2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $|\langle Te_\alpha,e_\alpha\rangle|\leq\|T\|$ 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, $\|E\|=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.