what is the reason of this of the following statement?

Question

in a paper i saw the following statement:

Let $\Phi:B(X)\longrightarrow B(X)$ is an additive and surjective map. If $T\in B(X)$ and for some $x\in X$

$Tx \otimes {\Phi(T)}^*f=\Phi(T)x\otimes T^*f$ ,implies that $\Phi(T)x$ and $Tx$ are linearly dependent. what is the reason?

($x\otimes f $ means a rank one operator )

thank you for your help.

What is $\Phi$? Is $\Phi \in B(X)$, or is $\Phi \in B(B(X))$? — Ben Grossmann, Jun 15 '16 at 18:23
Also, presumably $X$ is a Banach space. It's not also a Hilbert space by any chance, is it? — Ben Grossmann, Jun 15 '16 at 21:07
@joker: you could also accept these answers : (1), (2), (3), (4), (5), (6), (7), (8). Many thanks :-) ! — Watson, Jun 20 '16 at 09:24
And maybe these ones: (9), (10), (11), (12), (13), (14), (15), (16), (17). — Watson, Jun 20 '16 at 09:27

score 1 · Accepted Answer · answered Jun 15 '16 at 21:20

Note that $$ Tx \otimes \Phi(T)^*f = \Phi(T)x \otimes T^*f \implies\\ \langle (Tx \otimes \Phi(T)^*f)(y),g \rangle = \langle (\Phi(T)x \otimes T^*f)(y),g \rangle \quad \forall y \in X, g \in X^* \implies\\ g([\Phi(T)^*f(y)]Tx) = g([T^*f(y)]\Phi(T)x) \quad \forall y \in X, g \in X^* \implies\\ g([f(\Phi(T)y)]Tx) = g([f(Ty)]\Phi(T)x) \quad \forall y \in X, g \in X^* \implies\\ [f(\Phi(T)y)]g(Tx) = [f(Ty)]g(\Phi(T)x) \quad \forall y \in X, g \in X^* $$ Now, if we fix a $y$ such that either $\Phi(T)y$ or $Ty$ are not in $\ker f$, we see that there are constants $\alpha, \beta$ (not both zero) such that $$ \alpha \,g(Tx) = \beta \,g(\Phi(T)x) \quad \forall g \in X^* $$ which is to say that $Tx$ and $\Phi(T)x$ are linearly dependent.

what is the reason of this of the following statement?

1 Answers1