Square root lemma

This post is a continuation of Positive operators II, We begin by recalling the main fact proved there, then explain how a classical binomial series leads to a functional calculus for the square root of a positive operator.

Corollary 1 Let $A$ be a bounded self-adjoint positive operator on a pre-Hilbert space
$H$ . Then

(1) $\begin{equation*} \|A\| = \sup_{\|x\|=1}(x,Ax). \end{equation*}$

This equality between the operator norm and the supremum of quadratic forms will be used repeatedly below, in particular to control powers of $A$ and to justify convergence of operator series.

Binomial series

In this section we recall a classical binomial series expansion for $\sqrt{1-x}$ , discuss its convergence, and rewrite it in a form adapted to our later application to operators.
The philosophy is: first we understand the scalar series very well, then we simply replace the scalar variable by an operator and invoke a general result about Cauchy products in Banach algebras.

We want to “construct” a positive square root of a positive operator $A$ in a way that mimics the construction of the square root of a positive number. Historians of mathematics tell us that the Babylonians already had an efficient algorithm for extracting square roots; it was later described by Heron of Alexandria. Here we deliberately use a different, and in practice terribly ineffective, method. What matters to us is existence, not computational efficiency: our method uses a Taylor (binomial) series expansion.

One cannot comfortably use the Taylor series of $\sqrt{x}$ at $x=0$ , because the derivatives blow up there, but one can expand at $x=1$ or, equivalently, consider the Taylor series of $\sqrt{1-x}$ at $x=0$ .

Binomial series and its convergence

The series we need is discussed in many analysis textbooks, for instance in [1].
We have

(2) $\begin{equation*} \sqrt{1 - x} = \sum_{n=0}^{\infty} (-1)^n \binom{1/2}{n} x^n = 1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \cdots, \end{equation*}$

which can also be written as

(3) $\begin{equation*} \sqrt{1 - x} = 1 - \sum_{n=1}^{\infty} c_n x^n, \end{equation*}$

where

(4) $\begin{equation*} c_n = \frac{(2n-3)!!}{2^n n!}, \end{equation*}$

with the convention that $(-1)!!=1$ .

It is known (see \cite[pp.\ 56–57]{knapp}) that the series in \eqref{eq1mx0} is uniformly convergent for $x\in [-1,1]$ .
Notice that all $c_n\geq 0$ , and setting $x=1$ we obtain

(5) $\begin{equation*} \sum_{n=1}^{\infty} c_n = 1. \end{equation*}$

Thus, at the scalar level, the coefficients $c_n$ form a probability distribution on $\mathbb{N}$ (if one wishes to think probabilistically), and the corresponding power series converges uniformly on $[-1,1]$ .

We now state the main result we want to prove in this post, and then explain how the binomial series and the Cauchy product theorem in Banach algebras yield the desired square root of a positive operator.

Theorem 1. [Square Root Lemma]
Let $A$ be a bounded self-adjoint positive operator on a real or complex Hilbert space $H$ .
Then there is a unique bounded positive self-adjoint operator $B$ such that $A=B^2$ .
Furthermore, $B$ commutes with every bounded operator (linear or, if $H$ is complex, also antilinear) which commutes with $A$ .

Remark 1. In the proof below I essentially follow the method used in [2, Ch.\ VI.4, pp. 195–196]. As we shall see, this method works equally well for real Hilbert spaces.

Before we start the proof, let us make a couple of simple observations.
Let $A\in L(H)$ be such that $A=A^*\geq 0$ . Then, for every integer $n\geq 1$ , we have $A^n\geq 0$ . Indeed, if $n=2m$ is even, then

$(x,A^{2m}x)=(A^m x,A^m x)\geq 0,$

and if $n=2m+1$ is odd, then

$(x,A^{2m+1}x)=(A^m x,A A^m x)\geq 0.$

Suppose, additionally, that $A\leq I$ . Then also $A^n\leq I$ for all $n\geq 1$ . Indeed, from the corollary above we have, for $\Vert x\Vert=1$ , $(x,A^n x)\leq \|A^n\|$ , and $\|A^n\|\leq \|A\|^n\leq 1$ , since again $\|A\|\leq 1$ .

Proof. We first prove existence.

Reduction to the case $0\leq A\leq I$

The case $A=0$ is trivial, so we assume $\|A\|>0$ . It is enough to consider the case $0\leq A\leq I$ , or equivalently, since $A\geq 0$ , $\|A\|\leq 1$ . Indeed, in the general case we will define

$\sqrt{A}=\|A\|^{1/2}\,\sqrt{\frac{A}{\|A\|}},$

so the core of the argument is the construction for operators with spectrum contained in $[0,1]$ .

So let us assume $0\leq A\leq I$ .

Defining $\sqrt{I-A}$ via the binomial series

In the scalar formula \eqref{eq1mx} we replace the number $1$ by the identity operator $I$ and the scalar variable $x$ by the operator $A$ . Formally this gives

(6) $\begin{equation*} \sqrt{I-A} = I - \sum_{n=1}^{\infty} c_n A^n. \end{equation*}$

We first verify that the series on the right-hand side of \eqref{eqsqrA} is absolutely convergent in $L(H)$ to a bounded self-adjoint operator. Indeed, since $0\leq A\leq I$ , we have

$\|A^n\|\leq \|A\|^n\leq 1,$

which entails

$1+\sum_{n=1}^{\infty} \|c_n A^n\| \leq 1+\sum_{n=1}^{\infty} c_n =2,$

where we have used the fact that $c_n\geq 0$ and \eqref{eqcn1}.
Thus the series in \eqref{eqsqrA} converges absolutely in the Banach algebra $L(H)$ , and we can safely define

(7) $\begin{equation*} B := I - \sum_{n=1}^{\infty} c_n A^n. \end{equation*}$

The operator $B$ is clearly self-adjoint, since each $A^n$ is self-adjoint and the coefficients $c_n$ are real.

Showing that $B^2=I-A$

How do we know that $B^2$ is indeed equal to $I-A$ , given that we are now dealing with a series of operators rather than numbers, as in \eqref{eq1mx}?
The key point is the Cauchy product formula for absolutely convergent series in a Banach algebra, proved in the Appendix below.

In the scalar case we know that \eqref{eq1mx} holds for all $x\in[-1,1]$ . Taking the square of the series we formally compute

$\begin{split} \left(1-\frac12 x-\frac{1}{8}x^2+\cdots\right) \left(1-\frac12 x-\frac{1}{8}x^2+\cdots\right)\\ = 1-\frac12 x-\frac12 x-\frac{1}{8}x^2-\frac{1}{8}x^2+\frac{1}{4}x^2+\cdots\\ =1-x+0\cdot x^2+\cdots. \end{split}$

Already the first two terms give the desired result $1-x$ , so the coefficients in front of all powers $x^n$ with $n\geq 2$ must cancel to zero. These coefficients are exactly those that appear when we form the Cauchy product of the series

$1-\sum_{n=1}^{\infty}c_n x^n$

with itself.

By the Cauchy product theorem in Banach algebras (see the Appendix), the same identities hold when we replace the scalar variable $x$ by an operator $A$ with $\|A\|\leq 1$ , provided the series is absolutely convergent. Therefore, for our operator series,

$B^2 = (I-\sum_{n=1}^{\infty}c_n A^n)^2 = I-A.$

This shows that $B$ is indeed a square root of $I-A$ .

Positivity of $B$

We must now show that $B\geq 0$ . To this end we notice that the coefficients in front of $A^n$ , for $n\geq 1$ , are all negative. Taking into account the fact that $A^n\leq I$ for all $n\geq 1$ , we obtain

(8) $\begin{equation*} B = I - \sum_{n=1}^{\infty} c_n A^n \geq I - \sum_{n=1}^{\infty} c_n I = I-I = 0. \end{equation*}$

Hence $B$ is self-adjoint and positive.

From $\sqrt{I-A}$ to $\sqrt{A}$

We now replace, in Eq.\ \eqref{eqsqrA1}, the operator $A$ by $I-A$ , and thus $I-A$ by $A$ .
Notice that $A$ is self-adjoint if and only if $I-A$ is self-adjoint, and that $0\leq A\leq I$ if and only if $0\leq I-A\leq I$ . We obtain

(9) $\begin{equation} \sqrt{A} = I - \sum_{n=1}^{\infty} c_n (I-A)^n. \end{equation}$

This $B:=\sqrt{A}$ is again self-adjoint and positive, and satisfies $B^2=A$ when $0\leq A\leq I$ , hence in general, by the scaling reduction described earlier

(10) $\begin{equation} \sqrt{A} = \Vert A\Vert^{1/2}\left(I - \sum_{n=1}^{\infty} c_n (I-\frac{A}{\Vert A \Vert})^n\right). \end{equation}$

Commutation property

We now consider the \emph{commutation property} announced in the theorem.
It follows immediately from the construction above. The series in \eqref{eqsqrA2} is absolutely convergent. Therefore any bounded linear operator $T$ (or, if $H$ is complex, since the coefficients of the series are all real, also any bounded antilinear operator) that commutes with $A$ commutes also with all partial sums of the series, and therefore, by continuity of the operator product, also with $\sqrt{A}$ defined in \eqref{eqsqrA2}. Thus $T\sqrt{A}=\sqrt{A}\,T$ whenever $TA=AT$ .

Uniqueness

Finally we show \emph{uniqueness}. Let $B=\sqrt{A}$ be defined as in Eq.\ \eqref{eqsqrA2}, and assume that $B'$ is another bounded self-adjoint positive operator with ${B'}^{\,2}=A$ .

First we show that $B'$ commutes with $B$ . We have $B'A=B'B'^2=(B'^2)B'=AB'$ , so $B'$ commutes with $A$ . We already know that $B$ commutes with every bounded operator commuting with $A$ . Therefore $B$ commutes with $B'$ .

Consider now the two operators

$C := (B-B')B(B-B'),\qquad D := (B-B')B'(B-B').$

Since $B,B'$ are self-adjoint positive, and $B-B'$ is self-adjoint, both $C$ and $D$ are self-adjoint and positive. Moreover, using the commutativity of $B$ and $B'$ , one sees that

$C = B(B-B')^2,\qquad D = B'(B-B')^2.$

We compute their sum:

$\begin{split} (B-B')B(B-B') + (B-B')B'(B-B')=\\ 2A(B-B')+2A(B'-B)=0. \end{split}$

Since $B^2=B'^2=A$ and $BB'=B'B$ , we get

$(B^2 + B'^2) = 2A,\qquad BB' = B'B,$

and a short algebraic computation shows that

$C+D = 2A(B-B')+2A(B'-B)=0.$

Thus $C+D=0$ .
But $C$ and $D$ are self-adjoint positive, so their sum being zero implies that each of them must be zero separately, therefore their difference, which is $(B-B')^3$ is also zero. If $(B-B')^3=0$ , then also $(B-B')^4=0$ . But $\Vert (B-B')^4\Vert=\Vert (B-B')^2\Vert^2=\Vert B-B'\Vert^4.$
It follows that $B'=B$ .

This completes the proof of the theorem.

Corollary 2. Let $A,B$ be two commuting bounded positive self-adjoint operators on $H$ . Then their product $AB$ is self-adjoint positive.

Proof. $AB$ is evidently self-adjoint. We have $AB=A\sqrt{B}\sqrt{B}=\sqrt{B}A\sqrt{B}.$ Since $\sqrt{B}$ is self-adjoint and $A$ is positive, $AB$ is also positive.

Appendix: Cauchy Product of Absolutely Convergent Series in
a Banach Algebra

For completeness, and to make the “scalar-to-operator” transition completely transparent, we include here a standard result on Cauchy products in Banach algebras.
The proof is formally identical to the classical scalar case once $|\cdot|$ is replaced by the norm $\|\cdot\|$ and one uses completeness and continuity of multiplication.

Theorem. Let $(A,\|\cdot\|)$ be a Banach algebra over $\mathbb{K}$ , where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ .
Let $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$ be sequences in $A$ such that the series

$\sum_{n=0}^{\infty} a_n \qquad\text{and}\qquad \sum_{n=0}^{\infty} b_n$

are absolutely convergent, i.e.

$\sum_{n=0}^{\infty} \|a_n\| < \infty, \qquad \sum_{n=0}^{\infty} \|b_n\| < \infty.$

Define the Cauchy product $(c_n)_{n\ge 0}$ by

$c_n := \sum_{k=0}^{n} a_k b_{n-k}, \qquad n \ge 0.$

Then the series $\sum_{n=0}^{\infty} c_n$ is absolutely convergent in $A$ , and

$\sum_{n=0}^{\infty} c_n = \left(\sum_{n=0}^{\infty} a_n\right) \left(\sum_{n=0}^{\infty} b_n\right)$

with multiplication in $A$ .

Proof. Let

$A_N := \sum_{n=0}^{N} a_n,\qquad B_N := \sum_{n=0}^{N} b_n,\qquad C_N := \sum_{n=0}^{N} c_n$

denote the partial sums.
By completeness of $A$ and absolute convergence, there exist $A,B \in A$ such that

$A_N \to A,\qquad B_N \to B$

as $N\to\infty$ .

We proceed in two steps.

Step 1: Absolute convergence of $\sum c_n$ .
For each $n\ge 0$ we have

$\|c_n\| = \left\| \sum_{k=0}^{n} a_k b_{n-k} \right\| \le \sum_{k=0}^{n} \|a_k b_{n-k}\| \le \sum_{k=0}^{n} \|a_k\|\,\|b_{n-k}\|,$

using the triangle inequality and submultiplicativity of the norm.
Hence for each $N\ge 0$ ,

$\sum_{n=0}^{N} \|c_n\| \le \sum_{n=0}^{N} \sum_{k=0}^{n} \|a_k\|\,\|b_{n-k}\| = \sum_{k+\ell \le N} \|a_k\|\,\|b_{\ell}\|,$

where we have reindexed by $\ell = n-k$ .

Since $\sum_{k=0}^{\infty}\|a_k\| < \infty$ and
$\sum_{\ell=0}^{\infty}\|b_{\ell}\| < \infty$ , we obtain the uniform bound

$\sum_{n=0}^{N} \|c_n\| \le \left(\sum_{k=0}^{\infty}\|a_k\|\right) \left(\sum_{\ell=0}^{\infty}\|b_{\ell}\|\right) < \infty \qquad \text{for all } N.$

Thus the sequence of partial sums
$\left(\sum_{n=0}^{N} \|c_n\|\right)_{N\ge 0}$ is increasing and bounded,
hence convergent in $\mathbb{R}$ . This is exactly absolute convergence of
$\sum_{n=0}^{\infty} c_n$ in $A$ .

Step 2: Identification of the sum.
We first observe the combinatorial identity

(11) $\begin{equation*} A_N B_N = \sum_{k=0}^{N} a_k \sum_{\ell=0}^{N} b_{\ell} = \sum_{n=0}^{2N} d_{n}^{(N)}, \end{equation*}$

where

$d_{n}^{(N)} := \sum_{\substack{0 \le k,\ell \le N \\ k+\ell = n}} a_k b_{\ell}.$

On the other hand, by definition of $c_n$ we have

$c_n = \sum_{k=0}^{n} a_k b_{n-k},$

where the sum is finite for every $n$ . Thus the family $\{a_k b_{\ell}\}_{k,\ell\ge 0}$ is absolutely summable, and the finite sums in \eqref{eq:finite-cauchy} can be reorganized in terms of the $c_n$ and “tail” terms.
More precisely, for each $N$ we can write

$A_N B_N = \sum_{n=0}^{N} c_n + R_N,$

where $R_N$ is the sum of all terms $a_k b_{\ell}$ with
$0 \le k,\ell \le N$ and $k+\ell > N$ .
Equivalently,

$C_N = A_N B_N - R_N.$

We claim that $\|R_N\|\to 0$ as $N\to\infty$ . Indeed, by the triangle inequality and submultiplicativity,

$\|R_N\| \le \sum_{\substack{0 \le k,\ell \le N \\ k+\ell > N}} \|a_k\|\,\|b_{\ell}\|.$

Let $\varepsilon > 0$ be given.
By absolute convergence, choose $K$ such that

$\sum_{k\ge K} \|a_k\| < \varepsilon, \qquad \sum_{\ell\ge K} \|b_{\ell}\| < \varepsilon.$

Then, for all $N \ge 2K$ , any pair $(k,\ell)$ with $0\le k,\ell\le N$ and $k+\ell > N$ must satisfy at least one of $k \ge K$ or $\ell \ge K$ .
Hence

$\|R_N\| \le \sum_{\substack{0 \le k,\ell \le N \\ k+\ell > N}} \|a_k\|\,\|b_{\ell}\| \le \sum_{k\ge K} \|a_k\| \sum_{\ell\ge 0} \|b_{\ell}\| + \sum_{\ell\ge K} \|b_{\ell}\| \sum_{k\ge 0} \|a_k\|.$

The right-hand side is bounded by

$\varepsilon \sum_{\ell\ge 0} \|b_{\ell}\| + \varepsilon \sum_{k\ge 0} \|a_k\| = \varepsilon \left( \sum_{\ell\ge 0} \|b_{\ell}\| + \sum_{k\ge 0} \|a_k\| \right),$

which is independent of $N$ . Thus $\|R_N\|$ can be made arbitrarily small by choosing $K$ (and then $N$ )
sufficiently large; in particular,

$\lim_{N\to\infty} \|R_N\| = 0.$

Since $A_N \to A$ and $B_N \to B$ , and multiplication is continuous in the
Banach algebra $A$ , we have $A_N B_N \to AB$ as $N\to\infty$ . Using $C_N = A_N B_N - R_N$ and $\|R_N\|\to 0$ , we obtain

$\lim_{N\to\infty} C_N = AB.$

But the limit of $(C_N)$ is by definition the sum of the absolutely convergent series
$\sum_{n=0}^{\infty} c_n$ .
This proves that

$\sum_{n=0}^{\infty} c_n = AB = \left(\sum_{n=0}^{\infty} a_n\right) \left(\sum_{n=0}^{\infty} b_n\right),$

as claimed.

Remark The proof is formally identical to the classical scalar case once
$|\cdot|$ is replaced by the norm $\|\cdot\|$ and one uses completeness
and continuity of multiplication in the Banach algebra.

Afternotes.
20-06-26 Added Corollary 2.

References.
\[1] Knapp, A.~W., Basic Real Analysis. Digital Second Edition.
[2] Reed, M., Simon, B., Methods of Modern Mathematical Physics, I: Functional Analysis, Academic Press, 1980.

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Square root lemma

Binomial series

Binomial series and its convergence