[Math] compact and self adjoint square root of an operator

Question

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.

Answer 1

• 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}) $.