"Square root" of operator on a Banach space

Question

$\newcommand{\norm}[1]{\lVert #1\rVert}$Let $X$ be a Banach space and $A$ be a bounded linear operator on $X$ with $\norm{A}<1$. Show that there is a bounded linear operator $B$ on $X$ with $B^2=I+A$.

I'm working on this problem. I know I should try to define \begin{equation} B = I+ \sum_{n\geq 1} \frac{(-1)^{n-1}(2n)!}{(n!2^n)^2(2n-1)} A^n = \sum_{n\geq 0}a_n A^n, \end{equation} where the coefficients $a_n$ come from the power series expansion of $(1+x)^{1/2}$ at $0$ (assuming I haven't made some calculation mistakes). This sum converges in the space of bounded linear operators on $X$ because it's absolutely convergent (e.g., because $\norm{A}<1$ and $|a_n|$ is bounded). But I don't know how to show that $B^2 = I+A$.

If $B_k$ is the $(k+1)$th partial sum, then \begin{equation} \norm{B^2 - (I+A)}\leq \norm{B^2 - B_k^2}+\norm{B_k^2 - (I+A)}. \end{equation} The first norm is $\norm{B-B_k}\norm{B+B_k}$ since $B$ commutes with $B_k$; this goes to $0$ because the first factor goes to $0$ and the second factor is bounded. But I can't figure out what to do with the second norm in the sum above.

Intuitively, the partial sums of the power series of $(1+x)^{1/2}$ converge absolutely to $(1+x)^{1/2}$ for $|x|<1$, so their squares converge to $1+x$. But how can I make this rigorous with $A$ in place of $x$?

Answer 1

Consider the binomial series for the principal square root: $$(1 + z)^{1/2} = \sum_{n=0}^\infty a_n z^n, \quad a_n = \binom{1/2}{n},$$ which converges absolutely for $|z| < 1$. Define $$B = \sum_{n=0}^\infty a_n A^n.$$ Let $r = \|A\| < 1$. The series converges absolutely in the operator norm because $S(r) := \sum_{n=0}^\infty |a_n| r^n < \infty$ (the radius of convergence is 1). To verify $B^2 = I + A$, compute the square via the Cauchy product: $$B^2 = \left( \sum_{n=0}^\infty a_n A^n \right) \left( \sum_{m=0}^\infty a_m A^m \right) = \sum_{k=0}^\infty \left( \sum_{i+j=k} a_i a_j \right) A^k.$$ The double series $\sum_{i,j=0}^\infty |a_i| |a_j| r^{i+j} = S(r)^2 < \infty$ converges absolutely in the Banach algebra $\mathcal{L}(X)$, justifying the rearrangement by powers of $A$ (Tonelli for series). The coefficients satisfy $\sum_{i+j=k} a_i a_j = \binom{1}{k}$ (the coefficient of $z^k$ in $(1 + z)^{1/2} \cdot (1 + z)^{1/2} = 1 + z$) by the generalized Vandermonde identity: $$\sum_{i=0}^k \binom{1/2}{i} \binom{1/2}{k-i} = \binom{1}{k} = \begin{cases} 1 & k=0, \\ 1 & k=1, \\ 0 & k \geq 2, \end{cases}$$ matching the expansion of $1 + z$. Thus, $B^2 = I + A$.

Alternatively, via partial sums: Let $B_N = \sum_{n=0}^N a_n A^n$ and $T_N = \sum_{n=N+1}^\infty |a_n| r^n$, so $S(r) = S_N + T_N < \infty$ with $S_N = \sum_{n=0}^N |a_n| r^n$. Then $$B^2 - (I + A) = (B - B_N) B + B_N (B - B_N),$$ so $$\|B^2 - (I + A)\| \leq \|B - B_N\| (\|B\| + \|B_N\|) \leq 2 S(r) T_N \to 0 \quad \text{as } N \to \infty$$ (since $\|B - B_N\| \leq T_N$, $\|B\| \leq S(r)$, and $\|B_N\| \leq S(r)$), which yields $B^2 = I + A$.

For an explicit rate, note that $|a_n| \sim C n^{-3/2}$ for some $C > 0$, yielding $T_N \lesssim C' r^{N+1} / \sqrt{N}$ for some $C' > 0$, so $$\|B^2 - (I + A)\| \lesssim \frac{C'' r^{N+1}}{\sqrt{N}}$$ for some $C'' > 0$.

If $f(z) = \sum c_n z^n$ has radius $R > \|A\|$, then $f(A) = \sum c_n A^n$ converges, and products follow from Cauchy convolution. Here, $((1 + z)^{1/2})^2 = 1 + z$ implies $B^2 = I + A$.

If the spectral radius $\rho(A) < 1$ (not necessarily $\|A\| < 1$), then the binomial series defines $B = (I + A)^{1/2}$ in norm and $B^2 = I + A$. By Gelfand's formula $\rho(A) = \lim_{n \to \infty} \|A^n\|^{1/n}$, pick any $c$ with $\rho(A) < c < 1$; then there exists $N$ such that $\|A^n\|^{1/n} \leq c$ for all $n \geq N$, hence $\|A^n\| \leq c^n$ for $n \geq N$. Set $M := \max_{0 \leq n < N} \|A^n\| / c^n$; then $\|A^n\| \leq M c^n$ for all $n \geq 0$. This gives the norm convergence of $\sum a_n A^n$ immediately, and the same Cauchy-product argument gives $B^2 = I + A$.

Moreover, since $\rho(A) < 1$ implies $\sigma(I + A) \subset \{z : \operatorname{Re} z > 0\}$, there is a unique $B$ with $\sigma(B) \subset \{ \operatorname{Re} z > 0\}$ and $B = I + O(A)$ such that $B^2 = I + A$; the series defines exactly this principal square root. $B$ also commutes with $A$.

Newton's method:

$$B_{k+1} = \frac{1}{2} \left( B_k + (I + A) B_k^{-1} \right), \quad B_0 = I,$$ is well-defined—if $\|A\| < 1$ and $B_0 = I$, then $\|B_k - I\|$ stays small; whenever $\|B_k - I\| < 1$ we have $B_k^{-1} = \sum_{n \geq 0} (I - B_k)^n$ (Neumann series)—and converges quadratically in norm to the same $B$.

Answer 2

Are these finite-dimensional Banach spaces or just Banach spaces in general? For Banach algebras in general, the full holomorphic functional calculus would need to be constructed. Here's an exposition on a finite-dimensional variant of a holomorphic functional calculus that comes from Positive Operator Semigroups: From Finite to Infinite-dimensional by Bátkai, Fijavž, and Rhandi.

To rigorously define $f(A)$ where $A\in\mathcal{L}(X)$ on a Banach space $X$ (i.e. $X=\mathbb{C}^n$ with its norms), you begin by defining polynomials of $A\in\mathcal{L}(X)$ in the following way:

We denote the set of all polynomials of $A\in\mathcal{L}(X)$ by $\mathcal{P}_A$ given by \begin{equation*} \mathcal{P}_A :=\{\sum_{n=0}^m\alpha_nA^n: \alpha_n\in\mathbb{C},m\in\mathbb{N}_0\}\subseteq\mathcal{L}(X), \end{equation*} and if $p(x) = \sum_{n=0}^m\alpha_nx^n$ is a polynomial in $\mathbb{C}[x]$, we write \begin{equation*} p(A) := \sum_{n=0}^m\alpha_nA^n \end{equation*} and say that the operator $p(A)\in\mathcal{L}(X)$ is obtained by evaluating $p$ at A.

With this definition in mind, we seek to define $f(A)$ by restricting $f$ to be smooth functions in an open set that contains the spectrum of $A$. In other words,

Let $A\in\mathcal{L}(X)$ be an operator on the finite-dimensional vector space $X$ with spectrum $\sigma(A):=\{\lambda\in\mathbb{C}: Ax=\lambda x \}$ and respective multiplicities $\{\nu_i:1\leq i\leq n\}$. Then we define the set $C_A^\infty$ by \begin{equation*} C_A^\infty := \{f:D(f)\to\mathbb{C}: \exists U\subset D(f) \text{ open},\,\sigma(A)\subset U,\, f_{|_U}\in C^\infty\} \end{equation*} Then to define the functional calculus, let $f\in C_A^\infty$. Then for $A\in\mathcal{L}(X)$, we define \begin{equation*} f(A) :=\Phi_A(p_f) \equiv p_f(A) \end{equation*} where $p_f$ is an interpolation polynomial in the sense that \begin{equation*} f^{(\nu)}(\lambda_i)=p^{(\nu)}_f(\lambda_i) \end{equation*} for $i=1,\dots,m$ and $\nu=0,\dots,\nu_i-1.$ In other words, if you can create a polynomial $p_f(A)$ that matches the derivatives of the function $f\in C^\infty_A$ when evaluated on its spectrum, then you can define $f(A)$ to be the polynomial. This construction admits the following characterization:

Let $A\in\mathcal{L}(X)$ with eigenvalues $\{\lambda_i:1\leq i\leq m\}$, respective multiplicities $\{\nu_i:1\leq i\leq m\}$, and define projections $P_i$ such that \begin{equation*} P_i := \chi_{U_i}(A) \end{equation*} where $\chi_{U_i}(z)$ is the indicator function on $U_i$, open subsets of $\mathbb{C}$ that contain the corresponding eigenvalues $\lambda_i$.

Then for every $f\in C^\infty_A$, we have that \begin{equation*} f(A) := \sum_{i=1}^m\sum_{\nu=0}^{\nu_i-1}\frac{f^{(\nu)}(\lambda_i)}{\nu!}(A-\lambda_i)^\nu P_i \end{equation*}