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Let $H$ be a non-separable Hilbert space with an orthonormal basis $(e_\alpha)_{\alpha<\omega_1}$. To each $f=(f_\alpha)\in c_0(\omega_1)$ associate an operator on $H$ defined by $T_f (\sum_{\alpha<\omega_1} a_\alpha e_\alpha) = a_\alpha f_\alpha e_\alpha$. Is the linear subspace $\{T_f\colon f\in c_0(\omega_1)\}$ complemented in the space of compact operators on $H$? It looks like it is, but I don't how to calculate the norm of the obvious projection.

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You can compress to the diagonal the same as you would do in the countable case, and that operation yields a projection of norm one. Namely, for any compact $T$, define $$ E(T)x=\sum_\alpha\langle Te_\alpha,e_\alpha\rangle\,\langle x,e_\alpha\rangle\,e_\alpha $$ (this is well-defined because $|\langle Te_\alpha,e_\alpha\rangle\}_\alpha|\leq\|T\|$ for all $\alpha$). The compacity of $T$ implies that $\{\langle Te_\alpha,e_\alpha\rangle\}_\alpha\in c_0$. Also, $$ E(T_f)x=\sum_\alpha\langle T_fe_\alpha,e_\alpha\rangle\,\langle x,e_\alpha\rangle\,e_\alpha=\sum_\alpha f_\alpha\,\langle x,e_\alpha\rangle\,e_\alpha=T_fx, $$ so $E(T_f)=T_f$ and $E$ is a projection. Finally, $$ \|E(T)x\|^2=\sum_\alpha|\langle Te_\alpha,e_\alpha\rangle|^2\,|\langle x,e_\alpha\rangle|^2\leq\|T\|^2\,\sum_\alpha\,|\langle x,e_\alpha\rangle|^2=\|T\|^2\,\|x\|^2, $$ so $\|E(T)\|\leq\|T\|$, i.e. $\|E\|\leq1$. As $E$ is a projection, $\|E\|=1$.

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Thank you very much! –  Slavoj Žižek Nov 9 '12 at 20:45
    
You are welcome :) –  Martin Argerami Nov 9 '12 at 20:58
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