69.4. THE IMPLICIT CASE 2377

which requires easily that

Bgp =k

∑i=1

⟨Bgp,ei

⟩Bei,

the above holding for all k large enough. It follows that for any x ∈ span({gk}∞

k=1) , (finitelinear combination of vectors in {gk}∞

k=1)

Bx =∞

∑i=1⟨Bx,ei⟩Bei (69.4.19)

because for all k large enough,

Bx =k

∑i=1⟨Bx,ei⟩Bei

Also note that for such x ∈ span({gk}∞

k=1) ,

⟨Bx,x⟩ =

⟨k

∑i=1⟨Bx,ei⟩Bei,x

⟩=

k

∑i=1⟨Bx,ei⟩⟨Bx,ei⟩

=k

∑i=1|⟨Bx,ei⟩|2 =

∑i=1|⟨Bx,ei⟩|2

Now for x arbitrary, let xk→ x in W where xk ∈ span({gk}∞

k=1) . Then by Fatou’s lemma,

∑i=1|⟨Bx,ei⟩|2 ≤ lim inf

k→∞

∑i=1|⟨Bxk,ei⟩|2

= lim infk→∞

⟨Bxk,xk⟩= ⟨Bx,x⟩ (69.4.20)

≤ ∥Bx∥W ′ ∥x∥W ≤ ∥B∥∥x∥2W

Thus the series on the left converges. Then also, from the above inequality,∣∣∣∣∣⟨

q

∑i=p⟨Bx,ei⟩Bei,y

⟩∣∣∣∣∣≤ q

∑i=p|⟨Bx,ei⟩| |⟨Bei,y⟩|

(q

∑i=p|⟨Bx,ei⟩|2

)1/2( q

∑i=p|⟨By,ei⟩|2

)1/2

(q

∑i=p|⟨Bx,ei⟩|2

)1/2(∞

∑i=1|⟨By,ei⟩|2

)1/2

By 69.4.20,

(q

∑i=p|⟨Bx,ei⟩|2

)1/2(∥B∥∥y∥2

W

)1/2≤

(q

∑i=p|⟨Bx,ei⟩|2

)1/2

∥B∥1/2 ∥y∥W

69.4. THE IMPLICIT CASE 2377which requires easily thatkBgp = y (Bgp, ei) Bei,i=1the above holding for all k large enough. It follows that for any x € span ({gx};_1), (finitelinear combination of vectors in {g,};_1)coBx =)" (Bx,e;) Be; (69.4.19)because for all k large enough,Also note that for such x € span ({gx};_1)k k(Bx,x) = (x (Bx, e; sos) L( (Bx, e;) (Bx, e;)i=1k= | (Bx, e:) 7 =i=] i=8|(Bx, ei) |?1Now for x arbitrary, let x, — x in W where x, € span ({g,};_,). Then by Fatou’s lemma,Bx,e;)|? << lim inf | (Bxg,Pi l@se)|? < tim int i.e= lim inf (Bx,Xx) = (Bx, x) (69.4.20)—}oo2S< ||Bxllw: llellw << (IB Il MellyThus the series on the left converges. Then also, from the above inequality,(dam ej sews) |<4 1/2 1/2< (Zlasear) ( (axe?4 1/2 1/2< (Zes0?] ( (axedi=pBy 69.4.20,4 1/2 12 4 1/2< (Zeno?) (liBIl lvl) <(Zitane) ') BI? llvlhyl=pqY |(Bx, e:)| | (Bei,y)|i=pMs iMs