Steps for the proof

is closed, and its kernel Ker( ~grad)is the set of constant functions.

Proof. The proof of this quite dicult theorem is given in the next Ÿ.

And the open mapping theorem, cf. (8.10), then gives the needed result for the Stokes like problem, cf. (1.2):

Corollary 10.2

∃β > 0, ∀p ∈ L²(Ω), || ~gradp||_H−1 ≥ β||p||_L2(Ω)/R, (10.3) that is ∃β > 0, inf

p∈L²(Ω)

sup

v∈H₀¹(Ω)

|(div~v, p)_L2|

||~v||_H1

0(Ω)/Ker(div)||p||_L2 0(Ω)

≥ β (inf-sup condition).

10.2 Steps for the proof

10.2.1 Equivalent norms in H⁻¹(Ω)

Ωbeing bounded, the Poincaré inequality gives:

∃cΩ∈ R, ∀q ∈ H0¹(Ω), ||q||_L2 ≤ cΩ||q||_H¹

0. (10.4)

Let q ∈ L²(Ω), and let `q ∈ H⁻¹(Ω)be dened on H0¹(Ω)by

∀ψ ∈ H₀¹(Ω), h`q, ψi<sub>H</sub>−1,H<sup>1</sup><sub>0</sub> := (q, ψ)<sub>L</sub>2(Ω). (10.5) Thus `q is trivially linear, and, with (10.4),

∀ψ ∈ H₀¹(Ω), |h`_q, ψi_H−1,H₀¹| = |(q, ψ)_L2(Ω)| ≤ ||q||_L2||ψ||_L2 ≤ c_Ω||q||_L2||ψ||_H1

0. (10.6) Thus `q is continuous, thus `q ∈ H⁻¹(Ω), L²(Ω)is considered to be a subspace in H⁻¹(Ω).

Proposition 10.3 If q ∈ L²(Ω), then

||`q||_H−1≤ cΩ||q||_L2, || ~gradq||_H−1≤ ||q||_L2. (10.7)

Thus the injection

(L²(Ω) → H⁻¹(Ω) q → `_q

)

and the gradient operator( L²(Ω) → H⁻¹(Ω)ⁿ q →gradq~

)

are contin-uous. In particular, with (10.6),

if q ∈ L²(Ω)then `q denoted= q. (10.8) (The space L²(Ω) is the pivot space.)

Proof. (10.7)1is given by (10.6). Let q ∈ L²(Ω), with (10.2) we get ~gradq ∈ H⁻¹(Ω)ⁿ. We have, for all φ ∈ H~ ₀¹(Ω)ⁿ, cf. (10.2),

|h ~gradq, ~φi_H−1,H₀¹| = | − (q, div~φ)_L2| ≤ ||q||_L2||div~φ||_L2 ≤ ||q||_L2||grad~φ||_(L2)ⁿ² ≤ ||q||_L2||~φ||_(H1 0)ⁿ. Thus (10.7)2.

Let :

||.||+:( L²(Ω) → R

v → ||v||+= ||v||_H−1+ || ~gradv||_H−1. (10.9) Corollary 10.4 In L²(Ω) the norms ||.||L² and ||.||+are equivalent norms:

∃c1, c2> 0, ∀v ∈ L²(Ω), c1||v||+≤ ||v||_L2≤ c2||v||+. (10.10) Proof. (10.9) trivially denes a norm in L²(Ω), and (10.7) gives c1=_1+c¹

Ω.

Let Z = (L²(Ω), ||.||+). Thanks to _c¹₁, Z is a Banach space. Then consider the canonical injection I+: v ∈ (L²(Ω), ||.||_L2(Ω)) → I+(v) = v ∈ (L²(Ω), ||.||Z): it is the algebraic identity and thus is bijective.

And I+ is continuous (thanks to _c¹₁). Thus I+−1 : v ∈ (L²(Ω), ||.||_Z) → I₊(v) = v ∈ (L²(Ω), ||.||_L2(Ω))is continuous (open mapping theorem 8.3). Then let c2= ||I+−1||.

10.2.2 Rellich theorem L²(Ω) → H⁻¹(Ω)

Reminder: An operator κ ∈ L(E; F ) is compact i κ(BE(0, 1))is compact in F .

Lemma 10.5 Let E and F be Banach spaces. If κ ∈ L(E; F ) is compact, then it dual κ⁰ : F⁰ → E⁰ is compact.

Proof. Let (`n) ∈ BF<sup>0</sup>(0, 1). We have to prove that the sequence (T<sup>0</sup>.`n) ∈ T⁰(BF⁰) ⊂ E⁰has a converging subsequence. Let K = T (BE(0, 1)). K is a compact in F since T is compact. Then consider the restriction φn = `<sub>n|K</sub> : K → R. So (φ<sup>n</sup>)<sub>N</sub><sup>∗</sup> is a sequence in C<sup>0</sup>(K; R), and (φ<sup>n</sup>)<sub>N</sub><sup>∗</sup> ⊂ BF<sup>0</sup>(0, 1) is a bounded set in F<sup>0</sup>. Moreover (φn)<sub>N</sub><sup>∗</sup> is equicontinuous since `n is linear continuous ||`n(y)|| ≤ ||`n|| ||y||F ≤ ||y||F).

Thus the set (φn)_N^∗ is relatively compact in C⁰(K; R) (Ascoli theorem, see Brézis [6]). Thus we can extract a convergent subsequence (φnk)_k∈N^∗ in C⁰(K; R). Thus, T (BE) being relatively compact and thus bounded, we have

sup

x∈B_E

|h`<sub>nk</sub>− `_nm, T.xi| −→

k,m→∞0.

Thus ||T⁰.`n<sub>k</sub>− T<sup>0</sup>.`nm||E⁰ → 0. Thus E⁰ being a Banach space, since E is, (T⁰.`n<sub>k</sub>)<sub>k∈N</sub><sup>∗</sup>converges in E<sup>0</sup>. Thus the set (T<sup>0</sup>.`n_k)_k∈N^∗ is compact, thus T⁰ is compact.

Theorem 10.6 (Rellich) The canonical injection T : v ∈ L²(Ω) → v ∈ H⁻¹(Ω) is compact.

Proof. I10: v ∈ H₀¹(Ω) → v ∈ L²(Ω) is compact, Rellich theorem see Brézis [6], thus I10⁰ : v ∈ L²(Ω) → v ∈ H⁻¹(Ω)is compact, cf. Lemma 10.5.

10.2.3 PetreeTartar compactness theorem

Let E and F be two Banach spaces, and T ∈ L(E; F ) (linear and continuous). The purpose is to prove that the range of T is eventually closed. But to use theorem 8.3 and (8.8) to prove it, can be dicult. It can be easier to nd a compact operator κ : E → G, where G is a Banach space, s.t.

∃γ > 0, ∀x ∈ E, ||T.x||F+ ||κ.x||G≥ γ||x||E. (10.11) Theorem 10.7 Let E, F and G be three Banach spaces, let T ∈ L(E; F ) be injective (one-to-one), and κ ∈ L(E; G)be compact. If (10.11) holds then (8.8) holds, and thus Im(T ) is closed.

(If T is not injective, consider E/Ker(T ).)

Proof. Suppose (8.8) is false. Thus there exists a sequence (xn)_n∈N^∗ in E s.t. ||xn||E = 1 and

||T.xn|| −→n→∞0, cf. (8.10). And κ being compact and (xn)_n∈N^∗being bounded, the sequence (κ.xn)_n∈N^∗ has a convergent subsequence (κ.xn_k)_k∈N^∗ that converges in the Banach space G. With T continuous, κcompact and the hypothesis (10.11), we get

γ||xn_i− xn_j||E≤ ||T.xn_i− T.xn_j||F+ ||κ.xn_i− κ.xn_j||G −→

i,j→∞0 + 0 = 0.

Thus (xnk)_k∈N^∗ is a Cauchy sequence in the Banach space E, so converges to a limit x ∈ E. Since

||T.x_nk|| −→_k→∞0and T is continuous, we get ||T.x|| = 0, thus x = 0 since T is injective. But ||xnk||_E= 1 implies ||x||E= 1. Absurd, thus (8.8) is true.

10.2.4 The range of ~grad : L²(Ω) → H⁻¹(Ω)ⁿ is closed

We can now prove theorem 10.1. Let T = ~grad : L²(Ω) → H⁻¹(Ω)ⁿ, and κ the canonical injection L²(Ω) → H⁻¹(Ω). Since T is linear continuous, cf. (10.7), and κ is compact, cf. Rellich theorem 10.6, the PetreeTartar theorem 10.7 implies that the range of T is closed.

11 The closed range theorem

The results and full proofs can be found e.g. in Brézis [6].

11.0.1 The closed range theorem

Let T ∈ L(E; F ), so T⁰∈ L(F⁰; E⁰), cf. (8.5). We have

Ker(T⁰) = {m ∈ F⁰s.t. T⁰.m = 0} = {m ∈ F⁰s.t. m.T = 0} ⊂ F⁰, (11.1) since m ∈ Ker(T⁰) ⇔ T⁰.m = 0 ⇔ hT⁰.m, xiE⁰,E = 0 = hm, T.xiF⁰,F = m(T.x) = (m ◦ T )(x)for all x ∈ E

⇔ m.T = 0, with m.T the notation of m ◦ T when the maps are linear maps.

If M ⊂ E is a linear subspace in E then the dual orthogonal of M is

M^o:= {` ∈ E<sup>0</sup>: h`, xi_E⁰_,E = 0, ∀x ∈ M } (⊂ E⁰) (11.2) (the subspace of E⁰ of linear forms vanishing on M). M^o is a linear subspace in E⁰ (trivial). And M^ois closed in E⁰: Indeed if (`n)<sub>N</sub><sup>∗</sup> is a Cauchy sequence in M<sup>o</sup>, so ||`n− `<sub>m</sub>||<sub>E</sub><sup>0</sup>−→<sub>n,m→∞</sub>0, then, for x ∈ E the sequence (`n(x))is a Cauchy sequence in R, thus convergence toward a real named `(x); This denes a function ` : E → R. And `(x1+ λx2) = limn→∞`n(x1+ λx2) = limn→∞`n(x1) + λ limn→∞(x2) =

`(x1) + λ`(x2), thus ` is linear, and, for x ∈ E, ` is continuous at x since |`.x<sup>0</sup>− `.x| ≤ |(` − `n).x⁰+ (` −

`n).x| + |`n.x⁰− `n.x| ≤ (||` − `n|| + ||`n||)||x⁰− x||Ewith ||`n|| ≤ ||`N|| + 1for N large enough and n ≥ N.

If N ⊂ F⁰ is a linear subspace in F⁰ then let

N^⊥:= {y ∈ F : hm, yiF⁰,F = 0, ∀m ∈ N } (⊂ F ). (11.3) Then N^⊥ is a linear subspace in F (trivial) that is closed in F (similar proof than for M^o). To be compared with, cf. (11.2),

N^o= {y⁰⁰∈ F⁰⁰: hy⁰⁰, mi_F⁰⁰_,F⁰ = 0, ∀m ∈ N } (⊂ F⁰⁰). (11.4)

Remark 11.1 If F is reexive, that is F⁰⁰' F (identication) then N^o ' N^⊥ (identication). Indeed, with J the canonical isomorphism given in (8.7), if y ∈ N^⊥ then let y⁰⁰= J (y) ∈ F, so for all m ∈ N we have 0 = m.y = y⁰⁰.m, and thus y⁰⁰ ∈ N^o; And if y⁰⁰∈ N^o then let y ∈ F s.t. J(y) = y⁰⁰ (thanks to the reexivity), then for all m ∈ N we have 0 = y⁰⁰.m = m.y, and thus y ∈ N^⊥.

Theorem 11.2 (Closed range theorem) Let E and F be Banach spaces and T ∈ L(E; F ) (linear and continuous). Then the following properties are equivalent:

(i) Im(T ) is closed in F , (ii) Im(T⁰)is closed in E⁰, (iii) Im(T ) = Ker(T⁰)^⊥, (iv) Im(T⁰) = Ker(T )^o. We then deduce, with (8.16):

Corollary 11.3 If Im(T ) is closed in F then Im(T⁰)is closed in E', thus

∃γ⁰ > 0, ∀` ∈ F<sup>0</sup>, ||T<sup>0</sup>.`||_E⁰ ≥ γ⁰||`||_F0/Ker(T⁰). (11.5) Proof. The full proof of theorem 11.2 (even for unbounded operators with dense domain of denition) can be found e.g. in Brézis [6] or Yosida [34] (for locally convex spaces that are metrizable and complete).

We give here the proof in the simplied case of T a linear continuous mapping between two Banach spaces (sucient for our needs). We need some lemmas:

Lemma 11.4 If E is a Banach space and M is a linear subspace in E, then

M = (M^o)^⊥. (11.6)

Proof. If x ∈ M then h`, xiE<sup>0</sup>,E = 0for all ` ∈ M^o, thus x ∈ (M^o)^⊥, cf. (11.3). And (M^o)^⊥ being closed we get M ⊂ (M^o)^⊥.

Conversely: Suppose x0 ∈ (M^o)^⊥ and x0 ∈ M/ ; Then {x0} being compact and M being closed and convex (it is a linear subspace), there exists a hyperplane that strictly separates x0 and M (geometric form or the HahnBanach theorem), that is there exists ` ∈ E<sup>0</sup> and α ∈ R s.t. h`, xiE⁰,E< α < h`, x<sub>0</sub>i<sub>E</sub><sup>0</sup><sub>,E</sub> for all x ∈ M. And M being a linear space, taking −x ∈ M, it follows that h`, xiE⁰,E = 0 for all x ∈ M, thus ` ∈ M<sup>o</sup>. And h`, xiE⁰,E = 0 for all x ∈ M implies h`, x0iE<sup>0</sup>,E > 0 with x0 ∈ M/ , thus ` /∈ M^o. Absurd, thus (M^o)^⊥ ⊂ M.

Lemma 11.5 If G and L are two closed subspaces in a Banach space, then

G ∩ L = (G^o+ L^o)^⊥, and G^o∩ L^o= (G + L)^o. (11.7) Proof. (11.7)1: If x ∈ G ∩ L, and if m = g + ` ∈ G<sup>o</sup>+ L<sup>o</sup>, then m.x = g.x + `.x = 0 + 0, thus x ∈ (G^o+ L^o)^⊥.

Conversely, we have (G^o + L^o)^⊥ ⊂ (G^o)^⊥ (particular case of: If Y ⊂ Z then Z^⊥ ⊂ Y^⊥), and (G^o)^⊥= Gsince G is closed thus if x ∈ (G^o+ L^o)^⊥ then x ∈ G; And similarly x ∈ L,; Thus x ∈ G ∩ L.

Similar proof for (11.7)2.

Let E ×F be equipped with the (usual) norm ||(x, y)||E×F = max(||x||_E, ||y||_F), so E ×F is a Banach space.

Lemma 11.6 If T is continuous, then its graph

G(T ) = {(x, y) ∈ E × F s.t. ∃x ∈ E, y = T.x} = {(x, T.x) ∈ E × F } (11.8) is closed in E × F .

Proof. If ((xn, T.xn)))_N^∗is a Cauchy sequence in G(T ), then, E being a Banach space, (xn)_N^∗ converges toward a x ∈ E, thus, T being continuous, T.xn convergence toward T.x ∈ F , so (x, T.x) ∈ G(T ).

We have

G(T⁰) = {(m, T⁰.m) ∈ F⁰× E⁰}. (11.9)

Lemma 11.7 If T is continuous, then G(T⁰)is closed in F⁰× E⁰, and

(m, `) ∈ G(T<sup>0</sup>) ⇐⇒ (−`, m) ∈ G(T )^o. (11.10) Proof. Let ((mn, T⁰.mn))_N^∗ be a sequence in G(T ) s.t. (mn, T⁰.mn) −→n→∞(m, z) ∈ F⁰× E⁰. Thus mn−→ m ∈ F⁰ and T⁰.mn−→n→∞k ∈ E⁰, that is hT⁰.mn, xiE⁰,E−→n→∞hk, xiE⁰,E ∈ R for all x ∈ E.

And we have to check that k = T⁰.m. For all x ∈ E, we have hT⁰.mn, xiE⁰,E = hmn, T.xiF⁰,F, thus hmn, T.xiF⁰,F−→n→∞hk, xiE⁰,E, i.e. hT⁰mn, xiF⁰,F−→n→∞hk, xiE⁰,E, thus T⁰mn, −→n→∞k ∈ F⁰. So G(T⁰)is closed.

(m, `) ∈ G(T<sup>0</sup>) ⇔ ` = T⁰.m ⇔ h`, xiE<sup>0</sup>,E = hT<sup>0</sup>.m, xiE<sup>0</sup>,E = hm, T.xiE<sup>0</sup>,E for all x ∈ E ⇔ h`, xiE⁰,E− hm, T.xiE⁰,E= 0 for all x ∈ E ⇔ h(`, −m), (x, T.x)iE<sup>0</sup>×F<sup>0</sup>,E×F = 0for all x ∈ E ⇔ (`, −m) ∈ G(T )^o.

Dene

L := E × {0}. (11.11)

Lis closed in E × F since E and {0} are, and

L^o= {0} × F⁰. (11.12)

Indeed L^o= {(`, m) ∈ E<sup>0</sup>× F<sup>0</sup> : h(`, m), (x, 0)iE⁰×F⁰,E×F = 0, ∀x ∈ E} = {(`, m) ∈ E<sup>0</sup>× F<sup>0</sup>: h`, xiE⁰,E+ 0 = 0, ∀x ∈ E} = {0} × F⁰.

Lemma 11.8

Ker(T ) × {0} = G(T ) ∩ L (11.13)

E × Im(T ) = G(T ) + L (11.14)

{0} × Ker(T⁰) = G(T )^o∩ L^o (11.15)

Im(T⁰) × F⁰= G(T )^o+ L^o (11.16)

Proof. (x, y) ∈ Ker(T ) × {0} i T x = 0 and y = 0; And (x, y) ∈ G(T ) ∩ L i y = T x and y = 0, thus (11.13).

(x₁, y₁) ∈ E × Im(T ) i (x1, y₁) = (x₁, T.x⁰₁)for some x⁰1∈ E; And (x2, y₂) ∈ G(T ) + L i ∃x⁰2 ∈ E and ∃x⁰⁰2 ∈ E s.t. (x2, y2) = (x⁰₂, T.x⁰₂) + (x⁰⁰₂, 0) = (x⁰₂+ x⁰⁰₂, T.x⁰₂) = (x3, T.(x3− x⁰₂)), thus (11.14).

(`, m) ∈ {0} × Ker(T<sup>0</sup>) i ` = 0 and m ∈ KerT⁰; And (`, m) ∈ G(T )<sup>o</sup>∩ L<sup>o</sup> i (−m, `) ∈ G(T⁰), cf. (11.10), and (`, m) ∈ L<sup>o</sup>, i.e. i ` = −T⁰.mand ` = 0, i.e. i ` = 0 and m ∈ KerT⁰, thus (11.15).

(`, m) ∈ Im(T<sup>0</sup>) × F<sup>0</sup> i ∃k ∈ F<sup>0</sup> s.t. ` = T⁰.k and m ∈ F⁰; And (`, m) ∈ G(T )<sup>o</sup>+ L<sup>o</sup> i ∃(`1, m1) ∈ G(T )^o and ∃(`2, m2) ∈ L<sup>o</sup>s.t. ` = `1+ `2 and m = m1+ m2, i.e., with (11.10) and (11.12), i ∃m1∈ F⁰ (and then −`1 = T<sup>0</sup>.m1) and m2 ∈ F<sup>0</sup> s.t. ` = −T.m1+ 0 and m = m1+ m2, i.e. i ` ∈ Im(T⁰)and m ∈ F⁰, thus (11.16).

Corollary 11.9

Ker(T ) = Im(T⁰)^⊥, (11.17)

Ker(T⁰) = Im(T )^o, (11.18)

(Ker(T ))^o= Im(T⁰), (11.19)

Ker(T⁰)^⊥ = Im(T ). (11.20)

Proof. (11.16) gives R(T⁰)^⊥ × {0} = (G(T )^o+ L^o)^⊥ = G(T ) ∩ L, cf. (11.7), thus = Ker(T ) × {0}, cf. (11.13), thus (11.17). Thus (11.19).

(11.14) gives {0} × Im(T )^o = (G(T ) + L)^o = G^o∩ L^o, cf. (11.7), thus = {0} × Ker(T⁰), cf. (11.15), thus (11.18). Thus (11.20).

Proof of theorem 11.2: apply corollary 11.9.

12 A well-posed mixed problem

12.1 Notations

Let V and Q be two Banach spaces. Let b(·, ·) : V × Q → R be a bilinear form. b(·, ·) is said to be continuous (or bounded) i

∃c > 0, ∀(v, q) ∈ BV(0, 1) × BQ(0, 1), |b(v, q)| ≤ c. (12.1) Then let

||b|| := sup

v∈BV (0,1) q∈BQ(0,1)

|b(v, q)|. (12.2)

And let L(V, Q; R) be the space of bilinear and continuous forms with its (usual) norm given by (12.2).

If b(·, ·) ∈ L(V, Q; R) (bilinear and continuous), then dene

B :

(V → Q⁰ v → Bv

)

and B^t:

(Q → V⁰ q → B^tq

)

(12.3)

by

b(v, q) = hBv, qiQ⁰,Q= hB^tq, viV⁰,V. (12.4) Thus B and B^tare linear (trivial) and continuous with

||B|| = ||B^t|| = ||b||. (12.5)

Indeed ||Bv||V⁰ = sup_q∈B

Q(0,1)|hBv, qiQ⁰,Q| = sup_q∈BQ(0,1)|b(v, q)| ≤ sup_q∈BQ(0,1)||b|| ||v||V||q||Q =

||b|| ||v||V gives ||B|| ≤ ||b|| (continuity), and |b(x, y)| = |hBv, qiQ⁰,Q| ≤ ||Bx||Q⁰||y||Q ≤ ||B|| ||x||V||y||Q

gives ||b|| ≤ ||B||V⁰. So B ∈ L(V ; Q⁰). Idem pour B^t.

Suppose Q reexive, cf. denition 8.2, then the dual B⁰ ∈ L(Q⁰⁰; V⁰) of B ∈ L(V ; Q⁰), dened by hB⁰v, `i<sub>V</sub><sup>0</sup><sub>,V</sub> = hv, B`i_Q⁰⁰_,Q for all v ∈ Q⁰⁰and ` ∈ V⁰ cf. (8.5), is identied to B^t:

L(Q⁰⁰; V⁰) 3 B⁰ ' B^t∈ L(Q; V⁰). (12.6) Suppose V reexive, cf. denition 8.2, then the dual (B^t)⁰ ∈ L(V⁰⁰; Q⁰)of B^t∈ L(Q; V⁰), dened by h(B^t)⁰v, qi_Q⁰_,Q= hv, B^t`i<sub>V</sub><sup>00</sup><sub>,V</sub><sup>0</sup> for all v ∈ Q<sup>00</sup>and ` ∈ V⁰ cf. (8.5), is identied to B^t:

L(V⁰⁰; Q⁰) 3 (B^t)⁰ ' B ∈ L(V ; Q⁰). (12.7)

In document Contents. I Introduction 4. II Examples and preventions 5. Notes de cours de l'isima, troisième année (Page 32-37)