A generalization of Lax-Milgram’s Theorem

by J. Saint Raymond

Sunto : Lo scopo principale di questa nota è quello di provare il seguente risultato : Siano H uno spazio di Hilbert reale e A : H → H un operatore lineare tale che

inf_{∥x∥=1}(⟨Ax, x⟩ + ∥Ax∥) > 0.

Allora l’operatore A è continuo ed invertibile.

Si provano pure altri risultati concernenti perturbazioni compatte.

The main purpose of this note is to prove the following statement :

Theorem 1. Let H be a real Hilbert space and A a linear operator on H. If

inf_{∥x∥=1}(⟨Ax, x⟩ + ∥Ax∥) > 0

the operator A is continuous and invertible.

This is a generalization of the well-known theorem of Lax-Milgram since if

inf_{∥x∥=1}⟨Ax, x⟩ > 0

the previous inequality holds. In fact we shall even prove the following improvements of Theorem 1 :

Theorem 2. Let H be a real Hilbert space and A a linear operator on H. Let (y_{1}, y_{2}, …, y_{k}) be an orthonormal family in H and γ > 0. If

inf_{∥x∥=1} (⟨Ax, x⟩ + ∥Ax∥ + γ(∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2})^{1/2}) > 0

the operator A is continuous and invertible.

Theorem 3. Let H be a real Hilbert space and A a linear operator on H. Let K be a compact linear operator on H. If

inf_{∥x∥=1} (⟨Ax, x⟩ + ∥Ax∥ + ∥KAx∥) > 0

the operator A is continuous and invertible.

Both Theorem 1 and Theorem 2 are answers to questions asked by B. Ricceri.

If A satisfies the hypothesis of Theorem 2, we put

δ = inf_{∥x∥=1} (⟨Ax, x⟩ + ∥Ax∥ + γ(∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2})^{1/2}) > 0

Then we have for every x in the unit sphere

⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + γ(∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2})^{1/2}. ∥x∥ ≥ δ ∥x∥^{2}  (∗)

and by homogeneity this inequality holds for every x ∈ H. Notations. We will denote by V the linear space spanned by (y_{1}, y_{2}, …, y_{k}) and by F = V^{⊥} its orthogonal. V is a closed subspace of H of dimension k, and we denote by π the orthogonal projector on V. Then for every u in H we have

∑_{j=1}^{k}⟨u, y_{j}⟩^{2} = ∥πu∥^{2}

So (∗) becomes

⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + γ ∥πAx∥ . ∥x∥ ≥ δ ∥x∥^{2}  (∗∗)

We denote by I the identity operator on H. Lemma 4. Let A be a linear operator on H satisfying (∗∗). Then ker(A) = 0 Proof : We have

\begin{aligned}δ ∥x∥^{2} & ≤ ⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + γ ∥πAx∥ . ∥x∥ \\ & ≤ ∥Ax∥ . ∥x∥ + ∥Ax∥ . ∥x∥ + γ ∥Ax∥ . ∥x∥ = (γ + 2) ∥Ax∥ . ∥x∥ \end{aligned}

hence δ ∥x∥ ≤ (γ + 2) ∥Ax∥, and x = 0 if Ax = 0. ⋄ Lemma 5. Let A be a linear operator on H satisfying (∗∗). Then for any real t ≥ 0 and any x ∈ F the following inequality holds

∥(A + tI)x∥ ≥ \frac{δ}{γ + 2} ∥x∥

Proof : Since x ∈ F = V^{⊥},

π(A + tI)x = πAx + tπx = πAx
⟨(A + tI)x, x⟩ + ∥(A + tI)x∥ . ∥x∥ = ⟨Ax, x⟩ + t ∥x∥^{2} + ∥Ax + tx∥ . ∥x∥
≥ ⟨Ax, x⟩ + t ∥x∥^{2} + ∥Ax∥ . ∥x∥ − ∥tx∥ . ∥x∥
≥ ⟨Ax, x⟩ + ∥Ax∥ . ∥x∥

Hence

⟨(A + tI)x, x⟩ + ∥(A + tI)x∥ . ∥x∥ + γ ∥π(A + tI)x∥ . ∥x∥ ≥ ⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + γ ∥πAx∥ . ∥x∥ ≥ δ ∥x∥^{2} and since

⟨(A + tI)x, x⟩ + ∥(A + tI)x∥ . ∥x∥ + γ ∥π(A + tI)x∥ . ∥x∥ ≤ ∥(A + tI)x∥ . ∥x∥ + ∥(A + tI)x∥ . ∥x∥ + γ ∥(A + tI)x∥ . ∥x∥ = (γ + 2) ∥(A + tI)x∥ . ∥x∥

we get

(γ + 2) ∥(A + tI)x∥ . ∥x∥ ≥ δ ∥x∥^{2}

and this completes the proof of the Lemma. ⋄

Corollary 6. Let A be a continuous linear operator on H satisfying (∗∗). Then for every real t ≥ 0 the space (A + tI)(F) is closed in H. Moreover A + tI is one-to-one on F.

Proof : From Lemma 5 we get clearly F ∩ ker(A + tI) = {0}. Hence A + tI is one-to-one on F. Let (u_{n}) be a sequence in (A + tI)(F) converging to u in H. If u_{n} = (A + tI)x_{n} with x_{n} ∈ F, we have by Lemma 5

∥x_{n} − x_{m}∥ ≤ \frac{γ + 2}{δ} ∥u_{n} − u_{m}∥

and this shows that the sequence (x_{n}) is a Cauchy sequence. Since F is complete, the sequence (x_{n}) converges to some x ∈ F, and since A is continuous,

(A + tI)x = lim_{n→∞}(A + tI)x_{n} = lim_{n→∞} u_{n} = u

Thus u ∈ (A + tI)(F). ⋄

Lemma 7. Let A be a continuous linear operator on H satisfying (∗∗). Then for every real t ≥ 0 the space (A + tI)(H) has codimension k in H.

Proof : Denote by T the set

T = {t ≥ 0 : (A + tI)(F) is not of codimension k in H }

We want to prove T = ∅.

First of all, notice that, for t > ∥A∥, the operator A + tI is invertible on H. Hence the subspaces (A + tI)(F) and (A + tI)(V) are direct summands. And since dim((A + tI)(V)) = dim(V) = k, t does not belong to T. This shows that T ⊂ [0, ∥A∥]

Thus if T is not empty, it has an upper bound θ. Denote by E_{θ} the orthogonal subspace of (A + θI)(F) and by S_{t} the operator from F × E_{θ} to H defined by

S_{t}(x, u) = (A + tI)x + u

Then S_{t} is continuous for each t ≥ 0. Moreover S_{θ} is one-to-one since H = (A + θI)(F) ⊕ E_{θ} and since A + θI is one-to-one from F to (A + θI)(F) by Corollary 6.

Since S_{θ} is invertible, there exists some ρ > 0 such that S_{t} is invertible for |t − θ| < ρ. In particular for t = θ + \frac{ρ}{2} > θ, (A + tI)(F) has codimension k in H. And since S_{t} is invertible, (A + tI)(F) and E_{θ} are direct summands. Hence dim(E_{θ}) = k. Similarly for θ − ρ < t ≤ θ (A + tI)(F) and E_{θ} are direct summands. Hence (A + tI)(F) has codimension dim(E_{θ}) = k, and T ∩]θ − ρ, θ] = ∅. This shows that θ = sup T ≤ θ − ρ, a contradiction. This completes the proof. ⋄ Theorem 8. Let A be a continuous linear operator on H satisfying (∗∗). Then A is invertible.

Proof : By Lemma 7, A(F) has codimension k. By Lemma 4, A is one-to-one. Hence A(V) has dimension k and A(F) ∩ A(V) = {0}. Thus A(H) = A(F) + A(V) = H. Since A is continuous and one-to-one from H to H, it is invertible. ⋄

Theorem 9. Let A and K be continuous linear operators on H, K being compact. If

inf_{∥x∥=1}(⟨Ax, x⟩ + ∥Ax∥ + ∥KAx∥) > 0

then A is invertible.

Proof : Put γ = ∥K∥ and δ = inf_{∥x∥=1}(⟨Ax, x⟩ + ∥Ax∥ + ∥KAx∥). We have, as at the beginning of this paper,

⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + ∥KAx∥ . ∥x∥ ≥ δ ∥x∥^{2}  (∗∗∗)

Then K^{∗}K is compact, symmetric and positive, and so is K_{1} = (K^{∗}K)^{1/2}. Moreover γ = ∥K∥ = ∥K_{1}∥. For every y ∈ H we have

∥K_{1}y∥^{2} = ⟨K_{1}y, K_{1}y⟩ = ⟨K_{1}^{2}y, y⟩ = ⟨K^{∗}Ky, y⟩ = ⟨Ky, Ky⟩ = ∥Ky∥^{2}

So (∗∗∗) can be rewritten as

⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + ∥K_{1}Ax∥ . ∥x∥ ≥ δ ∥x∥^{2}

Hence we can and do assume that K is symmetric. There exists an orthonormal sequence (y_{j}) and a sequence of non-negative eigenvalues (λ_{j}) such that, for every y ∈ H

∥Ky∥^{2} = ∑_{j=1}^{∞} λ_{j}^{2}⟨y, y_{j}⟩^{2}

Moreover the sequence (λ_{j}) converges to 0. Put ε = \frac{δ}{2 ∥A∥}. Thus there exists some k such

that λ_{j} < ε for j > k. And λ_{j} ≤ γ for all j. Then for every x in H we have

∥KAx∥^{2} = ∑_{j=1}^{∞} λ_{j}^{2}⟨Ax, y_{j}⟩^{2}

≤ ∑_{j=1}^{k} γ^{2}⟨Ax, y_{j}⟩^{2} + ∑_{j=k+1}^{∞} ε^{2}⟨Ax, y_{j}⟩^{2}

≤ γ^{2} ∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2} + ε^{2} ∑_{j=1}^{∞}⟨Ax, y_{j}⟩^{2}

≤ γ^{2} ∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2} + ε^{2} ∥Ax∥^{2}

≤ γ^{2} ∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2} + ε^{2} ∥A∥^{2} ∥x∥^{2}

≤ γ^{2} ∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2} + \frac{δ^{2}}{4} ∥x∥^{2}

hence

∥KAx∥ ≤ \frac{δ}{2} ∥x∥ + γ\sqrt{∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2}}

and from (∗∗∗) we get

\begin{aligned}⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + γ\sqrt{∑_{j=1}^{k}⟨Ax, y_{j}⟩^{2}} & ≥ ⟨Ax, x⟩ + ∥Ax∥ . ∥x∥ + (∥KAx∥ − \frac{δ}{2} ∥x∥). ∥x∥ \\ & ≥ \frac{δ}{2} ∥x∥^{2} \end{aligned}

This shows that A satisfies (∗∗) (with δ/2 instead of δ). Then the conclusion follows from Theorem 8.

From now on, the operator A is not assumed to be continuous. We denote by W the closed linear subspace of H defined by

W = {w ∈ H : there exists a sequence (x_{n}) such that lim_{n→∞} x_{n} = 0 and lim_{n→∞} Ax_{n} = w} .

Lemma 10. Let X and Y be two Banach spaces and Φ a continuous linear mapping from X onto Y. Then for every closed linear subspace G of X containing the kernel of Φ, Φ(G) is closed in Y.

Proof : By Banach’s Theorem, Φ is an open mapping. Thus Φ(X \ G) is an open subset of Y. Now it is enough to notice that

Y \ Φ(G) = Φ(X \ G)

for completing the proof. ⋄

For A a linear operator on H, we denote by q the orthogonal projector on W^{⊥}, and we put A′ = qA.

Lemma 11. A′ is continuous.

Proof : Denote by G_{A} (resp. G_{A′}) the graph of A (resp. A′) in H ×H. If Φ(x, y) = (x, qy), we have G_{A′} = Φ(G_{A}).

By the definition of W, we have W_{0} = \{0\} × W ⊂ \overline{G_{A}}, hence G_{A} + W_{0} ⊂ \overline{G_{A}}. Conversely, if (x, y) ∈ \overline{G_{A}}, there exists a sequence ((x_{n}, y_{n})) in G_{A} converging to (x, y). Then (x_{n} − x, Ax_{n} − Ax) ∈ G_{A}, x_{n} − x → 0 and Ax_{n} − Ax → y − Ax. We conclude that y − Ax ∈ W and that (x, y) = (x, Ax) + (0, y − Ax) ∈ G_{A} + W_{0}. Hence \overline{G_{A}} = G_{A} + W_{0}.

Since \overline{G_{A}} ⊃ W_{0} = ker(Φ), Lemma 10 implies that Φ(\overline{G_{A}}) is closed, but

Φ(\overline{G_{A}}) = Φ(G_{A} + (\{0\} × W)) = Φ(G_{A}) + Φ(W_{0}) = Φ(G_{A}) = G_{A′}

Hence G_{A′} is closed, and A′ is continuous. ⋄ Lemma 12. Let H_{1} and H_{2} be Hilbert spaces, A a linear operator from H_{1} to H_{2}, K a compact linear operator on H_{2}. If H_{1} is infinite-dimensional, then for every ε > 0 there exists some u in the unit sphere of H_{1} such that ∥KAu∥ < ε.

Proof : If not we have ∥KAx∥ ≥ ε ∥x∥ for any x ∈ H_{1}. Then for x ∈ H_{1}, w = A′x − Ax ∈ W, and there exists some sequence (x_{n}) converging to 0 such that Ax_{n} → w. Then the sequence (KA(x + x_{n})) converges to K(Ax + w) = KA′x, and since

∥KA(x + x_{n})∥ ≥ ε ∥x + x_{n}∥

we conclude that

∥KA′x∥ ≥ ε ∥x∥  (1)

but, since A′ is continuous by Lemma 11, KA′ is compact, and this contradicts (1). ⋄

Lemma 13. If A satisfies (∗∗) or (∗∗∗), so does A′.

Proof : Since γπ is a compact operator, it is enough to prove the lemma for (∗∗∗). Let x ∈ H. Then w = A′x − Ax ∈ W. There exists a sequence (x_{n}) converging to 0 such that w = lim_{n→∞} Ax_{n}. Applying (∗∗∗) to x + x_{n} we get

⟨A(x + x_{n}), x + x_{n}⟩ + ∥A(x + x_{n})∥ . ∥x + x_{n}∥ + ∥KA(x + x_{n})∥ . ∥x + x_{n}∥ ≥ δ ∥x + x_{n}∥^{2}

and by letting n go to the infinity

⟨Ax + w, x⟩ + ∥Ax + w∥ . ∥x∥ + ∥K(Ax + w)∥ . ∥x∥ ≥ δ ∥x∥^{2}
⟨A′x, x⟩ + ∥A′x∥ . ∥x∥ + ∥KA′x∥ . ∥x∥ ≥ δ ∥x∥^{2}

This last inequality shows that A′ satisfies (∗∗∗). ⋄

Proof of Theorem 3. If A satisfies the hypothesis of Theorem 3, it satisfies (∗∗∗). Then A′ is continuous by Lemma 11 and satisfies (∗∗∗) by Lemma 13. Thus A′ is invertible by Theorem 9. In particular

H = A′(H) = q(A(H)) ⊂ W^{⊥}

Hence W^{⊥} = H, q = I and A′ = qA = A. This proves that A is continuous and invertible. ⋄

Proof of Theorem 2. If A satisfies the hypothesis of Theorem 2, it satisfies (∗∗). And since γπ is a compact operator, the conclusion follows from Theorem 3. ⋄

We finish this note by reproving an earlier (unpublished) result of the author :

Theorem 14. Let A and K be linear operators on the Hilbert space H, K being compact. If

inf_{∥x∥=1}(|⟨Ax, x⟩| + ∥KAx∥) > 0

then A is continuous and invertible.

Proof : If A satisfies the previous hypothesis, it is clearly one-to-one. So we can assume H is infinite-dimensional. It is enough to prove that either A or −A satisfies the hypothesis of Theorem 3.

Put δ = inf_{∥x∥=1}(|⟨Ax, x⟩| + ∥KAx∥), γ = ∥K∥ and ε = \frac{δ}{γ + 2}. Then

δ ∥x∥^{2} ≤ |⟨Ax, x⟩| + ∥KAx∥ . ∥x∥ ≤ ∥Ax∥ . ∥x∥ + ∥K∥ . ∥Ax∥ ≤ (γ + 1) ∥Ax∥ . ∥x∥ and

∥Ax∥ ≥ \frac{δ}{γ + 1} ∥x∥

If A does not satisfy the hypothesis of Theorem 3, there is some x_{1} in the unit sphere of H such that

⟨Ax_{1}, x_{1}⟩ + ∥Ax_{1}∥ + ∥KAx_{1}∥ < ε

and thus

∥KAx_{1}∥ < ε  and  ⟨Ax_{1}, x_{1}⟩ < ε − ∥Ax_{1}∥ ≤ \frac{δ}{γ + 2} − \frac{δ}{γ + 1} < 0  (2)

Similarly if −A does not satisfy the hypothesis of Theorem 3, there is some x_{2} in the unit sphere such that

⟨−Ax_{2}, x_{2}⟩ + ∥−Ax_{2}∥ + ∥−KAx_{2}∥ < ε

and thus

∥KAx_{2}∥ < ε  and  ⟨Ax_{2}, x_{2}⟩ > −ε + ∥Ax_{2}∥ ≥ −\frac{δ}{γ + 2} + \frac{δ}{γ + 1} > 0  (3)

Let V be the linear space spanned by x_{1} and x_{2}. By Lemma 12 applied to H_{1} = V^{⊥} and

H_{2} = H, there exists some u ∈ V^{⊥} such that ∥u∥ = 1 and ∥KAu∥ ≤ \frac{δ}{6}. Then we put for t ∈ [0, 1]

ψ(t) = tx_{2} + (1 − t)x_{1} + (1 − ∥tx_{2} + (1 − t)x_{1}∥^{2})^{1/2} u

We have ∥ψ(t)∥ = 1 for all t, and ⟨Aψ(0), ψ(0)⟩ = ⟨Ax_{1}, x_{1}⟩ < 0 < ⟨Aψ(1), ψ(1)⟩ = ⟨Ax_{2}, x_{2}⟩

Thus since the restriction of A to the space V ⊕ ℝu is continuous, there is some t^{∗} ∈ [0, 1] such that ⟨Aψ(t^{∗}), ψ(t^{∗})⟩ = 0. And

\begin{aligned}∥KAψ(t^{∗})∥ & ≤ t^{∗} ∥KAx_{2}∥ + (1 − t^{∗}) ∥KAx_{1}∥ + (1 − ∥t^{∗}x_{2} + (1 − t^{∗})x_{1}∥^{2})^{1/2} ∥KAu∥ \\ & ≤ t^{∗}ε + (1 − t^{∗})ε + ∥KAu∥ ≤ ε + \frac{δ}{6} ≤ \frac{δ}{2} + \frac{δ}{6} < δ \end{aligned}

hence

|⟨Aψ(t^{∗}), ψ(t^{∗})⟩| + ∥KAψ(t^{∗})∥ < δ

a contradiction. Thus either A or −A satisfies the hypothesis of Theorem 3 and the proof is complete. ⋄

Jean Saint Raymond

Equipe d’Analyse — Boîte 186 Université Paris 6 4, place Jussieu F 75252 Paris CEDEX 05 E-mail : jsr@ccr.jussieu.fr