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Ben has answered your question, but here is an even easier route to a counter-example (modulo the hard part of the existence of a space without the AP). Take your favourite separable space without the AP. You may embed it into $C[0,1]$, which actually has a basis (the Schauder system, a prototypical example of a Schauder basis), so in particular it has the ...


By this answer, you are looking at the set $$ B_{w^{\ast}}(X) = \{T^{\ast} : T \in B(X_{\ast})\} $$ which is convex. Now if $\sigma$ denotes the weak-$\ast$-topology on $B(X)$, then $(B(X),\sigma)$ is a topological vector space whose topology is generated by the family of semi-norms given by $$ p_{f,y}(T) := |T(f)(y)| $$ for $f\in X, y\in X_{\ast}$. Hence, ...


Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$

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