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Let H a Hilbert space and $T:H\rightarrow H$ a linear bounded, self-adjoint, positive and compact operator. How can i prove that the square root of T, $\ T^{1/2}:H\rightarrow H$ is also compact and self-adjoint; thanks.

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  • As $ T $ is a self-adjoint operator, we can apply the Continuous Functional Calculus (C.F.C.) to $ T $.

  • As $ T $ is a positive operator, i.e., $ {\sigma_{B(\mathcal{H})}}(T) \subseteq [0,\infty) $, we can apply the $ \sqrt{\bullet} $-function to $ T $ to obtain $ \sqrt{T} $.

  • The range of the C.F.C. of $ T $ is the $ \| \cdot \|_{B(\mathcal{H})} $-closed $ ^{\ast} $-subalgebra of $ B(\mathcal{H}) $ generated by $ T $.

  • As involution is a continuous operation on $ B(\mathcal{H}) $, it follows that $ \sqrt{T} $ is a self-adjoint operator.

  • As $ T \in K(\mathcal{H}) $ and $ K(\mathcal{H}) $ is a $ \| \cdot \|_{B(\mathcal{H})} $-closed ideal of $ B(\mathcal{H}) $, it follows that the range of the C.F.C. of $ T $ is contained in $ K(\mathcal{H}) $.

  • Therefore, $ \sqrt{T} \in K(\mathcal{H}) $.

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  • $\begingroup$ i was looking for an answer whithout using $C^*-$algebra tools. the compactness can be derived from spectral representation but what about self-adjointness; $\endgroup$ – nikosp Feb 23 '13 at 9:39
  • $\begingroup$ The range of C.F.C is not in the ideal of compacts. Take 1+T for example. $\endgroup$ – user207135 Jan 11 '15 at 14:56
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Hint: the diagonalization of compact self-adjoint operators makes this easy. See here for instance.

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