣⟨ f ′ (t) , fn (t)− f (t)

⟩∣∣)φ (t)dt.

43.1. SOME STANDARD TECHNIQUES IN EVOLUTION EQUATIONS 1421provided n is large enough. This follows from continuity of translation in L’? with Lebesguemeasure. Since € > 0 is arbitrary, it follows f, > f in L?(R;V). Similarly, f, > f inL? (R;H). This follows because p > 2 and the norm in V and norm in H are related by|x| <C||x||y for some constant, C. NowW(t) F(t) ifr € (0,7),a, J W(t) f(2T -¢) ift € [7,27],f= VF itt € [0.7],ifr ¢ [7,27].An easy modification of the argument of Lemma 43.1.13 yieldsSt W(y) f’ (t) if € [0,7],f (t)= W(t) f (27-1) —¥ (0) f (27 —1) ifr € [7,27],W(t) f(—1)—B (0) f (1) ifr € [-7,0),Oift ¢ [—-T,2T].Recall=SO=all l|aSYSYSs =>5 |_” a| aa >Soaal|o—>)—=|mna>=NoTherefore,a—~~“—l| |mo ——3S)LSSy) =a —> |= ia~ “—Q| aa lI> lol|——>)“—wn~—"Ss==>|=QBH|o—Sy=|a>=—ienAllcm==—)—~~|a>=—w“—Qwnand it follows from the first line above that f/ is continuous with values in V for allt € R.Also note that both f/ and f;, equal zero if t ¢ [-T,2T] whenever n is large enough. Exactlysimilar reasoning to the above shows that f’ — f’ in L”’ (IR;V’).Now let @ € C? (0,T).2 al[elie ar=~ [ 2H.) oatI (fn (t)sfa(t)) 9 (t)dt (43.1.7)— [2(.0).fn()) 0 (eatRNowhiaoore (ar [ (rons @ (t) dt< [no- (0), fn (t)) | +P (0. OF) |) o ar.