1170 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

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 (34.2.15)

=−∫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.

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 34.2.16,∫

R

∣∣∣ 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. Let