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Let $E$ be a Banach space and $A,B \in \mathcal L (E).$ Prove that $\lim\limits_{n \to \infty} \left (e^{\frac {A} {n}} e^{\frac {B} {n}} \right )^n = e^{A + B}.$

I have managed to prove two things $:$

$(1)$ If $\lim\limits_{n \to \infty} A_n = A$ then by DCT we have $$\lim\limits_{n \to \infty} \left (I + \frac {A_n} {n} \right )^n = e^A.$$

$(2)$ $\displaystyle {\lim\limits_{n \to \infty} n \left (e^{\frac {A} {n}} - I \right )} = A.$

In the lecture notes it has been claimed that the required limit can be computed easily using $(1)$ and $(2).$ But I can't able to get that hint. Can anybody please help me in this regard?

Thanks for your time.

ACB
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  • Note that if you in (2) put $A_n = n(e^{A/n}-1)$ then $e^{A/n} = 1 + \frac{A_n}{n}$ and you know $A_n\to A$. You can use this with (1) in the original product. – Winther Dec 17 '20 at 19:05
  • Similarly if I define $B_n = n \left (e^{\frac {B} {n}} - I \right )$ then I get $e^{\frac {B} {n}} = I + \frac {B_n} {n}.$ So we have $$\left (e^{\frac {A} {n}} e^{\frac {B} {n}} \right )^n = \left [ \left ( I + \frac {A_n} {n} \right ) \left (I + \frac {B_n} {n} \right ) \right ]^n.$$ Now what should I do? Here the problem is that $\left (I + \frac {A_n} {n} \right )$ and $\left ( I + \frac {B_n} {n} \right )$ do not always commute unless $A_n$'s and $B_n$'s commute. Am I right @Winther? – ACB Dec 17 '20 at 19:20
  • Multiply out the stuff inside the bracket. This gives you $I + \frac{C_n}{n}$ for some $C_n$. Use $(1)$ again to conclude.. – Winther Dec 17 '20 at 19:20
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    Oh! I see now. We have $$\left (e^{\frac {A} {n}} e^{\frac {B} {n}} \right )^n = \left [ \left ( I + \frac {A_n} {n} \right ) \left (I + \frac {B_n} {n} \right ) \right ]^n = \left (I + \frac {A_n + B_n + \frac {A_n B_n} {n}} {n} \right )^n.$$ Now since $\lim\limits_{n \to \infty} \left ( A_n + B_n + \frac {A_n B_n} {n} \right ) = A + B$ thus by $(1)$ it follows that $$\lim\limits_{n \to \infty} \left (e^{\frac {A} {n}} e^{\frac {B} {n}} \right )^n = e^{A + B}.$$ Am I right @Winther? – ACB Dec 17 '20 at 19:26
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    Does this answer your question? Proving the Lie-Product formula – user159517 Dec 17 '20 at 20:57

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