$T$ selfadjoint with nonnegative spectrum implies $T=S^2$ for some selfadjoint $S$

Question

I have question regarding the following problem: Let $H$ be a Hilbert space and $T$ a bounded linear operator from $H$ to $H$. If we assume that $T$ is selfadjoint with spectrum in $[0,\infty)$ can we conclude that $T=S^2$ for some selfadjoint bounded operator $S$ (I know that this implies that $T$ is positive but I want to show the implication directly)?

I started to define $X = T-aI$ where $a \in (-\infty,0)$. Then $X$ is selfadjoint and invertible but I don't know how to proceed.

Answer 1

The answer is yes. The Theorem on page 265 in Frigyes Riesz and Béla Sz.-Nagy's Functional Analysis (Dover paperback 1990) states:

Every positive symmetric transformation $A$ possesses a positive symmetric square root, and only one, which we denote by $A^{1/2}$. It can be represented as the limit (in the strong sense) of a sequence of polynomials in $A$, and consequently is permutable with all transformations which are permutable with $A$.

What I like about their proof is that it is constructive, there is no compactness assumption, it works in real or complex Hilbert space and does not use the spectral theorem.

Answer 2

Let me summarize what I said in the comments.

If your self-adjoint operator $T$ is on a finite dimensional space, or otherwise is a compact operator, then the spectral theorem for compact operators says it has eigenvalues, real eigenvalues due to self-adjointness. And non-negative by hypothesis. Then we may define $S\colon e_i\mapsto \sqrt{\lambda_i}e_i,$ and we will have $S^2=T.$

If $T$ is assumed to be bounded, but not compact, then we are not guaranteed to have eigenvalues, but we still have a complete spectral decomposition. In this case the spectral theorem can take a few forms, in terms of spectral measure, projection-valued measure, and a multiplication operator.

In terms of the projection-valued measure, if we have $T=\int_{\sigma(T)}\lambda\,dE_\lambda$, we may construct $S=\int_{\sigma(T)}\sqrt{\lambda}\,dE_\lambda$ as our square root.

This is the general method by which you can apply any continuous function to an operator, via the Borel functional calculus.