Show that the operator sequence $ A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ converges strongly, $A_0 = I+T$, where $T$ is compact and $||T|| \le 1/2$. I'm studying for an analysis prelim and am stumped on an old exam problem for which there are no solutions given.
The full question is as follows: Let $X$ denote a Hilbert space, and $T$ a self-adjoint compact operator such that $||T|| \le 1/2$. Define $A_0 = I + T$, and set $A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ for $n \ge 1$. Prove that the sequence $(A_n)_{n=1}^{\infty}$ converges strongly.  
I can show that the sequence is well-defined (all the necessary inverses exist) and that each $A_n$ is self-adjoint. I'm unsure where to go from there, though. It seems likely that the operator the sequence converges to is the identity operator. I don't know how to show this, though; in particular, I don't see the implications of the compactness and self-adjointness of $T$. Two facts that may (?) be helpful that I know would be fair game for use in the proof: Compact operators (on a Hilbert space) map weakly convergent sequences into strongly convergent sequences, and the spectral representation theorem for compact, self-adjoint operators. (Don't know how the spectral theorem would be helpful, but seeing the words "self-adjoint" and "compact" triggered it as a possibility in my mind).  
Thanks in advance!
 A: Hint: Define $K = \{S:X \to X \mid S = S^* \text{ and } \|S - I\| \leq \frac 12 \}$.
Define the map $\phi:K \to K$ by $\phi: A \mapsto \frac 12 (A + A^{-1})$.  Define $f: \Bbb R \to \Bbb R$ by $f(x) = \frac 12 (x + \frac 1x)$.  Note that 
$$
\sigma(\phi(A)) = f(\sigma(A))
$$
Note that $f$ is a contractive mapping on $[1/2,3/2]$, since $|f'(x)| < 1$ over the interval $[1/2,3/2]$. Note also that $f$ has the unique fixed point $x = 1$.

More thorough answer: (using compactness)
Let $N$ be the kernel of $T$, and let $\{e_n\}$ be an orthonormal basis of the complement consisting of eigenvectors of $T$.  For any $n$, we have
$$
\begin{cases}
A_n(x) = x & x \in N\\
A_n(e_k) = \lambda_k^{(n)} e_n
\end{cases}
$$
Moreover, we have $\lambda_k^{(n+1)} = f(\lambda_k^{(n)})$.  So, we have
$$
\|A_n - I\| = \sup_{k \in \Bbb N} |\lambda_k^{(n)} - 1| = 
\max_{k \in \Bbb N} |\lambda_k^{(n)} - 1|
$$
It suffices to show that this sequence approaches zero.
A: Using the spectral theorem, you can assume without loss of generality that $T$ is a diagonal matrix. Let the $\{\lambda_n\}$ be the diagonal values. Then $A$ is diagonal with diagonal values $\{1+\lambda_n\}$. Since $\|T\|<1/2$, you know $|\lambda_n|<1/2$ for all $n$. 
Now you've reduced it to a system of decoupled single-variable limit problems. The map $x\mapsto \frac{1}{2}(x+1/x)$ is contractive on the necessary range, so all these limits should tend to $1$ uniformly. (See, for example, the contraction mapping theorem for details.)
A: This is a classic algorithm used to find the square root of a number. Suppose for example that you want to find $\sqrt{2}$. Then you start with a guess $x_0 > 0$, divide it into $2$ and average with the original $x$, i.e., $x_1 = \frac{1}{2}(x_0+2/x_0)$. In general, $x_{n+1}=\frac{1}{2}(x_{n}+2/x_{n})$. Such an algorithm has very good convergence properties.
Your algorithm is the special case of finding $\sqrt{1}=1$. So I would expect $A_{n}$ to converge strongly to $I$ because all you have to do is look at the non-zero eigenvalues of $I+T$ and apply the algorithm to the non-zero eigenvalues because $A_{n}=I$ on $\mathcal{N}(T)$. Start by noticing that
$$
              x_{n+1}-1= \frac{1}{2}(x_{n}+1/x_{n})-1 = \frac{2}{x_n}(x_{n}-1)^{2}.
$$
You end up with quadratic convergence of $\{ x_{n} \}$ to $1$; i.e., if $x_{n}-1 \approx 0.001$, then $x_{n+1}-1 \approx 0.000001$. I don't see why you won't get uniform operator topology convergence to $I$ because $\|T\| \le 1/2$.
https://en.wikipedia.org/wiki/Methods_of_computing_square_roots
