Unique square root of a positive operator on a Hilbert space

Question

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?

Answer 1

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 $S^2=T$.

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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 $\|I-T\|\leq1$.