Let $x,a$ be bounded positive operators on a Hilbert space $\mathcal H$ such that $x\leq a.$ How can we prove that for some contraction $u$ on $\mathcal H$ we have that $x^{\frac{1}{2}}=ua^{\frac{1}{2}}$?
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You define $u$ explicitly on the range of $a^{1/2}$: $$ u(a^{1/2}\xi):=x^{1/2}\xi. $$ This is well-defined, because if $a^{1/2}\xi=0$, then $$\tag1\|x^{1/2}\xi\|^2=\langle x\xi,\xi\rangle\leq\langle a\xi,\xi\rangle=\|a^{1/2}\xi\|^2=0.$$ To define $u$ on $[a^{1/2}H]^\perp$, you notice that $$ [a^{1/2}H]^\perp=\ker a^{1/2}=\ker a. $$ By the above, you can define $u$ as $0$ on $[a^{1/2}H]^\perp$.
Finally, the estimate $(1)$ also shows that $u$ is bounded with $\|u\|\leq1$, and so it is a contraction and it extends to the closure of $a^{1/2}H$.
Martin Argerami
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