In idempotent form: ( T = T_1 \mathbfe_1 + T_2 \mathbfe_2 ), where ( T_1, T_2 ) are complex linear operators between ( X_1, Y_1 ) and ( X_2, Y_2 ).
with componentwise addition and multiplication. Equivalently, introduce an independent imaginary unit ( \mathbfj ) (where ( \mathbfj^2 = -1 ), commuting with ( i )), and write:
Solution: Define a as a map ( | \cdot | : X \to \mathbbR_+ ) satisfying standard Banach space axioms, but with scalar multiplication by bicomplex numbers respecting: Basics of Functional Analysis with Bicomplex Sc...
Any bicomplex Banach space ( X ) is isomorphic (as a real Banach space) to ( X_1 \oplus X_2 ), where ( X_1, X_2 ) are complex Banach spaces, and bicomplex scalars act by: [ (z_1 + z_2 \mathbfj) (x_1 \mathbfe_1 + x_2 \mathbfe_2) = (z_1 - i z_2) x_1 \mathbfe_1 + (z_1 + i z_2) x_2 \mathbfe_2. ]
[ | \lambda x | = |\lambda| \mathbbC | x | \quad \textor more generally \quad | \lambda x | = |\lambda| \mathbbBC | x | ? ] But ( |\lambda|_\mathbbBC = \sqrtz_2 ) works, giving a real norm. However, to preserve the bicomplex structure, one uses : In idempotent form: ( T = T_1 \mathbfe_1
[ \mathbbBC = z_1 + z_2 \mathbfj \mid z_1, z_2 \in \mathbbC ]
( T ) is bounded if there exists ( M > 0 ) such that ( | T x | \leq M | x | ) for all ( x ). This is equivalent to ( T_1 ) and ( T_2 ) being bounded complex operators. ] [ | \lambda x | = |\lambda|
The bicomplex spectrum of ( T ) is: [ \sigma_\mathbbBC(T) = \lambda \in \mathbbBC : \lambda I - T \text is not invertible . ] In idempotent form: [ \sigma_\mathbbBC(T) = \sigma_\mathbbC(T_1) \mathbfe 1 + \sigma \mathbbC(T_2) \mathbfe_2 ] where the sum is in the sense of idempotent decomposition: ( \alpha \mathbfe_1 + \beta \mathbfe_2 : \alpha \in \sigma(T_1), \beta \in \sigma(T_2) ).
