Here is a simple problem that has stumped me for some time; sharing with the community, as I suspect it has been solved somewhere, or is immediately implied by the correct theorem.
Let $\textbf{diag}: \mathbb{R}^n_+ \to \mathbb{R}^{n \times n}$ represent the operation of taking a vector $v$ and constructing a matrix with $v$ on the diagonal. We'll assume we only operate on vectors with positive elements (this is the meaning of the notation $\mathbb{R}^n_+$), so that the resulting matrix is in particular positive definite.
Let $\textbf{diagpart}: \mathbb{R}^{n \times n} \to \mathbb{R}^n$ represent the operation of extracting the diagonal elements of a matrix as a vector. Let $A$ be some symmetric positive-definite matrix--from the application which serves as the source of this problem, we may take $A$ to have all elements at least 1 if desired, but I think this is likely unnecessary.
Define $ \phi: \mathbb{R}^n_+ \to \mathbb{R}^n_+$ by
$$ \phi(v) = \textbf{diagpart}\left(\sqrt{\textbf{diag}(v)^{1/2} A \textbf{diag}(v)^{1/2}}\right) $$
Conjecture: $\phi$ has a unique fixed point on $\mathbb{R}^n_+$, and iterates of $\phi$ converge to this fixed point from any initial vector $v_0$ (having all positive elements of course).
Numerically, this convergence seems to hold independently of both initial point and $A$, and be extremely rapid--I'll post some code presently, but I suspect this audience may not be particularly interested in empirics.
Is a result like this (or implying this) known? I'll put a little commentary with sketches of solution attempts below.
Commentary and attempted solutions:
First, notice that in the one-dimensional case this essentially reduces to iterating the map $x \to \alpha\sqrt{x}$ for $\alpha > 0$. In fact, if $A$ is diagonal and positive definite, the iterations above reduce to iterating such a mapping elementwise. These parallel iterations can be shown to converge to their unique fixed point by a number of techniques, each of those I've tried becoming somewhat problematic in higher dimensions:
Considering $\widetilde{\phi} = \gamma \phi$ restricted away from 0 and using the Banach fixed-point theorem. In 1d, this comes down to balancing the values of the derivative against the values of the function itself (to ensure that $\widetilde{\phi}$ maps the appropriate restricted space to itself). This balancing can be done, but relies heavily the 2 in the denominator of $\frac{d}{dx} \sqrt{x}$. I've taken this to mean any matrix-calculus approach going through this strategy must be extremely precise--losing even a factor of 2 in the analysis would ruin the approach. It may be possible to pursue this approach to the end and prove the conjecture, though the calculations are a little hairy.
Using monotonicity. One of the simplest ways of showing convergence in 1D is just noting the monotonicity of the mapping--go up if you're less than 1 and down if you're bigger. In conjunction with boundedness, this monotonicity at least gives us convergence. However, I have yet to determine a basis in which the action described by $\phi$ is monotonic in any particular sense (besides seeming to always make progress towards the fixed point).
Using concavity. Given existence of some high-powered convergence results for iterates of concave mappings, and looking at $\phi$, might lead one to believe that surely this mapping must be concave. Alas, this does not seem to be the case in any natural sense (concave as a mapping to the matrix Loewner order once after ditching the $\textbf{diagpart}$ in $\phi$, or elementwise concave as a mapping from vectors to vectors). This conclusion is numerical (wanted to verify the concavity numerically before attempting to prove), and though I have no reason to be suspicious of it, it still strikes me as quite strange and possibly misleading.
None of these three approaches have yielded me any fruit. I suspect Brouwer's fixed point theorem may be used to establish the existence of a fixed point, in conjunction with some direct analysis to identify a compact set which is fixed by the mapping $\phi$, but does not immediately imply uniqueness or convergence of the iterates. It is also worth noting that the matrix square root does indeed converge to the identity when iterated--but given the interplay of the $\textbf{diagpart}$ and the square root above, the proofs I have come up with of this statement (e.g. via the logarithm) fail to be immediately applicable.
Any pointers would be appreciated!!
Edit: partial answer
This fixed-point formulation came from an optimization problem, to which we can show existence of a unique solution. Lemma 3.4 here provides the explicit mapping, and Corollary 3.2 shows that $\phi$ does indeed have a unique fixed point.
However, we have still not shown that iterating $\phi$ converges to this fixed point. An explicit formula for the fixed point of $\phi$ in terms of $A$ would be ideal, but may not be possible. There is a little discussion in the 'future work' section of some partial progress on fixed-points of $\phi$ that might be interesting. my co-authors and I would definitely still be interested in any fruitful directions.
Edit 1/25/23: convergence of iterates in a neighborhood of the fixed point
In the most recent version of the resulting paper, we show in Theorem 3.3 that iterates of this mapping converge in some (quantifiable) neighborhood of the fixed point. Global convergence is still neither known nor disproved.
