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Let $T$ be a positive operator on a Hilbert space $\mathcal{H}$. That is $\langle Tx,x\rangle\geq0$ for all $x$. Then there is a unique positive operator $S$ such that $S^2=T$ and this $S$ is called the square root of $T$ and is denoted by $S=T^{\frac{1}{2}}$. Can anyone suggest a proof of this with out using functional calculus and spectral theorem?

budi
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  • Look at Problem 95 and its solution in 'A Hilbert Space Problem Book' by Paul Halmos. – Kavi Rama Murthy Apr 01 '22 at 07:30
  • Thanks. I could find one proof. There it is mentioned one sentence which I couldnt understand. Since $(\alpha T)^{\frac{1}{2}}=\alpha^{\frac{1}{2}}T^{\frac{1}{2}}$ for $\alpha\geq 0$, we may assume that $T\leq I$. – budi Apr 01 '22 at 07:38
  • Take $\alpha =\frac 1 n$ with $n$ so large that $0 \leq \frac 1 n T \leq I$. This is possible because $ \langle (I-\frac 1 n T)x, x \rangle =|x|^{2}-\frac 1n \langle Tx, x\rangle \geq |x|^{2}-\frac 1n |T||x|^{2} \geq 0$ for $n\geq |T|$. – Kavi Rama Murthy Apr 01 '22 at 07:50

1 Answers1

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If $\|T\|\leq1$, you can define $$\tag1 S=\sum_{n=0}^\infty (-1)^n\binom{1/2}n\,(I-T)^n. $$ This converges in norm as long as $\|I-T\|\leq1$ (this requires some work to check$^*$), and this is guaranteed by $\|T\|\leq1$ since $T\geq0$.

For arbitrary positive $T$, use the above to find $(cT)^{1/2}$ for $c=\frac1{\|T\|}$ and take $S=c^{-1/2}(cT)^{1/2}$; then $S^2=T$.

About the convergence in $(1)$: the coefficients of the series in $(1)$ are not alternating. Rather, they are of the form, for $n\geq1$, $$ a_n=\frac{2(2n-2)!}{4^nn!(n-1)!}. $$ From here one shows that $\sum_n|a_n|<\infty$, which guarantees convergence in $(1)$ when $\|I-T\|\leq1$.

Martin Argerami
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    For one's information, $T\ge 0$, $||T||\le 1\Longrightarrow ||I-T||\le 1$ is discussed here, where it is stated that $T\ge 0$, $||T||\le 1\Longleftrightarrow 0\le\langle Tu,u\rangle\le||u||^2$, $\forall u\in H$. – Jianing Song Dec 15 '23 at 23:40
  • If anyone, like me, is looking for a reference for this formula: this can be found, e.g., in "Methods of Modern Mathematical Physics I: Functional Analysis" (1980) by Reed & Simon as Theorem VI.9 (together with the preceding lemma) – Frederik vom Ende Feb 13 '24 at 11:29