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$\mathbf {The \ Problem \ is}:$ If $A$ and $B$ are two bdd linear operators on a Banach space, then show that

$\lim_{n \to \infty} (e^\frac{A}{\sqrt n}e^\frac{B}{\sqrt n}e^\frac{-A}{\sqrt n}e^\frac{-B}{\sqrt n})^n = e^{(AB-BA)}$

$\mathbf {My \ approach}:$ Actually, we know $e^\frac{T}{\sqrt n} = I + \frac{T}{\sqrt n} + o(\frac{1}{\sqrt n}) $ where $T$ is a bdd linear operator and $o$ is Landau's little order notation , but the multiplication is getting bigger and confusing .

Is there any other better way to prove this thing except this multiplication .

A small hint is warmly appreciated .

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  • $\begingroup$ Actually, multiplying out up to $1\over n$ looks like a reasonable strategy. $\endgroup$
    – user619894
    Dec 22, 2020 at 19:07
  • $\begingroup$ Remember that if $x_n\to x$ then $(1+ x_n/n)^n\to e^{x}$ in any Banach algebra. $\endgroup$
    – s.harp
    Dec 22, 2020 at 19:14
  • $\begingroup$ Perhaps it could be proven as a consequence of the Lie product formula $\endgroup$ Dec 22, 2020 at 19:39
  • $\begingroup$ An attempt: Denote $E_n = \exp[A/\sqrt{n}]$. We note that $$ E_n \exp[B/\sqrt{n}]E_n^{-1} = \exp[[E_nBE_n^{-1}]/\sqrt{n}]. $$ Note that we have $$ E_nBE_n^{-1} = B + [AB - BA]/\sqrt{n} + o(1/\sqrt{n}). $$ Now, write $$ \exp(A/\sqrt{n})\exp(B/\sqrt{n})\exp(-A/\sqrt{n})\exp(-B/\sqrt{n}) = \\ \exp(B/\sqrt{n} + [AB - BA]/n + o(1/n))\exp(-B/\sqrt{n}). $$ Via the Lie product formula, we have $$ [\exp(B/\sqrt{n} + [AB - BA]/n + o(1/n))\exp(-B/\sqrt{n})]^{\sqrt{n}} = \\ \exp(B - B + [AB - BA]/\sqrt{n}) + o(1) = \\ \exp([AB - BA]/\sqrt{n}) + o(1) $$ $\endgroup$ Dec 22, 2020 at 20:02
  • $\begingroup$ @Ben Grossman, yes , I wad trying to apply that formula, but I couldn't do it up to last . $\endgroup$ Dec 22, 2020 at 20:04

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