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Estimation optimale du gradient du semi-groupe de la chaleur sur le ...

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The minimum is realized for<br />

S. Albeverio, A. Kosyak / Journal of Functional Analysis 236 (2006) 634–681 645<br />

xk = bk<br />

<br />

n<br />

ak<br />

k=1<br />

For any subset I ⊂ N <strong>le</strong>t us <strong>de</strong>note as before by 〈fn | n ∈ I〉 the clo<strong>sur</strong>e of the linear space<br />

generated by the set of vectors (fn | n ∈ I) in a Hilbert space H .<br />

We note that the distance d(fn+1;〈f1,...,fn〉) of the vector fn+1 in H from the hyperp<strong>la</strong>ne<br />

〈f1,...,fn〉 may be calcu<strong>la</strong>ted in terms of the Gram <strong>de</strong>terminants Γ(f1,f2,...,fk) corresponding<br />

to the set of vectors f1,f2,...,fk (see [10]):<br />

d fn+1;〈f1,...,fn〉 = min<br />

t=(tk)∈R n<br />

<br />

<br />

<br />

<br />

fn+1 +<br />

b 2 k<br />

ak<br />

n<br />

k=1<br />

−1<br />

.<br />

tkfk<br />

<br />

<br />

<br />

<br />

<br />

2<br />

= Γ(f1,f2,...,fn+1)<br />

, (6)<br />

Γ(f1,f2,...,fn)<br />

where the Gram <strong>de</strong>terminant is <strong>de</strong>fined by Γ(f1,f2,...,fn) = <strong>de</strong>t γ(f1,f2,...,fn) and<br />

γ(f1,f2,...,fn) =: γn is the Gram matrix<br />

Lemma 10. We have<br />

⎛<br />

(f1,f1) (f1,f2) ...<br />

⎞<br />

(f1,fn)<br />

⎜ (f2,f1)<br />

γ(f1,f2,...,fn) = ⎜<br />

⎝<br />

(f2,f2) ...<br />

. ..<br />

(f2,fn) ⎟<br />

⎠<br />

(fn,f1) (fn,f2) ... (fn,fn)<br />

.<br />

d fn+1;〈f1,...,fn〉 =<br />

<strong>de</strong>t γn+1<br />

<strong>de</strong>t γn<br />

where dn+1 = ((f1,fn+1), (f2,fn+1),...,(fn,fn+1)) ∈ R n .<br />

Proof. We may write<br />

<br />

n<br />

<br />

<br />

<br />

k=1<br />

tkfk − fn+1<br />

<br />

2<br />

<br />

=<br />

<br />

n<br />

k,m=1<br />

tktm(fk,fk) − 2<br />

= (fn+1,fn+1) − γ −1<br />

n dn+1,dn+1<br />

<br />

,<br />

n<br />

k=1<br />

= (γnt,t)− 2(t, dn+1) + (fn+1,fn+1),<br />

tk(fk,fn+1) + (fn+1,fn+1)<br />

where t = (t1,t2,...,tn) ∈ Rn . Using (58) for An = γn we get<br />

(γnt,t)− 2(t, dn+1) = γn(t − t0), (t − t0) − γ −1<br />

n dn+1,dn+1<br />

<br />

,<br />

where t0 = γ −1<br />

n dn. Hence we get (see (6))<br />

min<br />

t=(tk)∈R n<br />

<br />

<br />

<br />

<br />

fn+1 −<br />

n<br />

k=1<br />

tkfk<br />

<br />

<br />

<br />

<br />

<br />

2<br />

= min<br />

t=(tk)∈Rn <br />

(γnt,t)− 2(t, dn+1) + (fn+1,fn+1)

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