Positive square root of a compact positive operator

Construction of the root. Positivity forces $\lambda \ge 0$ for every eigenvalue $\lambda$ of $T$, because for a unit eigenvector $e$ one has $\lambda =⟨Te,e⟩\ge 0$; hence $\mathrm{\Sigma }$ consists of positive numbers and $0$ is an eigenvalue only in the form of the kernel [A1]. For $x\in H$ set $Sx:={\sum }_{\lambda \in \mathrm{\Sigma }}\sqrt{\lambda }{P}_{\lambda }x$, a finite-subset net over the at most countable index set $\mathrm{\Sigma }$. The family $(\sqrt{\lambda }{P}_{\lambda }x{)}_{\lambda }$ is orthogonal with ${\sum }_{\lambda }\mathrm{\parallel }\sqrt{\lambda }{P}_{\lambda }x{\mathrm{\parallel }}^2={\sum }_{\lambda }\lambda \mathrm{\parallel }{P}_{\lambda }x{\mathrm{\parallel }}^2=⟨Tx,x⟩\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x{\mathrm{\parallel }}^2<+\mathrm{\infty }$, using the expansion, orthogonality and Bessel [A1, A2]; so [A3] makes the net converge, $S$ is well defined with $\mathrm{\parallel }Sx{\mathrm{\parallel }}^2={\sum }_{\lambda }\lambda \mathrm{\parallel }{P}_{\lambda }x{\mathrm{\parallel }}^2=⟨Tx,x⟩\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x{\mathrm{\parallel }}^2$, and $S$ is linear with $\mathrm{\parallel }S\mathrm{\parallel }\le \sqrt{\mathrm{\parallel }T\mathrm{\parallel }}$. Moreover $⟨Sx,x⟩={\sum }_{\lambda }\sqrt{\lambda }⟨P_{\lambda }x,x⟩={\sum }_{\lambda }\sqrt{\lambda }\mathrm{\parallel }{P}_{\lambda }x{\mathrm{\parallel }}^2\ge 0$, because $P_{\lambda }$ is self-adjoint and $P_{\lambda }x\perp (x-P_{\lambda }x)$ [A2]; and $S$ is self-adjoint, since $⟨Sx,y⟩={\sum }_{\lambda }\sqrt{\lambda }⟨P_{\lambda }x,y⟩={\sum }_{\lambda }\sqrt{\lambda }⟨x,P_{\lambda }y⟩=⟨x,Sy⟩$ by absolute convergence and self-adjointness of each $P_{\lambda }$. Finally $S^2=T$: on $E_{\lambda }$ one has $S=\sqrt{\lambda }I$ whence $S^2=\lambda I=T$, both operators are continuous and agree on the linear span of $\bigcup E_{\lambda }$, hence on $M$ by continuity, and both vanish on $\ker T$ (for $x\in \ker T$, $P_{\lambda }x=0$ for all $\lambda$ because $E_{\lambda }\perp \ker T$ and $x$ is in every $E_{\lambda }^{\perp }$), so $S^2=T$ on $H=M\oplus \ker T$ by [A4].

Every positive square root kills the kernel. Let $R$ be compact and positive with $R^2=T$, and let $x\in \ker T$. Put $y:=Rx$, so $Ry=R^{2}x=0$. For every real $t$, positivity at $x+ty$ gives $0\le ⟨R(x+ty),x+ty⟩=⟨y,x⟩+t\mathrm{\parallel }y{\mathrm{\parallel }}^2$. Here $⟨y,x⟩=⟨Rx,x⟩$ is real and nonnegative. If $y\mathrm{\ne }0$, choosing $t=-(⟨y,x⟩+1)\mathrm{/}\mathrm{\parallel }y{\mathrm{\parallel }}^2$ makes the right side $-1$, impossible. Thus $Rx=0$ for every $x\in \ker T$. This works over both scalar fields and uses neither self-adjointness nor compactness of $R^{\ast }$.

The root is compact. For each $n\in \mathbb{N}$ let $F_n:=\{\lambda \in \mathrm{\Sigma }:\lambda \ge 1\mathrm{/}(n+1)\}$, finite by [A7], and put $S_n:={\sum }_{\lambda \in F_n}\sqrt{\lambda }{P}_{\lambda }$. This operator has finite-dimensional range by [A1], so is compact by [A5], including when $F_n$ is empty. For every $x\in H$, the orthogonal summation identity [A3] and Bessel [A2] give $\mathrm{\parallel }(S-S_n)x{\mathrm{\parallel }}^2={\sum }_{\lambda \mathrm{\notin }{F}_n}\lambda \mathrm{\parallel }{P}_{\lambda }x{\mathrm{\parallel }}^2\le (n+1{)}^{-1}{\sum }_{\lambda \mathrm{\notin }{F}_n}\mathrm{\parallel }{P}_{\lambda }x{\mathrm{\parallel }}^2\le (n+1{)}^{-1}\mathrm{\parallel }x{\mathrm{\parallel }}^2$. Thus $\mathrm{\parallel }S-S_n\mathrm{\parallel }\le (n+1{)}^{-1\mathrm{/}2}\to 0$. Since $H$ is Banach, [A5] implies that $S$ is compact. The same zero-based sequence handles empty, finite and infinite $\mathrm{\Sigma }$.

A positive square root acts diagonally. Let $R$ be as in [step 1.2], let $\lambda \in \mathrm{\Sigma }$, and put $a:=\sqrt{\lambda }>0$. For $x\in E_{\lambda }(T)$ set $v:=(R-aI)x$. The identity $R^{2}x=Tx=\lambda x=a^{2}x$ gives $Rv=a^{2}x-aRx=-av$. Positivity therefore implies $0\le ⟨Rv,v⟩=-a\mathrm{\parallel }v{\mathrm{\parallel }}^2$, forcing $v=0$. Hence $Rx=\sqrt{\lambda }x$ on the entire eigenspace, without a diagonalization of $R$ or an assumption that $R$ is self-adjoint.

Existence. By [step 1.1] and [step 1.3] the operator $S$ is a compact self-adjoint positive operator with $S^2=T$, and by construction it acts as $\sqrt{\lambda }I$ on each $E_{\lambda }(T)$, $\lambda \in \mathrm{\Sigma }$, and as $0$ on $\ker T$.

Uniqueness. Let $R$ be compact and positive with $R^2=T$. By [step 1.2] $R$ vanishes on $\ker T$, by [step 2.1] it equals $\sqrt{\lambda }I$ on each $E_{\lambda }(T)$, and by [step 2.2] the same two descriptions hold for $S$; hence the bounded operator $R-S$ vanishes on $\ker T$ and on each $E_{\lambda }(T)$. Since $H$ is the closed linear span of $\ker T\cup {\bigcup }_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }(T)$ by [A1], continuity gives $R-S=0$.

Conclusion. The operator $S$ of [step 1.1] is compact, self-adjoint and positive with $S^2=T$ by [step 2.2], and [step 3.1] shows that every compact positive $R$ with $R^2=T$ equals $S$; the action of $S$ on the positive eigenspaces and on the kernel is stated in [step 2.2]. This proves the lemma, including the uniqueness among compact positive square roots.