1
$\begingroup$

$H$ is a Hilbert space, $M$ is a self-adjoint bounded linear operator on $H$ with $M \leq I$, and $S_0 = 0$; $S_{n+1} = (1/2)(M + S^2_n)$ for $n = 0, 1, 2, ...$. For all $n$, both $S_n$ and $S_n - S_{n-1}$ are polynomials in $M$ with nonnegative, real coefficients, $S_n \geq 0$ and $S_n - S_{n-1} \geq 0$, and $||S_n|| \leq 1$. Show that for each $x$ in $H$, the sequence $\lbrace S_n x \rbrace$ converges.

This is a piece of a problem whose goal is to find a "square root" for positive self-adjoint operators, and many of the facts above come from earlier parts of the problem. I'm just not quite sure what approach to take. I know that since $||S_n|| \leq 1$, $||S_n x|| \leq ||x||$ for every $x$ in $H$, so the sequence is bounded, so it has a convergent subsequence $\lbrace S_{n_k} \rbrace$, say with limit $S$. I'm having trouble coming up with a good bound on $||Sx - S_n x||$ at this point to get the rest of the convergence, or else simply bounding $||S_m x- S_n x||$ to show the sequence is Cauchy.

I would just like a hint on which of the several facts about the sequence should be used, and how, to get the rest of the way to the sequence converging.

$\endgroup$
1
$\begingroup$

We have to show that if $(U_n)_{n\geqslant 1}$ is a sequence of self-adjoint operators such that $\lVert U_n\rVert\leqslant 1$ for each $n$ and $U_n-U_{n-1}$ is positive for each $n$, then $(U_nx)_{n\geqslant 1}$ is (strongly) convergent for each $x$.

We start from a Cauchy-Schwarz type inequality: $$|\langle Ax,y\rangle|^2\leqslant \langle Ax,x\rangle\cdot \langle Ay,y\rangle$$ for each $x,y\in H$ where $A$ is a non-negative bounded self-adjoint operator. We then use this inequality with $A:=U_{m+n}-U_n$, some fixed $x$ and $y:=U_{m+n}x-U_nx$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you please explain this further? I'm working on the same problem, but I don't know how to proceed. $\endgroup$ – HCS Nov 22 '19 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.