Self-Adjoint Operators: Monotone Convergence Theorem, Existence of Square Roots

One of the many irons in the fire that is my ongoing mathematical learning is a review of functional analysis motivated by my study of probability theory and stochastic processes. As I’ve mentioned before, I’ve been using the great text by Adam Bobrowski. In his discussion of discrete-time martingales, Bobrowski introduces a convergence theorem for monotone sequences of nonnegative (sometimes called positive semi-definite) operators. I would to like to give an exposition of the theorem, including a full proof. I would then like to give an application of this theorem, proving the existence of nonnegative square root for a nonnegative self-adjoint operator.

We say that a self-adjoint operator A: \mathbb{H}\rightarrow\mathbb{H} is nonnegative if \langle{Ax,x}\rangle \geq 0 for all x\in\mathbb{H}; we write A \geq 0. If A,B are two self-adjoint operators such that A-B\geq 0, then we write A\geq B.

Lemma 1. If A is nonnegative, then A^{n} is nonnegative for any n\in\mathbb{Z}^{\geq 0}.

Proof. If n is even, then

\displaystyle\langle{A^{n}x,x}\rangle=\langle{A^{\frac{n}{2}}x,A^{\frac{n}{2}}x}\rangle\geq 0,\indent\forall x\in\mathbb{H}

Similarly, if n is odd and n > 1, then

\displaystyle\langle{A^{n}x,x}\rangle=\langle{A^{\frac{n-1}{2}}A^{\frac{n-1}{2}+1}x,x}\rangle=\langle{A(A^{\frac{n-1}{2}}x),A^{\frac{n-1}{2}}x}\rangle\geq 0,\indent\forall x\in\mathbb{H}

\Box

Lemma 2. (Nonnegative Operator Inequality) Let A be a nonnegative, self-adjoint operator on a Hilbert space \mathbb{H}. Then \langle{A^{2}x,x}\rangle\leq\left\|A\right\|\langle{Ax,x}\rangle for all x\in\mathbb{H}.

Proof. If \left\|A\right\|=0, then there is nothing to prove, so assume otherwise. Observe that proving the inequality above is equivalent to showing \langle{B^{2}x,x}\rangle\leq\langle{Bx,x}\rangle, where B:=\frac{1}{\left\|A\right\|}A. Since A is nonnegative, \langle{Bx,x}\rangle\geq 0. Observe that I-B is self-adjoint, being the difference of self-adjoint operators. By Cauchy-Schwarz,

\displaystyle\langle{(I-B)x,x}\rangle=\left\|x\right\|^{2}-\langle{Bx,x}\rangle \geq 0,\indent\forall x\in\mathbb{H}

Since B-B^{2}=B(I-B)B+(I-B)B(I-B), we see that

\begin{array}{lcl}\displaystyle\langle{(B-B^{2})x,x}\rangle&=&\displaystyle\langle{B(I-B)Bx,x}\rangle+\langle{(I-B)B(I-B)x,x}\rangle\\&=&\displaystyle\langle{(I-B)Bx,Bx}\rangle+\langle{B(I-B)x,(I-B)x}\rangle\\&\geq&\displaystyle 0\end{array},

which completes the proof. \Box

We now prove a convergence theorem for monotone sequences of self-adjoint operators.

Theorem 3. Suppose (A_{n}) is a sequence of self-adjoint operators on a Hilbert space \mathbb{H} such that A_{n}\leq A_{n+1}\leq MI, for all n \in \mathbb{Z}^{\geq 0}, where M \in\mathbb{R} is some constant. Then there exists a self-adjoint operator A: \mathbb{H} \rightarrow\mathbb{H} such that

\displaystyle Ax=\lim_{n\rightarrow\infty}A_{n}x,\indent\forall x\in\mathbb{H}

In other words, A_{n} converges to A strongly (but not necessarily in the operator norm).

Proof. For any x\in \mathbb{H}, the real sequence (\langle{A_{n}x,x}\rangle)_{n=1}^{\infty} is nondecreasing and bounded from above by M\left\|x\right\|^{2}, hence converges to a real number F(x). For any x,y\in\mathbb{H}, the polarization identity lets us write

\displaystyle\langle{A_{n}x,y}\rangle=\dfrac{1}{4}\left[\langle{A_{n}(x+y),x+y}\rangle-\langle{A_{n}(x-y),x-y}\rangle+i\langle{A_{n}(x+iy),x+iy}\rangle-i\langle{A_{n}(x-iy),x-iy}\rangle\right]

so the limit G(x,y) := \lim_{n \rightarrow\infty}\langle{A_{n}x,y}\rangle exists.

For any n\in\mathbb{Z}^{\geq 1}, we have by Cauchy-Schwarz that

\displaystyle -\left\|A_{1}\right\|\left\|x\right\|^{2}\leq\langle{A_{n}x,x}\rangle\leq M\left\|x\right\|^{2},\indent\forall x\in\mathbb{H}

so by Lemma 2, \left\|A_{n}\right\|\leq M':=\max\left\{M,\left\|A_{1}\right\|\right\}. By another application of Cauchy-Schwarz, we see that

\displaystyle\left|G(x,y)\right|\leq M'\left\|x\right\|\left\|y\right\|,\indent\forall x,y\in\mathbb{H}

Hence, for x fixed, \overline{G(x,y)} is a bounded complex-linear functional with respect to y\in\mathbb{H}. By the Riesz representation theorem, there exists a unique element Ax\in\mathbb{H} such that

\displaystyle\langle{y,Ax}\rangle=\overline{G(x,y)} \Longleftrightarrow G(x,y)=\overline{\langle{y,Ax}\rangle}=\langle{Ax,y}\rangle,\indent\forall y\in\mathbb{H}

A is complex-linear, since

\begin{array}{lcl}\displaystyle\langle{A(\alpha x+\beta x'),y}\rangle=G(\alpha x+\beta x',y)=\lim_{n\rightarrow\infty}\langle{A_{n}(\alpha x+\beta x'),y}\rangle&=&\displaystyle\lim_{n\rightarrow\infty}\alpha\langle{A_{n}x,y}\rangle+\beta\langle{A_{n}x',y}\rangle\\&=&\displaystyle\alpha G(x,y)+\beta G(x',y)\\&=&\displaystyle\langle{\alpha Ax+\beta Ax',y}\rangle\end{array}

for all y\in\mathbb{H}. Cauchy-Schwarz implies that \left\|Ax\right\|\leq M'\left\|x\right\|, hence \left\|A\right\|\leq M'. A is self-adjoint, since A_{n} is self-adjoint for all n and

\displaystyle\langle{Ax,y}\rangle=\lim_{n\rightarrow\infty}\langle{A_{n}x,y}\rangle=\lim_{n\rightarrow\infty}\langle{x,A_{n}y}\rangle=\langle{x,Ay}\rangle

for all x,y\in\mathbb{H}. Being the difference of two self-adjoint operators, A-A_{n} is self-adjoint and by Lemma 2,

\displaystyle\left\|(A-A_{n})x\right\|^{2}=\langle{(A-A_{n})^{2}x,x}\rangle\leq\left\|A-A_{n}\right\|\langle{(A-A_{n})x,x}\rangle\leq 2M'\langle{(A-A_{n})x,x}\rangle

This last quantity converges to 0 as n\rightarrow\infty, for x fixed, which completes the proof. \Box

The preceding convergence theorem also applies to decreasing sequences of self-adjoint operators. If (A_{n})_{n=1}^{\infty} is a sequence of self-adjoint operators such that \langle{A_{n}x,x}\rangle \geq \langle{A_{n+1}x,x}\rangle \geq M\left\|x\right\|^{2}, for some constant M, then the A_{n} converge strongly to a self-adjoint operator A. This result follows immediately from applying the preceding theorem to the sequence of operators B_{n}:=-A_{n}.

We use the preceding convergence theorem for self-adjoint operators to prove the existence of a square root of self-adjoint operators A.

Theorem 4. If A is a self-adjoint operator such that \langle{Ax,x}\rangle \geq 0 for all x\geq 0, then there exists a self-adjoint operator B such that B^{2}=A, and B commutes with all operators that commute with A.

Proof. Without loss of generality, we may assume that 0 \leq A \leq I. Otherwise, A=0, in which case the theorem is obvious, or the operator A':=\frac{1}{\left\|A\right\|}A satisfies 0\leq A'\leq I and B= \sqrt{\left\|A\right\|}B', where {B'}^{2}=A'. Set C:=I-A, and observe that 0\leq C\leq I. We will find B by an interative argument.

Define a sequence of operators inductively by A_{0}:=0 and A_{n+1}:=\dfrac{1}{2}(C+A_{n}^{2}), for n \geq 1. The intuition behind this recursive inductive definition is that if A_{\infty}=\lim_{n\rightarrow\infty}A_{n} exists (in the strong operator topology), then for any x\in\mathbb{H},

\displaystyle A_{\infty}x=\lim_{n\rightarrow\infty}\dfrac{1}{2}\left(C+A_{n}^{2}\right)x=\dfrac{1}{2}\left(C+A_{\infty}^{2}\right)x,

so that

\displaystyle Ax=A_{\infty}^{2}x-2A_{\infty}x+Ix=(I-A_{\infty})^{2}x

Thus, B=I-A_{\infty} is a positive square root of A.

Induction shows that each A_{n} is self-adjoint, nonnegative, and commutes with A. Furthermore, induction shows that A_{n}+A_{n-1} is polynomial in C with positive coefficients and the observation

\begin{array}{lcl}\displaystyle A_{n+1}-A_{n}=\dfrac{1}{2}\left(C+A_{n}^{2}\right)-\dfrac{1}{2}\left(C+A_{n-1}^{2}\right)&=&\displaystyle\dfrac{1}{2}A_{n}^{2}-A_{n-1}^{2}\\&=&\displaystyle\dfrac{1}{2}\left(A_{n}-A_{n-1}\right)\left(A_{n}+A_{n-1}\right)\end{array},

so that A_{n+1}-A_{n} is polynomial in C with positive coefficients. Hence, A_{n+1}\geq A_{n}, for every n\geq 0. I claim that A_{n}\leq I, for each n\geq 0. Clearly, A_{0} \leq 0. If A_{n}\leq I, then self-adjoint implies that \left\|A_{n}\right\|\leq 1, so that A_{n}^{2}\leq \left\|A_{n}\right\|A_{n}\leq A_{n}. Hence,

\displaystyle A_{n+1}=\dfrac{1}{2}\left(C+A_{n}^{2}\right)\leq\dfrac{1}{2}\left(C+A_{n}\right)\leq I

By the monotone convergence theorem, there exists a strong limit A_{\infty}. As shown above, B^{2}=A, where B=I-A_{\infty}, and since A_{n} commutes with A_{n} for each n, taking the limit shows that A_{\infty}, and therefore B, commutes with A as well. \Box

We now derive some useful corollaries from Theorem 4, which have applications to the convergence of discrete-time martingales.

Corollary 5. Suppose \left\{\mathbb{H}_{n}\right\}_{n=1}^{\infty} is an increasing sequence of closed subspaces of a Hilbert space \mathbb{H}. Then the projections P_{n} onto \mathbb{H}_{n} converge strongly to the projection P_{\infty} onto the closed subspace \mathbb{H}_{\infty}:=\overline{\bigcup_{n\geq 1}\mathbb{H}_{n}}.

If \mathbb{H}=L^{2}(\Omega,\mathcal{F},\mathbb{P}) and \mathbb{H}_{n}=L^{2}(\Omega,\mathcal{F}_{n},\mathbb{P}), where \left\{\mathcal{F}_{n}\right\}_{n=1}^{\infty} is a filtration, then \mathbb{H}_{\infty}=L^{2}(\Omega,\mathcal{F}_{\infty},\mathbb{P}), where \mathcal{F}_{\infty}=\sigma\left(\bigcup_{n\geq 1}\mathcal{F}_{n}\right).

Proof. I first claim that P_{n}\leq I. Indeed,

\displaystyle\langle{P_{n}x,x}\rangle=\langle{P_{n}x,(x-P_{n}x)+P_{n}x}\rangle=\left\|P_{n}x\right\|^{2}+\left\|x-P_{n}x\right\|^{2}=\left\|x\right\|^{2}

By the monotone convergence theorem for self-adjoint adjoint operators, P_{n} converges strongly to an operator A_{\infty}:\mathbb{H}\rightarrow\mathbb{H}. We now show that A_{\infty} is the projection operator P_{\infty}.

Since P_{n}^{2}=P_{n}, \left\|P_{n}\right\|\leq 1 and therefore

\displaystyle\left\|P_{n}^{2}x-P_{n}A_{\infty}x\right\|\leq\left\|P_{n}x-A_{\infty}x\right\|\rightarrow 0,n\rightarrow\infty

we see that A_{\infty}x=\lim_{n\rightarrow\infty}P_{n}x=\lim_{n\rightarrow\infty}P_{n}A_{\infty}x=A_{\infty}^{2}x. Denote the range of A_{\infty} by \tilde{\mathbb{H}}_{\infty}. Since A_{\infty}^{2}=A_{\infty}, we see that

\displaystyle\lim_{n\rightarrow\infty}A_{\infty}x_{n}=y \Longrightarrow A_{\infty}y=\lim_{n\rightarrow\infty}A_{\infty}^{2}x_{n}=\lim_{n\rightarrow\infty}A_{\infty}x_{n}=y

and therefore \tilde{\mathbb{H}}_{\infty} is closed. Since A_{\infty} is self-adjoint, A_{\infty} is the orthogonal projection onto the subspace \tilde{\mathbb{H}}_{\infty}. We need to show that \tilde{\mathbb{H}}_{\infty}=\mathbb{H}_{\infty}. I claim that \mathbb{H}_{n}\subset\tilde{\mathbb{H}}_{\infty} for each n\geq 1. Indeed, for x\in \mathbb{H}_{n} and m \geq n, P_{m}x=P_{n}x\in\mathbb{H}_{n}, so A_{\infty}x=P_{n}x\in\tilde{\mathbb{H}}_{\infty}, since the latter space is closed. By definition of the closure, \mathbb{H}_{\infty}\subset\tilde{\mathbb{H}}_{\infty}. The reverse inclusion follows from observing that x=\lim_{n\rightarrow\infty}P_{n}x, where P_{n}x\in\mathbb{H}_{n}.

We know that \mathbb{H}_{\infty}\subset L^{2}(\Omega,\mathcal{F}_{\infty},\mathbb{P}) since the pointwise limit of \mathcal{F}_{n}-measurable functions is \mathcal{F}_{\infty}-measurable. For the reverse inclusion, I claim that \mathcal{G}:=\left\{A\in\mathcal{F}:\mathbf{1}_{A}\in\mathbb{H}_{\infty}\right\} is a Dynkin system. Indeed, \mathbf{1}_{\Omega}\in\mathbb{H}_{\infty}. If A\in\mathcal{G}, then

\displaystyle\mathbf{1}_{A^{c}}=\mathbf{1}_{\Omega}-\mathbf{1}_{A}\in\mathbb{H}_{\infty}\Longrightarrow A^{c}\in\mathcal{G}

If \left\{A_{n}\right\}_{n=1}^{\infty}\subset\mathcal{G} is a countable collection of pairwise disjoint sets, then by the dominated convergence theorem \mathbf{1}_{A}=\sum_{n=1}^{\infty}\mathbf{1}_{A_{n}}\in \mathbb{H}_{\infty}, where A:=\bigcup_{n=1}^{\infty}A_{n}. \mathbb{H}_{n}\subset \mathbb{H}_{\infty}, for each n\geq 1, so that \bigcup_{n=1}^{\infty}\mathcal{F}_{n}\subset\mathcal{G}. Since \bigcup_{n=1}^{\infty}\mathcal{F}_{n} is closed under finite intersection, by Dynkin’s \pi\lambda lemma (see my notes), \mathcal{F}_{\infty}\subset\mathcal{G}, which implies that L^{2}(\Omega,\mathcal{F}_{\infty},\mathbb{P})\subset\mathbb{H}_{\infty}. \Box

Corollary 6. Let (\mathcal{F}_{n})_{n=1}^{\infty} be a filtration of a probability space (\Omega,\mathcal{F},\mathbb{P}). For X\in L^{1}(\Omega,\mathcal{F},\mathbb{P}), the sequence defined by X_{n}:=\mathbb{E}[X\mid\mathcal{F}_{n}] converges in L^{1}-norm to X_{\infty}:=\mathbb{E}[X\mid\mathcal{F}_{\infty}], where \mathcal{F}_{\infty}:=\sigma\left(\bigcup_{n=1}^{\infty}\mathcal{F}_{n}\right).

Proof. Suppose X\in L^{2}(\Omega,\mathcal{F}_{\infty}). Then by Hölder’s inequality,

\displaystyle\left\|X_{n}-X_{\infty}\right\|_{L^{1}}\leq\left\|X_{n}-X_{\infty}\right\|_{L^{2}} \rightarrow 0,

by Corollary 5. Since L^{2}(\Omega,\mathcal{F},\mathbb{P}) is dense in L^{1}(\Omega,\mathcal{F},\mathbb{P}) and conditional expectation is a Markov operator with norm 1, it follows that convergence holds on the entirety of L^{1}(\Omega,\mathcal{F},\mathbb{P}) (this is a standard 3\epsilon-type argument, which I encourage the confused reader to work out). \Box

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3 Responses to Self-Adjoint Operators: Monotone Convergence Theorem, Existence of Square Roots

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