Potential and monotone homeomorphisms in Banach spaces

Theorem 1

(Ekeland variational principle, [6]) Let ( M , d ) be a complete metric space. Assume that I : M ⟶ R is a continuous functional bounded from below. Then, for every x ∈ M and δ > 0 satisfying

there exists a point y ∈ M such that

Theorem 2

(Mountain pass lemma, [16]) Assume that J : X ⟶ R is a locally Lipschitz functional satisfying the Palais-Smale condition. If there exist an open neighbourhood Ω of u ∈ X and a point v ∈ X \ Ω such that

then J has a critical point w satisfying

where Γ ≔ { γ ∈ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = u , γ ( 1 ) = v } .

Theorem 3

(Ky Fan’s Min-Max theorem) Assume that Y and V are the Hausdorff topological vector spaces. Take convex sets C ⊂ Y , K ⊂ V and a function φ : C × K ⟶ R such that

1. φ ( y , ⋅ ) is convex and continuous for all y ∈ C ;

2. φ ( ⋅ , v ) is concave and continuous for all v ∈ K ;

3. the set K is compact.

Proposition 1

Let K be a closed, convex, and bounded subset of a real and reflexive Banach space X. Assume that A : K ⟶ X * is radially continuous and strictly monotone. Then, there exists a unique u ∈ K satisfying

Proposition 2

Let X be a real and reflexive Banach space. Assume that A : X ⟶ X * is continuous, coercive, monotone and that it satisfies condition (S). Then, A is a homeomorphism.

Assumption 1

X is a real Banach space and J : X ⟶ R is a locally Lipschitz functional satisfying the Palais-Smale condition.

Lemma 1

Let Assumption 1  hold. Assume additionally that there exist an open set U and a closed set C with C ⊂ U such that dist ( C , ∂ U ) > 0 and

Then, J has a critical point in C.

Proof

Denote r = dist ( C , ∂ U ) and fix ε < r 2 . Then, M = U ¯ equipped with a distance function

becomes a complete metric space. Since inf C J = inf U J = inf U ¯ J , then for every n ∈ N , there is u n ∈ C such that J ( u n ) < inf M J + ε n . Taking δ = ε n , M = U ¯ , and I = J in the Ekeland variational principle, we obtain v n ∈ U ¯ such that

Consequently, v n ∈ U and

Taking w = v n + τ h , where h ∈ S 1 and τ > 0 is sufficiently small, we obtain

Therefore, again for every h ∈ S 1 , we have J ∘ ( v n ; h ) ≥ − 1 n , and by the properties of Clarke subdifferential, sup u * ∈ ∂ J ( v n ) ⟨ u * , h ⟩ ≥ − 1 n . Consequently,

which, due to Theorem 3, gives

Hence, ( v n ) is a Palais-Smale sequence. Therefore, it possesses a convergent subsequence, denoted also by ( v n ) , such that v n → w ε . Moreover,

Hence, for large n , we obtain a sequence ( w n ) satisfying (3) with ε = 1 n and w ε = w n . Note that ( w n ) is again a Palais-Smale sequence. Therefore, up to a subsequence, w n → w for some w ∈ U ¯ . Then, J ( w ) = inf U J , w ∈ C , and 0 ∈ ∂ J ( w ) .□

Example 1

Let J : ℓ 2 ⟶ R be given by the formula:

The functional J satisfies the Palais-Smale condition. Denote by ( e n ) n ∈ N a standard base in ℓ 2 , i.e.

Take C = n n + 1 e n : n ∈ N and let U = B 1 . Then, C = C ¯ ⊂ U and inf C J = inf U J . However, the unique critical point of J is 0. Note that dist ( C , ∂ U ) = 0 .

Proposition 3

Let Assumption 1  hold.

1. If 0 is a local minimum of J and if there exists v such that J ( v ) < J ( 0 ) , then there exists a critical point w, with J ( w ) ≥ J ( 0 ) , which is not a local minimum;

2. If 0 is a strict local minimum of J and if there exists v distinct from 0 such that J ( v ) ≤ J ( 0 ) , then there exists a critical point w, with J ( w ) > J ( 0 ) , which is not a local minimum;

3. If 0 is a local minimum of J, then either there exists a second critical point, which is not a local minimum, or 0 is a global minimum and the set of global minima is connected;

4. If J has two local minima, then there is a third critical point;

5. Let λ be a fixed real number, and suppose that each critical point with a critical value greater than λ is a local minimum. Then, each one is a global minimum and the set of those points is connected.

Proof

1. Let us denote

   W ≔ ⋃ { V : V is open and inf V J ≥ J ( 0 ) } and K ≔ { u ∈ W ¯ : J ′ ( u ) = 0 and J ( u ) = J ( 0 ) } .

   Since J ( v ) < J ( 0 ) for some v ∈ X , the set ∂ W is nonempty. Moreover, the set K is compact by the Palais-Smale condition. Hence, one of the following holds:

   • There exists w ∈ ∂ W ∩ K . Then, such a w is a critical point of J and J ( w ) = J ( 0 ) since w ∈ K . Now, suppose that w is an argument of a minimum. Then, inf V J ≥ J ( w ) = J ( 0 ) for some open neighbourhood V of w . Hence, w ∈ V ⊂ W , which means that w is an interior point of W and that contradicts w ∈ ∂ W .

   • Sets ∂ W and K are disjoint. By compactness of K , it means that r ≔ dist ( ∂ W , K ) > 0 . Let us put U ≔ K + B r ∕ 2 . Then, dist ( ∂ U , ∂ W ) > 0 , 0 ∈ U ⊂ U ¯ ⊂ W , and ∂ U ∩ K = ∅ . Therefore, by Lemma 1, we obtain inf ∂ U J > J ( 0 ) . Applying Theorem 2, we obtain the assertion.

2. If 0 is a strict local minimum, then we need to have inf S r J > J ( 0 ) for sufficiently small r . Otherwise we can use Lemma 1 to obtain contradiction. Hence, it is enough to use the mountain pass lemma.

3. Assume that J has a local minimum at 0 and that all critical points are arguments of a local minimum. Then, the set of critical points needs to be a subset of some level set of J and every critical point needs to be an argument of a global minimum. Otherwise, we will have a contradiction with (1). However, it means that the set M of critical points of J and the set of arguments of a global minima of J coincide. Now, suppose that M is not connected. This means that there exists two disjoint open sets V , W ⊂ X such that M = ( M ∩ V ) ∪ ( M ∩ W ) . Therefore, in particular, M ∩ ∂ V = ∅ . Hence, taking any u ∈ V , C = ∂ V , and U = X in Lemma 1, we obtain inf ∂ V J > J ( u ) . Using Theorem 2, we obtain the existence of a critical point, which is not a global minimum, so we have a contradiction.

4. It is an immediate consequence of (a) and (c).

5. Take a critical point u such that J ( u ) > λ . Since each critical point with critical value greater than λ is a local minimum, then, by (a), u needs to be an argument of a global minimum of J . Now, it is enough to use (c) for J ( ⋅ − u ) .□

Corollary 1

Let Assumption 1  hold and assume that u is an argument of a local minimum of J. Then, J has a strict global minimum at u or there exists another critical point of J.

Assumption 2

X is a real Banach space and J : X ⟶ R is a functional of class C 1 such that J ′ : X ⟶ X * is a homeomorphism.

Theorem 4

Let Assumption 2  hold. Then, the following conditions are equivalent:

1. J is strictly convex;

2. There exists an open and convex set U such that J is convex on U;

3. There exists u * ∈ X * such that J − u * has a local minimum;

4. There exists u * ∈ X * such that J − u * is bounded from below.

Corollary 2

Let Assumption 2  holds. Then, the following conditions are equivalent:

1. J is strictly convex;

2. J is convex on some open set;

3. J is bounded from below;

4. J has a local minimum.

Proof of Theorem 4

Note that for every u * ∈ X * , the functional J − u * satisfies the Palais-Smale condition. Indeed, since J ′ is a homeomorphism, then ( J ′ − u * ) ′ ( u n ) → 0 gives J ′ ( u n ) → u * , and consequently, u n → ( J ′ ) − 1 ( u * ) . Therefore, every Palais-Smale sequence is convergent.

It is clear that (a) ⇒ (b).

Let us show that (b) ⇒ (c). Assume that (b) holds and take u * = J ′ ( u ) for some u ∈ U . Then, J − u * is convex on U . Since every critical point of a convex functional is an argument of a minimum, (c) holds.

Now, we show the equivalence (c) ⇔ (d). If (c) holds, then J − u * is necessarily bounded from below by Corollary 1, and hence, (d) holds. On the other hand, if (d) is satisfied, then J − u * has a global minimum by Lemma 1. Therefore, (c) is satisfied.

Finally, let us prove (c) ⇒ (a). Assume that (c) holds. Denote

Σ is clearly closed. We show that it is also open. Denote for any u the following ψ u ≔ J − J ′ ( u ) . Let w ∈ Σ be fixed. By Assumption 2, ψ w satisfies the Palais-Smale condition. Moreover, by what we have already shown, w is the unique critical point of ψ w . Hence,

Indeed, if it is not the case, we can apply Lemma 1 to obtain the existence of another critical point of ψ w , which contradicts the bijectivity of J ′ . By continuity of J ′ , we can find an open neighbourhood W of w such that

where ε ≔ inf B 2 ( w ) \ B 1 ( w ) ψ w − ψ w ( w ) . Take y ∈ B 2 ( w ) \ B 1 ( w ) and v ∈ W ∩ B 1 ( w ) . Then,

Since y and v were taken arbitrary, we obtain

The functional ψ v is bounded from below on B 2 ( w ) , it satisfies the Palais-Smale condition and v is a unique critical point of ψ v . Moreover,

Hence, by Lemma 1, ψ v has a local minimum at v , which is a strict global minimum by Corollary 1. Thus,

and hence, v ∈ Σ . Since v and w were taken arbitrary, Σ is open. Finally, note that Σ is nonempty by (c). Hence, Σ is a connected component of X . Since X is connected, Σ is an entire space. Consequently, (c) ⇒ (a).□

Example 2

Denote by M θ a rotation in R 2 , i.e.

Moreover, let φ be a smooth function satisfying φ ≡ 0 on ( − ∞ , 1 ] ∪ [ 4 , ∞ ) and φ ≡ π on [ 2 , 3 ] . Then, f ( x ) = M φ ( ∣ x ∣ 2 ) x is an example of a smooth diffeomorphism, which is strictly monotone on B 1 and coercive on R 2 . However, the function f is not monotone on entire R 2 , since, for instance,

Theorem 5

Let Assumption 2  hold and assume that J is convex. Then,

Proof

Since J ′ is bijective, J has a unique critical point w ∈ X . Denote I ( u ) ≔ J ( u + w ) for all u ∈ X . Note that I is coercive iff J is so. Moreover,

and 0 is a global minimum of I . By Lemma 1 and Corollary 1, we need to have

Put a ≔ inf S 1 I − I ( 0 ) . Since I ′ is monotone, we have ⟨ I ′ ( α u ) , u ⟩ ≥ ⟨ I ′ ( β u ) , u ⟩ whenever α ≥ β . Therefore, for every u ∈ X with ‖ u ‖ ≥ 1 , one has

Hence, I and consequently J are coercive. Moreover, since ⟨ J ′ ( u ) , u ⟩ ≥ ⟨ J ′ ( τ u ) , u ⟩ for every τ ∈ [ 0 , 1 ] by monotonicity, we have

Hence, the assertion holds.□

Example 3

An operator A : R 2 ⟶ R 2 given by A ( x , y ) = ( − y , x ) is a monotone homeomorphism, which is not coercive.

Example 4

Take X = R 2 , A ( x , y ) = ( e x − y , x ) , and B ( x , y ) = ( y , − x ) . Then, both A and B are homeomorphisms. However, they are not potential. Moreover, ( A + B ) ( x , y ) = ( e x , 0 ) is clearly not a bijection.

Theorem 6

Let X be a real and reflexive Banach space. Take two monotone and radially continuous operators A , B : X ⟶ X * . If A is a potential homeomorphism, then A + B is bijective.

Proof

Fix u * ∈ X * and define T ( u ) ≔ A ( u ) + B ( u ) − u * . Then, A − u * + B ( 0 ) is also a potential homeomorphism. By Theorem 5, we have

Hence, for every u ∈ X , one has

Therefore, there exists R > 0 such that ⟨ T ( u ) , u ⟩ > 0 for all u ∈ ∂ B R . Proposition 1 applied to T and B R ¯ yields existence of u ∈ B R ¯ satisfying

Taking v = 0 , we see that u ∈ B R . Therefore, B r ( u ) ⊂ B R for sufficiently small r > 0 . Hence, taking v = u + w in (6) with w ∈ B r , we obtain ⟨ T ( u ) , w ⟩ ≥ 0 for all w ∈ B r . Consequently, T ( u ) = 0 , which means A ( u ) + B ( u ) = u * . Finally, since A is strictly monotone by Theorem 4, T is also strictly monotone. Therefore, u is determined uniquely, and since u * was taken arbitrary, the assertion follows.□

Lemma 2

Let X be a real Banach space. If J : X ⟶ R is Gâteaux differentiable and J ′ is strong-to-weak* continuous, then J is continuous.

Remark 1

We provide a short proof while being aware that some other proofs can be known but we have not found any in the known literature. The aforementioned lemma also says why we cannot use fully the content of Proposition 2, i.e., examine inversions of radially continuous maps.

Proof of Lemma 2

Let u n → u 0 . Then, the set { τ u n : n ∈ N 0 , 0 ≤ τ ≤ 1 } is compact and since J ′ is strong-to-weak* continuous, there exists R > 0 such that

for all n ∈ N 0 . Moreover, for every τ ∈ [ 0 , 1 ] , we have

Consequently, the Lebesgue dominated convergence theorem yields

Theorem 7

Assume that J : X ⟶ R is Gâteaux differentiable satisfying the weak Palais-Smale condition and that J ′ : X ⟶ X * is strong-to-weak* continuous. If there exist an open set Ω , u ∈ Ω , and v ∈ X \ Ω such that

then J has a critical point w satisfying

where Γ ≔ { γ ∈ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = u , γ ( 1 ) = v } .

Proposition 4

Assume that J : X ⟶ R is Gâteaux differentiable with strong-to-weak* continuous derivative and that it satisfies the weak Palais-Smale condition. If there exists a strong-open set U, its strong-closed subset C satisfying dist ( ∂ U , C ) > 0 and if

then J has a critical point in C.

Proposition 5

Assume that F : R 2 ⟶ R is of class C 2 and that F ′ is a diffeomorphism. Then, one of the following holds:

1. F is convex;

2. F is concave;

3. for every x ∈ R 2 , there exist a real orthogonal matrix S x and real numbers λ x 1 < 0 < λ x 2 satisfying (7).

Hypothesis 1

Assume that F : R 2 ⟶ R is of class C 2 and that F ′ is a diffeomorphism. Then, there exist closed linear subspaces U , V ⊂ X satisfying U ⊕ V = R 2 such that U ∋ w ⟼ F ( w + v ) is convex and V ∋ w ⟼ F ( u + w ) is concave for all u ∈ U and v ∈ V .

Hypothesis 2

Assume that J is a functional of class C 1 and that J ′ : X ⟶ X * is a homeomorphism. Then, there exist closed linear subspaces U , V ⊂ X satisfying U ⊕ V = X such that U ∋ w ⟼ J ( w + v ) is convex and V ∋ w ⟼ J ( u + w ) is concave for all u ∈ U and v ∈ V .