Bound $\|(A+B)-(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2\|$ in terms of commutator $\|AB-BA\|$

Question

For positive definite matrices $A$ and $B$, can $$ \|(A+B)-(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2\| $$ be bounded in terms of $\|AB-BA\|$?

Note that if the matrices commute, then both norms are zero.

Answer 1

Let $[F,G] := FG-GF$. $$ %(A^{1/4}C^{1/2}A^{1/4})^{2} = A^{1/4}C^{1/2}A^{1/2}C^{1/2}A^{1/4}\\ %= A^{3/4}C^{1}A^{1/4}+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}\\ %= AC+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}+A^{3/4}[C,A^{1/4}] %$$

Easy algebra shows that $$ (A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2= A+A^{1/2}BA^{-1/2}+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}+A^{3/4}[C,A^{1/4}]\\ = A+B+[A^{1/2},B]A^{-1/2}+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}+A^{3/4}[C,A^{1/4}]. $$ where $C:=(1+A^{-1/2}BA^{-1/2})$. Since $$ [C,A] = CA-AC = A^{-1/2}[A,B]A^{-1/2} $$ we get a a result if we can bound $[F^{r},G^{s}]$ in terms of $[F,G]$.

I don't have an elegant way to do this. One way is to note that there are polynomials $p$ and $q$ of degree $n$ such that $F^{r} = p(F)$ and $G^{s} = q(G)$, hence $$ F^{r}G^{s} = p(F)q(G) - q(G)p(F) = \sum_{i,j=1}^{n}[F^{i},G^{j}] $$ and we can focus on integer exponents. For those, repeatedly swapping $F$ and $G$ and collecting the commutators shows that $$ \|[F^{i},G^{j}]\|\leq \|F\|^{n}\|G\|^{n}\|[F,G]\|. $$

In summary, this gives us $$ \|(A+B)-(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2\|\\ \leq \|[A,B]\| ( \|A\|^{n}\|B\|^{n}\|A^{-1}\|^{1/2}+\|A\|^{1/2}\|A\|^{n}\|C\|^{n}\|A^{-1}\|+\|A\|^{3/4}\|A\|^{n}\|C\|^{n}\|[A,B\|) $$ with $\|C\|^{n}\leq \|A^{-1}\|^n2^{n}(\|A\|^{n}+\|B\|^{n})$