Since {(1, 1), (1,−1), (−1, 1), (−1,−1)} is the set of all extreme points of the unit ball of l∞2 and T is bilinear,

Note that if ||T|| = 1, then |a| ≤ 1, |b| ≤ 1, |c| ≤ 1 and |d| ≤ 1.

It follows from Theorem 2.1 and the above remark of Theorem 2.1.

Let

We call Norm(T) the norming set of T.

Suppose that T∈L(l2∞2) with ||T|| = 1. By Theorem 2.2, we may assume that a ≥ |b| and c ≥ |d|.

(⇐): Obviously,

Hence, 1 = |a + b + c + d| = |a + b − c − d| = |a − b − c + d| = |a − b + c − d|, so 1 = a + b + c + d = a − b + c − d, so c = 1− a, d = −b. Since 1 = |a + b − c − d| = |a−b−c+d|, 1 = |(2a−1)+2b| = |(2a−1)−2b|. Hence, (2a−1)2b = 0, so a=12 or b = 0. If a=12, then b=±12 and T=(12,12,12,-12) or T=(12,-12,12,12). Note that (12,12,12,-12) and T=(12,-12,12,12) are extreme. If b = 0, then a = 0 or a = 1, so T = (0, 0, 1, 0) or T = (1, 0, 0, 0). Note that (0, 0, 1, 0) and (1, 0, 0, 0) are extreme. By Theorem 2.2, (-12,12,12,12),(12,12,-12,12), (0, 1, 0, 0), (0, 0, 0, 1) are extreme.

(⇒): Suppose that T∈extBL(l2∞2) and the necessary condition is not true.

Then

Without loss of generality we may assume that

Case 1: Norm(T) = {((1, 1), (1, 1)), ((1,−1), (1,−1)), ((1, 1), (1,−1))}

Let T1=(a-1n,b+1n,c-1n,d+1n) and T2=(a+1n,b-1n,c+1n,d-1n) for a sufficiently large n ∈ ℕ so that ||T_i|| = 1 for i = 1, 2. Therefore, T is not extreme, which is a contradiction.

Case 2: Norm(T) = {((1, 1), (1, 1)), ((1,−1), (1,−1)), ((1,−1), (1, 1))}

Let T1=(a+1n,b-1n,c-1n,d+1n) and T2=(a-1n,b+1n,c+1n,d-1n) for a sufficiently large n ∈ ℕ so that ||T_i|| = 1 for i = 1, 2. Therefore, T is not extreme, which is a contradiction.

Case 3: Norm(T) = {((1, 1), (1, 1)), ((1, 1), (1,−1)), ((1,−1), (1, 1))}

Let T1=(a-1n,b-1n,c+1n,d+1n) and T2=(a+1n,b+1n,c-1n,d-1n) for a sufficiently large n ∈ ℕ so that ||T_i|| = 1 for i = 1, 2. Therefore, T is not extreme, which is a contradiction.

Case 4: Norm(T) = {((1,−1), (1,−1)), ((1, 1), (1,−1)), ((1,−1), (1, 1))}

Let T1=(a+1n,b+1n,c+1n,d+1n) and T2=(a-1n,b-1n,c-1n,d-1n) for a sufficiently large n ∈ ℕ so that ||T_i|| = 1 for i = 1, 2. Therefore, T is not extreme, which is a contradiction.

It follows from the proof of Theorem 2.3.

It follows from Theorem 2.3 and the fact that

Note that if ||f|| = 1, then |α| ≤ 1, |β| ≤ 1, |δ| ≤ 1, |γ| ≤ 1.

It follows from Theorem 2.1 and the above remark of Theorem 2.1.

Now we are in position to describe all the exposed points of the unit ball of L(l2∞2).

It is enough to show that extBL(l2∞2)⊂expBL(l2∞2).

Claim: T = (1, 0, 0, 0) is exposed.

Let f∈L(l2∞2)* be such that α = 1,0 = β = δ = γ. Then f(T) = 1, |f(S)| < 1 for every S∈extBL(l2∞2){±T}. By Theorem 3.2, T is exposed. By Theorem 3.3, (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) are exposed.

Claim: T=12(-1,1,1,1) is exposed.

Let f∈L(l2∞2)* be such that α = 1,0 = β, δ=γ=12. Then f(T) = 1, |f(S)| < 1 for every S∈extBL(l2∞2){±T}. By Theorem 3.2, T is exposed. By Theorem 3.3, 12(1,-1,1,1),12(1,1,-1,1),12(1,1,1,-1) are exposed.

It follows from Theorem 2.1 and the above remark of Theorem 2.1.

By Theorem 4.1, we may assume that a ≥ |b| and c ≥ |d|.

(⇒): Suppose that T is smooth. For (u1,v1),(u2,v2)∈Sl∞2, let δ(u1,v1),(u2,v2)∈L(l2∞2)* such that

δ_{(u1,v1),(u2,v2)}(L) = L((u₁, v₁), (u₂, v₂)) for L∈L(l2∞2). Then ||δ_{(u1,v1),(u2,v2)}|| ≤ 1. Note that, by Theorem 3.1, 1 = ||δ_{(1,1),(1,±1)}|| = ||δ_{(1,−1),(−1,1)}|| = ||δ_{(−1,1),(1,1)}||.

Obviously,

Hence, if T∈smBL(l2∞2), then ||T|| = 1, so, by Theorem 2.1, (|a+b|+|c+d| = 1, |a − b| + |c − d| < 1) or (|a − b| + |c − d| = 1, |a + b| + |c + d| < 1). Suppose that |a+b|+|c+d| = 1, |a−b|+|c−d| < 1. We will show that 0 < |a+b| < 1. Otherwise. Then a + b = 1 or a + b = 0. Suppose that a + b = 1. Let f1∈L(l2∞2)* such that f₁(x₁x₂) = 1 = f₁(y₁y₂), f₁(x₁y₂) = 0 = f₁(x₂y₁), and let f2∈L(l2∞2)* such that 1 = f₂(x₁x₂) = f₂(y₁y₂) = f₂(x₁y₂) = f₂(x₂y₁). Since f₁ ≠ f₂,1 = ||f_j|| = f_j(T) for j = 1, 2, T is not smooth, which is a contradiction. Suppose that a + b = 0. Then c + d = 1. Let g1∈L(l2∞2)* such that g₁(x₁x₂) = 0 = g₁(y₁y₂), g₁(x₁y₂) = 1 = g₁(x₂y₁). Since g₁ ≠ f₂,1 = ||g₁|| = g₁(T), T is not smooth, which is a contradiction. Therefore, if |a + b| + |c + d| = 1, |a − b| + |c − d| < 1, then 0 < |a + b| < 1. Similarly as the case that |a + b| + |c + d| = 1, |a − b| + |c − d| < 1, if |a − b| + |c − d| = 1, |a + b| + |c + d| < 1, then we should have 0 < |a − b| < 1.

(⇐): Let f∈L(l2∞2)* such that 1 = ||f|| = f(T) with α = f(x₁x₂), β = f(y₁y₂), δ = f(x₁y₂), γ = f(x₂y₁).

Case 1: 0 < |a + b| < 1, |a + b| + |c + d| = 1, |a − b| + |c − d| < 1

By Theorem 2.1, ||T|| = 1. Note that

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

From (**),1≥f(ax1x2+(b±1n)y1y2+(c∓1n)x1y2+dx2y1)=1+1n∣β-δ∣, hence β = δ and 1≥f(ax1x2+(b±1n)y1y2+cx1y2+(d∓1n)x2y1)=1+1n∣β-γ∣, hence β = γ. 1≥f((a±1n)x1x2+by1y2+(c∓1n)x1y2+dx2y1)=1+1n∣α-δ∣, hence α = δ. Therefore, by (*), 1 = α = β = δ = γ, hence f is uniquely determined. Therefore, T is smooth.

Case 2: 0 < |a − b| < 1, |a − b| + |c − d| = 1, |a + b| + |c + d| < 1

By Case 1 and Theorem 4.1, T is smooth.