On the Geometry of the Unit Ball of a JB^*-Triple

Definition 1. In any JB^*-triple, an element x is BP-quasi invertible with BP-quasi inverse y if (and only if) the Bergman operator B(x, y) = 0.

Theorem 2. For any fixed unitary element u of a JB^*-triple 𝒥 and for all x, y, z ∈ 𝒥, one has the following:

• (i)

  the Jordan triple product $$ in 𝒥^[u] coincides with {xy^*z};

• (ii)

  P_u(x) = P(x);

• (iii)

  L_u(x, y) = L(x, y);

• (iv)

  B_u(x, y) = B(x, y);

• (v)

  $$.

Proof. (i) By using the basic Jordan triple identity, we get that

(ii) By the part (i), $$ for all x ∈ 𝒥.

(iii) For any a, b ∈ 𝒥, P(a, b) = P(a + b) − P(a) − P(b) (cf. [12]). This together with the part (ii) gives L_u(x, y)z = P_u(x, z)y = P_u(x + z)y − P_u(x)y − P_u(z)y = P(x + z)y − P(x)y − P(z)y = P(x, z)y = L(x, y)z for all x, y, z ∈ 𝒥.

(iv) Follows from parts (ii) and (iii) since B(x, y) = I − 2L(x, y) + P(x)P(y).

(v) $$ (by (iv)) $$.

Theorem 3. For any closed ideal V in JB^*-triple 𝒥, the triple homomorphism ϕ : 𝒥 → 𝒥/V given by ϕ(x) = x + V preserves the BP-quasi invertibility.

Proof. V being closed ideal of 𝒥 is a JB^*-triple, and so is the quotient 𝒥/V. The quotient JB^*-triple 𝒥/V admits the canonical surjective triple homomorphism ϕ : 𝒥 → 𝒥/V defined by ϕ(x) = x + V such that ϕ({xyz}) = {ϕ(x)ϕ(y^*)ϕ(z)} for all x, y, z ∈ 𝒥 [8, Proposition 5.5].

Now, for any fixed $$ with inverse y ∈ 𝒥 and for all z ∈ 𝒥, B(ϕ(x), ϕ(y))ϕ(z) = ϕ(z) − 2{ϕ(x)(ϕ(y)) ^*ϕ(z)} + {ϕ(x){ϕ(y)(ϕ(z)) ^*ϕ(y)} ^*ϕ(x)} = ϕ(z) − 2ϕ({xy^*z}) + ϕ({x{yz^*y} ^*x}) = ϕ(B(x, y)(z)) = ϕ(0) = 0. Hence, $$.

Theorem 4. For each von Neumann regular element x of a JB^*-triple 𝒥, there exists a unique tripotent e_x ∈ 𝒥 such that x is positive invertible in 𝒥₁(e_x).

Proof. Since x is von Neumann regular, there exists a unique tripotent e_x ∈ 𝒥, called the range tripotent of x (cf. [10]) such that x is invertible in the Peirce space 𝒥₁(e_x) (cf. [10, Lemma  3.2], [14, page 540], and [15, Theorem  1]). Let 𝒥(x) denotes the norm closure of P(x)(𝒥) = {x𝒥x}, which is the smallest norm closed inner ideal of 𝒥 containing x (cf. [16, pages 19-20]). From [10, Lemma  3.2], we see that P(x)(𝒥) is norm closed in 𝒥. Hence, 𝒥(x) = P(x)(𝒥) = 𝒥₁(e_x) since x is von Neumann regular in 𝒥 (cf. [10, page 572]). Thus, x is positive in 𝒥₁(e_x) by [16, Proposition  2.1].

Remark 5. The range tripotent e_x of x is defined as the least tripotent for which x is a positive element in the JB^*-algebra 𝒥₁(e_x). If u and v are two tripotents in 𝒥 such that x is positive invertible in both the Peirce spaces 𝒥₁(u) and 𝒥₁(v), then u = v; for details, see [17, Lemma  3.3] and [16, Proposition  2.1].

Theorem 6. Let 𝒥 be a JB^*-triple and $$. Then the range tripotent e_x is a unique extreme point of (𝒥) ₁ such that x is positive invertible in 𝒥₁(e_x). Moreover, there exists a unique $$ satisfying P(x)y = x, P(y)x = y, P(x)P(y) = P(y)P(x), and L(x, y) = L(e_x, e_x) = L(y, x).

Proof. Since x being BP-quasi invertible is a von Neumann regular element, Theorem 4 gives the existence of a unique tripotent e_x in 𝒥 such that x is positive invertible in 𝒥₁(e_x). By [10, Lemma  3.2], there exists a unique generalized inverse y ∈ 𝒥 satisfying all the conditions of the theorem; in fact, this generalized inverse is called the Moore-Penrose inverse, usually denoted by x^†. Since {xx^†x} = x, {zx^†x} = {zx^†{xx^†x}} = 2{{zx^†x}x^†x}−{x{x^†zx^†}x} for all z ∈ 𝒥. So that P(x)P(x^†) = 2L(x, x^†) ² − L(x, x^†). Hence, B(x, x^†) = 1 − 3L(x, x^†) + 2L(x, x^†) ² = (2L(x, x^†) − 1)(L(x, x^†) − 1). In view of [18, Lemma  2.1] and [10, page 573], we have L(x, x^†) = L(e_x, e_x) = L(x^†, x) since e_x is the unique von Neumann regular element in (𝒥) ₁(e_x). Therefore, B(x, x^†) = B(e_x, e_x) = Π₀, where Π_μ stands for the Peirce projection associated with the tripotent e_x onto the subtriple 𝒥_μ(e_x) = {z : L(e_x, e_x)z = μz} (cf. [12]). In [19, page 192], authors show that e_x is unitary in P(x)𝒥, hence P(x)𝒥 = 𝒥₁(e_x). Therefore, e_x ∈ ℰ(𝒥) ₁ by [20, Lemma  4]. Thus, B(x, x^†) = B(e_x, e_x) = 0. We conclude that $$.

Remark 7. The unique BP-quasi inverse of any BP-quasi invertible matrix x in the JB^*-triple ℳ₂₃(ℂ), satisfying the conditions of Theorem 6, is precisely the Moore-Penrose inverse x^† (cf. [10, Lemma  3.2]), which can be calculated according to the rank of the matrix (cf. [21]). If rank (x) = 2 then xx^t is invertible in ℳ₂(ℂ), and the Moore-Penrose inverse of x is given by x^† = x^t(xx^t) ⁻¹. For example, let $$. Then x is BP-quasi invertible and its unique BP-quasi inverse, via Theorem 6, is given by $$. Since x(x^†) ^t = (x^†) ^tx = e (the unit in ℳ₂(ℂ)), x(x^†) ^ty = (x^†) ^txy = y for all y ∈ M₂₃ and so L(x, x^†) = I. Thus, B(x, x^†) = 0.

Theorem 8. For any JB^*-triple 𝒥, the set $$ is open in the norm topology.

Proof. Assume B(a, c) = 0 where c = a^† + z for some nonzero z ∈ 𝒥, and a^† denotes the Moore-Penrose type inverse of a (cf. [4, 10]). Consequently, P(a)c = {ac^*a} = a, and so a = $$ = $$ = a + {az^*a}. Thus, P(a)z = {az^*a} = 0, which means P(a^†)P(a)z = 0. Recall that Π₁ = P²(v) = P(a^†)P(a) (see [10] and [19, page 192]). Also, since $$ implies that Π₁h = 0, the nonzero element z ∉ range(Π₁). Moreover, as a projection, Π₀ is the identity operator on 𝒥₀(v), so Π₀B(a, c)Π₀ = IΠ₀ = Π₀ = 0 since B(a, c) = 0. Thus, range(Π₀) = 0 and z ∈ range(Π_1/2) = 𝒥_1/2(v). Conversely, if a is BP-quasi invertible with the unique BP-quasi inverse a^† and z ∈ range(Π_1/2) = 𝒥_1/2(v), then B(a^†, a  +  z) = $$ = 1 − 3L(a^†, a)  + $$ + 2(L(a^†, a)L(a^†, z) + L(a^†, z)L(a^†, a) + $$ = B(a^†, a) + 2(L(a^†, a)L(a^†, z) + L(a^†, z)L(a^†, a) + $$. Since B(a^†, a) = 0 and {𝒥₁𝒥₀𝒥} = 0 [12, Theorem  5.4 (9)], L(a^†, z) = 0, and we get the value 0 for the right hand side. Hence a + z is BP-quasi invertible. The elements x ∈ 𝒥₁(v): = range(Π₁) with spectrum $$ form an open subset of 𝒥₁. It follows that $$ is an open subset of 𝒥. Thus, $$ is an open set.

Theorem 9. Let x be an invertible element in a unital JB^*-algebra 𝒥. Then there exists unique u ∈ 𝒰(𝒥) such that x is positive and invertible in 𝒥^[u].

Proof. Since x ∈ 𝒥⁻¹, there is a unique element y ∈ 𝒥 with x∘y = e and x²∘y = x where “∘” denotes the Jordan binary product and e is the unit in 𝒥. Since $$ and that any JB^*-algebra is a JB^*-triple under the triple product {ab^*c}: = a∘(b^*∘c) − b^*∘(a∘c) + c∘(a∘b^*), so, by Theorem 6, there exists a unique u ∈ ℰ(𝒥) ₁ (viz, the range tripotent e_x) such that x is positive and invertible in 𝒥₁(u) with the inverse of the same y in 𝒥₁(u) (cf. [15, Theorem  6]). Moreover, x^† = x⁻¹, and so $$. Hence, the Peirce decomposition of 𝒥 reduces to 𝒥 = 𝒥₁(u). Of course, the product and the involution defined on both the JB^*-algebra 𝒥₁(u) and the unitary isotope 𝒥^[u] of 𝒥 are the same. Further, e = e ∘_u u = u ∘_u e = {uu^*e} = u∘u^* and u = u∘_uu = {uu^*u} = 2u∘(u∘u^*) − u²∘u^*, so that u²∘u^* = u. Hence, u ∈ 𝒥⁻¹ with inverse u^*; that is, u ∈ 𝒰(𝒥). We conclude that 𝒥₁(u) and 𝒥^[u] coincide as JB^*-algebras.

Theorem 10. Let 𝒥 be a JB^*-triple and $$. Then x is not positive invertible in the Peirce 1-space 𝒥₁(v), for all v ∈ ℰ(𝒥) ₁.

Proof. Suppose x is positive invertible in 𝒥₁(v) for some v ∈ ℰ(𝒥) ₁. Then there exists a unique inverse y ∈ 𝒥₁(v) satisfying x ·_v y = y ·_v x = v and x² ·_v y = x. Since $$, x is von Neumann regular with generalized inverse y satisfying P_v(x)y = x and P_v(y)x = y, where $$. Being positive, x is self-adjoint in 𝒥₁(v); that is, $$. However, the invertibility is invariant under the involution (see above). Therefore, x ∈ (𝒥₁(v)) ⁻¹ with inverse y such that $$ = $$ = $$ = $$ by Theorem 2. This together with the von Neumann regularity of x in 𝒥₁(v) gives x = P_v(x)y = P(x)P(v)y = P(x)y. Thus, the element x is von Neumann regular in 𝒥.

Now, by Theorem 4, there exists a (unique) tripotent e_x ∈ 𝒥 such that x is positive invertible in 𝒥₁(e_x). Hence, by Remark 5 and our supposition on x, we get the coincidence 𝒥₁(e_x) = 𝒥₁(v).

Next, by Theorem 6, y is the unique generalized inverse of x satisfying P(x)y = x, P(y)x = y, and L(x, y) = L(v, v) = L(y, x). Hence, B(x, y) = B(v, v) = 0. This means $$, a contradiction to the hypothesis.

Theorem 11. An element x in a JB^*-triple 𝒥 is BP-quasi invertible if and only if x is positive and invertible in the Peirce 1-space 𝒥₁(v) for some extreme point v ∈ ℰ(𝒥) ₁.

Theorem 12. Let 𝒥 be a JB^*-triple with v ∈ ℰ(𝒥) ₁ and $$ (the open unit ball of 𝒥₁(v)). Then for any positive integer n, $$ with v_i ∈ ℰ(𝒥) ₁ for each i = 1, …, n.

Proof. Since v ∈ ℰ(𝒥) ₁, 𝒥₁(v) is a JB^*-algebra with unit v. Since ∥  − s∥ < 1 in 𝒥₁(v), v + s and so (1/2)(v + s) are invertible elements in 𝒥₁(v) (cf. [22, Lemma  2.1(iii)]). Clearly, ∥(1/2)(v + s)∥ < 1. Hence, by [7, Lemma  2.1], there exist unitaries v₁, v₂ ∈ 𝒰(𝒥₁(v)) with v  +  s = v₁  +  v₂. Moreover, v₁, v₂ ∈ ℰ(𝒥) ₁ by [20, Lemma  4]. The required result now follows by induction on n.

Definition 13. For any element x in a JB^*-triple 𝒥, the number e_m(x) is defined by $$. If x has no such decomposition, then e_m(x): = ∞.

Theorem 14. Let 𝒥 be a JB^*-triple, v ∈ ℰ(𝒥) ₁ and x ∈ 𝒥 with nx − v ∈ 𝒥₁(v) satisfying ∥nx − v∥ < n − 1 for some n ≥ 2. Then e_m(x) ≤ n. On the other hand, if e_m(x) ≤ n, then dist (nx, ℰ(𝒥) ₁) ≤ n − 1.

Proof. Clearly, $$. Replacing s by (n − 1) ⁻¹(nx − v) in Theorem 12, we get $$ with v_i ∈ ℰ(𝒥) ₁ for i = 1, …, n. Thus, e_m(x) ≤ n. On the other hand, suppose e_m(x) ≤ n. Then $$ for some 1 ≤ r ≤ n with v_is in ℰ(𝒥) ₁. Then $$. Further, ∥rx  −  v₁∥ = $$. Hence, ∥nx  −  v₁∥ = ∥(n − r)x  +  rx  −  v₁∥ ≤ ∥(n  −  r)x∥  +  ∥rx  −  v₁∥ ≤ n  −  r  +  r  −  1 = n  −  1 since ∥x∥ ≤ 1. Thus, dist (nx, ℰ(𝒥) ₁) = $$.

Example 15. Let 𝒥 be the JB^*-triple ℳ₁₂(ℂ), and let v denotes the matrix [1      0]. Then ∥v∥ = 1 and vv^tv = v. So, v is a norm one tripotent matrix. Moreover, for any matrix $$, we have $$. Also note that $$, where $$ denotes the complex conjugate of z_i. So that, $$. Hence, $$. Thus, v ∈ ℰ(𝒥) ₁.

Next, consider the Peirce spaces of 𝒥, relative to the extreme point matrix v = [1      0], given by 𝒥₁(v) = {z : z = [z₁      0], z₁ ∈ ℂ}, 𝒥_1/2(v) = {z : z = [0      z₂], z₂ ∈ ℂ} and 𝒥₀(v) = {[0      0]}. Clearly, both the balls (𝒥₁(v)) ₁ and (𝒥_1/2(v)) ₁ are nontrivial. It is easily seen that the matrix [0      1] ∈ ℰ(𝒥) ₁∖ℳ₁(v).

Theorem 16 (A Russo-Dye Theorem type for JB*-triples). Let 𝒥 be a JB^*-triple, v ∈ ℰ(𝒥) ₁.

• (i)

  For any x ∈ 𝒥₁(v) with ∥x∥ < 1 − 2n⁻¹ for some n ≥ 3, there exists v_i ∈ ℰ(𝒥) ₁ for all i = 1,2, 3, …, n such that $$.

• (ii)

  $$.

• (iii)

  $$.

Proof. (i) Since ∥x∥ < 1 − 2n⁻¹, we have ∥nx∥ < n − 2, so that ∥(n−1)⁻¹(nx − v)∥ < 1. Also, note that (n − 1) ⁻¹(nx − v) ∈ 𝒥₁(v); recall that x is in the JB^*-algebra 𝒥₁(v). Hence, by taking v the same as in Theorem 12 while s = (n − 1) ⁻¹(nx − v), we get $$ for some extremes v_is in 𝒥.

(ii) Suppose $$. Then ∥x∥ < 1 − 2n⁻¹ for some positive integer n ≥ 3. Therefore, x ∈ co  ℰ(𝒥) ₁ by the part (i).

(iii) Using the part (ii), we get $$.

Remark 17. If a JB^*-triple 𝒥 has a BP-quasi invertible element then ℰ(𝒥) ₁ is a nonempty set by Theorem 6. Hence, the above theorem holds in this case.

Corollary 18. For any extreme point v of the closed unit ball in a JB^*-triple 𝒥, $$. If v is a unitary element in 𝒥, then $$. Moreover, $$ = $$ = $$ = (𝒥) ₁.

Proof. We know from [20, Lemma  4] that 𝒰(𝒥₁(v))⊆ℰ(𝒥) ₁. So, by the Russo-Dye Theorem for JB^*-algebras [7, Theorem  2.3], $$. By Example 15, there exists a JB^*-triple 𝒥 with u, v ∈ ℰ(𝒥) ₁ such that u ∉ 𝒥₁(v). Hence, we have $$.

If v ∈ 𝒰(𝒥), then L_v,v = I (the identity operator on 𝒥)  (cf. [3]). So, 𝒥 = 𝒥₁(v) which is a unital JB^*-algebra, and hence 𝒥 is a JB^*-algebra. Then, $$ by [7, Theorem  2.3]. Hence, $$.

As mentioned above, every unitary element v in a JB^*-triple 𝒥 satisfies L_v,v = I, and so P(v)P(v) = I (cf. [10, page  582]). Hence, v is an extreme point of (𝒥) ₁ by [23, Lemma  3.2 and Proposition  3.5]. Thus, for any v ∈ 𝒰(𝒥), (𝒥) ₁ = $$ = $$. Therefore, $$ = $$ = $$ = (𝒥) ₁.

Corollary 19. Each element of the Peirce 1-space 𝒥₁(v) in a JB^*-triple 𝒥 with v ∈ ℰ(𝒥) ₁ is a positive multiple of the sum of three extreme points of (𝒥) ₁.

Proof. Let x ∈ 𝒥₁(v) and ϵ > 0. Let y = (3∥x∥ + ϵ) ⁻¹x. Then ∥y∥ < 1/3. Hence, by Theorem 16(i), there exist three extreme points u₁, u₂, u₃ in 𝒥 such that y = (1/3)(u₁  +  u₂  +  u₃). Thus, x = (∥x∥ + ϵ/3)(u₁ + u₂ + u₃).

Theorem 20. In any JB^*-triple 𝒥, $$.

Proof. Let $$. Then, there is a unique v ∈ ℰ(𝒥) ₁ such that x is positive and invertible in 𝒥₁(v) by Theorem 6. In particular, x is self-adjoint in the JB^*-algebra 𝒥₁(v). This together with ∥x∥  ≤  1 gives the existence of unitaries v₁, v₂ ∈ 𝒥₁(v) satisfying x = (1/2)(v₁ + v₂) by [24, Theorem  2.11]. The result now follows from [20, Lemma  4].

Theorem 21. Let 𝒥 be a JB^*-triple and v ∈ ℰ(𝒥) ₁. Let x ∈ (𝒥₁(v)) ₁ with dist (x, 𝒰(𝒥₁(v)) < 2α and α < 1/2. Then x ∈ αℰ(𝒥) ₁ + (1 − α)ℰ(𝒥) ₁.

Proof. Since 𝒥₁(v) is a JB^*-algebra with unit v, x ∈ α𝒰(𝒥₁(v))+(1 − α)𝒰(𝒥₁(v)) by [26, Corollary  3.4]. Hence, x ∈ αℰ(𝒥) ₁ + (1 − α)ℰ(𝒥) ₁ by [20, Lemma  4].

Lemma 22. Let 𝒥 be a JB^*-triple. Let x ∈ 𝒥 and n ∈ ℕ (the set of positive integers). Then

• (i)

  x ∈ co _nℰ(𝒥) ₁⇔e_m(x) ≤ n.

• (ii)

  n ∈ V(x)⇔e_m(x) ≤ n.

• (iii)

  e_m(x) = min (V(x)∩ℕ).

Theorem 23. Let 𝒥 be a JB^*-triple, v ∈ ℰ(𝒥) ₁ and x ∈ (𝒥) ₁∩𝒥₁(v).

• (a)

  Let ∥γx − v₀∥  ≤  γ − 1 for some γ ≥ 1 and some v₀ ∈ 𝒰(𝒥₁(v)). Let (δ₂, …, δ_m) ∈ ℝ^m−1 such that 0 ≤ δ_j < γ − 1 for all j = 2, …, m − 1 and $$. Then there exist v₁, …, v_m ∈ ℰ(𝒥) ₁ satisfying $$. Moreover, ]γ, ∞[ ⊆ V(x).

• (b)

  On the other hand, if ]γ, ∞[ ⊆V(x) holds, then for each r > γ, there is v₁ ∈ ℰ(𝒥) ₁ such that ∥rx − v₁∥ ≤ r  −  1.

Proof. Since 𝒥₁(v) is a unital JB^*-algebra, the part (a) follows from [20, Theorem  2.2] and [20, Lemma  4].

(b) Suppose ]γ, ∞[ ⊆V(x). If r > γ then x ∈ co _rℰ(𝒥) ₁ with x = r⁻¹(v₁ + ⋯+v_n−1 + (1  +  r  −  n)v_n) for some v₁, …, v_n ∈ ℰ(𝒥) ₁ and positive integer n satisfying n − 1 < r ≤ n. Hence, $$.

Corollary 24. For any JB^*-triple 𝒥, co _γℰ(𝒥) ₁⊆co _δℰ(𝒥) ₁ if 1 ≤ γ ≤ δ. In particular, for each x ∈ (𝒥) ₁∩𝒥₁(v), V(x) is either empty or equal to [γ, ∞) or (γ, ∞) for some γ ≥ 1.

Lemma 25. Let 𝒥 be a JB^*-triple with nonempty ℰ(𝒥) ₁ and x ∈ 𝒥. Then

• (i)

  α_q(rx) = |r | α_q(x) for all r ∈ ℂ;

• (ii)

  α_q(x) ≤ ∥x∥;

• (iii)

  |α_q(x) − α_q(y)| ≤ ∥x − y∥ for all y ∈ 𝒥;

• (iv)

  if 𝒥 is a JB^*-algebra, then α_q(x) = α_q(x^*).

Proof. (i) For any complex number r ≠ 0, $$ if and only if $$. So, the part (i) is clear for all nonzero complex numbers. If r = 0, then α_q(rx) = α_q(0) = 0, one may consider the BP-quasi invertible element (1/n)v with v ∈ ℰ(𝒥) ₁ and positive integer n → 0.

(ii) Since for any v ∈ ℰ(𝒥) ₁ and any positive integer n, $$; hence, α_q(x) = $$. It follows that α_q(x) ≤ ∥x∥.

(iii) Since the set ℰ(𝒥) ₁ is nonempty, and since any extreme point is BP-quasi invertible, α(x) < ∞. Now, by definition of α_q(x), for each n ∈ ℕ, there exists an element $$ such that ∥x − x_n∥ ≤ α_q(x) + 1/n. Hence, α_q(y) ≤ ∥y − x_n∥ ≤ ∥y − x∥ + ∥x − x_n∥ ≤ ∥x − y∥ + α_q(x) + 1/n for all n ∈ ℕ, so that α_q(y) ≤ ∥x − y∥ + α_q(x). Interchanging x and y, we get α_q(x) ≤ ∥x − y∥ + α_q(y). Thus, |α_q(x) − α_q(y)| ≤ ∥x − y∥ for all y ∈ 𝒥.

(iv) Now, let the 𝒥 be a JB^*-algebra. If $$, then $$ by [4, Theorem  3.2], and hence α_q(x) = 0 = α_q(x^*). Next, if x is not BP-quasi invertible, then α_q(x) = $$ = $$ = $$ = α_q(x^*) since the involution “*” is an isometry.

Theorem 26. Let 𝒥 be a JB^*-triple, v ∈ ℰ(𝒥) ₁ and $$. Then dist (x, ℰ(𝒥) ₁) ≥ max {α_q(x) + 1, ∥x∥ − 1}.

Proof. Since v ∈ ℰ(𝒥) ₁, ∥x − v∥ ≥ ∥x∥ − 1. We assume ∥x − v∥ < 1, then the fact that v is the unit of the JB^*-algebra 𝒥₁(v) gives the invertibility of x in 𝒥₁(v), and so x is positive and invertible in the unitary isotope of 𝒥₁(v) induced by certain unitary element u in 𝒥₁(v); hence, u ∈ ℰ(𝒥) ₁ by [20, Lemma  4]. For any z ∈ 𝒥₁(v) ^[u], we have L(v, v)z = z and L_v(u, u) = I on 𝒥₁(v) since u is unitary in 𝒥₁(v), so that L(u, u)z = L_v(u, u)z = z by Theorem 2. Hence, 𝒥₁(v) ^[u]⊆𝒥₁(u). For the reverse inclusion, recall that u ∈ ℰ(𝒥) ₁, and so 𝒥₁(u) = P(u)(𝒥) (see the above proof of Theorem 4); similarly, 𝒥₁(v) = P(v)(𝒥), and so u = P(v)w for some w ∈ 𝒥. Hence, for any fixed z ∈ 𝒥₁(u), there exists s ∈ 𝒥 with z = P(u)s = P(P(v)w)s = P(v)P(w)P(v)s ∈ P(v)𝒥 = 𝒥₁(v). Thus, 𝒥₁(u)⊆𝒥₁(v) ^[u]. Therefore, 𝒥₁(v) ^[u] = 𝒥₁(u) as sets. Moreover, we observe that $$ and $$ by Theorem 2. We conclude that (𝒥₁(v)) ^[u] and 𝒥₁(u) coincide as JB^*-triples and x is positive invertible in the Peirce 1-space 𝒥₁(u). So, $$ by Theorem 10; a contradiction to the hypothesis. Thus, ∥x − v∥ ≥ 1.

By setting y = ∥x−v∥⁻¹(x − v) + v, we get y ∈ 𝒥₁(v) since x, v ∈ 𝒥₁(v). Then ∥y − v∥ = ∥x−v∥⁻¹∥x − v∥ = 1. However, the open unit ball 𝔅(v; 1) in 𝒥₁(v) about v is included in (𝒥₁(v)) ⁻¹ (cf. [22, Lemma  2.1]). Therefore, $$. From our above discussion, we also have $$. Thus, α_q(x) = $$ = ∥x − ∥x−v∥⁻¹(x − v) − v∥ = ∥x − v∥(1 − ∥x−v∥⁻¹) = ∥x − v∥ − 1 as ∥x − v∥  ≥  1; which means α_q(x) + 1  ≤  ∥x − v∥. However, ∥x∥ − 1 ≤ ∥x − v∥, as seen above. We conclude that dist (x, ℰ(𝒥) ₁) ≥ max {α_q(x) + 1, ∥x∥ − 1}.

Theorem 27. If 𝒥 is a JB^*-triple, then dist (x, ℰ(𝒥) ₁) ≤ max {1, ∥x∥ − 1} for all $$.

Proof. Let $$. Then by Theorem 6, x is positive invertible in the JB^*-algebra 𝒥₁(v) for some v ∈ ℰ(𝒥) ₁, which is a JB^*-algebra with unit v. Hence, x being positive satisfies ∥x − v∥ = dist (x, ℰ(𝒥) ₁) by [26, Proposition  3.2]. By the functional calculus for positive elements in JB^*-algebras,

Hence, dist (x, ℰ(𝒥) ₁) ≤ max {1, ∥x∥ − 1} for all $$. Next, suppose (x_n) is a sequence in $$ such that ∥x_n − x∥ < (1/n) for all n ∈ ℕ. By Theorem 6, there exists a unique v_n ∈ ℰ(𝒥) ₁ corresponding to each x_n. Hence, dist (x_n, ℰ(𝒥)₁) ≤ ∥x_n − v_n∥ for all n. By (2), ∥x_n − v_n∥ ≤ max {1, ∥x_n∥ − 1} for all n. Hence, for large n, we get dist (x, ℰ(𝒥)₁) ≤ ∥x − v_n∥ = ∥x_n − v_n∥ + ∥x − x_n∥ ≤ max {1, ∥x_n∥ − 1} + (1/n) by using (2). It follows that dist (x, ℰ(𝒥) ₁) ≤ max {1, ∥x∥ − 1}.

Theorem 28. Let 𝒥 be a JB^*-triple, v ∈ ℰ(𝒥) ₁, and $$.

• (i)

  If α_q(x) = ∥x∥ then dist (x, ℰ(𝒥)₁) = α_q(x) + 1. In particular, α_q(x) = ∥x∥ = 1 implies dist (x, ℰ(𝒥)₁) = 2.

• (ii)

  If α_q(x) < ∥x∥ = 1, then dist (x, ℰ(𝒥) ₁) = α_q(x) + 1.

Proof. (i) Since $$ ≤  ∥x − 0∥ + dist (0, ℰ(𝒥) ₁) ≤ α_q(x) + max {1, ∥0∥ −  1} = α_q(x) + 1 by Theorem 27. On the other hand, dist (x, ℰ(𝒥) ₁) ≥ α_q(x) + 1 by Theorem 26. Hence, the part (i) is proved.

(ii) Let ϵ > 0 with α_q(x)  +  ϵ < 1. Then there exists $$ such that ∥y − x∥  < α_q(x) + ϵ. So, ∥y∥ ≤ ∥y − x∥ + ∥x∥ < α_q(x) + ϵ + 1 < 2. Since $$, y is positive and invertible in 𝒥₁(v) for some v ∈ ℰ(𝒥)₁ by Theorem 6; 𝒥₁(v) is a JB^*-algebra with unit v. By the continuous functional calculus, we get ∥y − v∥  ≤  ∥y∥ − 1 < 2 − 1 = 1 since ∥y∥ < 2. Therefore, ∥x − v∥ ≤ ∥x − y∥ + ∥y − v∥ < α_q(x) + ϵ + 1. Hence, dist (x, ℰ(𝒥) ₁) ≤ α_q(x) + 1. Moreover, dist (x, ℰ(𝒥) ₁) ≥ α_q(x) + 1 by Theorem 26.

Corollary 29. Let 𝒥 be a JB^*-triple, v ∈ ℰ(𝒥) ₁ and x ∈ 𝒥 with ∥x∥ ≥ 2.

• (i)

  If x is positive in 𝒥₁(v), then ∥x − v∥ = ∥x∥ − 1.

• (ii)

  If $$, then there is u ∈ ℰ(𝒥) ₁ such that ∥x − u∥ = ∥x∥ − 1.

Proof. (i) Since v being an extreme point of the closed unit ball, v is a tripotent in 𝒥, v = P_vv = {vv^*v}. Hence, ∥v∥ = ∥{vv^*v}∥ = ∥v∥³, so that ∥v∥ = 1. Now, since x is positive in 𝒥₁(v) and since ∥x∥ ≥ 2, we get by the functional calculus for positive elements that ∥x − v∥ = ∥x∥ − 1.

(ii) Suppose $$. By Theorem 6, x is positive in 𝒥₁(v) for some v ∈ ℰ(𝒥) ₁. Hence, ∥x − u∥ = ∥x∥ − 1 by Part (i).