# On a limiting value of a matrix exponential

$$\mathbf {The \ Problem \ is}:$$ 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 .

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