If $t\mapsto \rho(t)=e^{(1-t)A}\cdot e^{tB},$ then $e^{A}-e^{B}=\int^{t}_{0}e^{(1-t)A}(A-B)e^{tB}dt$

Question

Suppose \begin{align}A,B\in M_{n\times n}(\Bbb{R}),\;\forall\;\;n\geq 2\end{align} We define \begin{align}\rho:\Bbb{R}\to M_{n\times n}(\Bbb{R})\end{align} \begin{align}t\mapsto \rho(t)=e^{(1-t)A}\cdot e^{tB}\end{align} I want to prove two statements

1. \begin{align}e^{A}-e^{B}=\int^{1}_{0}e^{(1-t)A}(A-B)e^{tB}dt\end{align} 2. \begin{align}\forall\,r>0,\exists\,c(r)>0\;\text{such that}\end{align} \begin{align}\Vert e^{A}-e^{B}\Vert \leq c(r)\Vert A-B\Vert,\;\text{for}\;\Vert A\Vert\leq r,\Vert B\Vert\leq r\end{align}

MY THOUGHTS:

Based on the fact that I'm currently studying Basic Theory of ODE, I'm thinking this might have something to do with

1. Gronwall's Lemma
2. Mean Value Theorem
3. Local or global Lipschitzian

MY TRIAL

Number 2:

\begin{align}A,B\in M_{n\times n}\end{align} implies that $A,B$ are continuous since $M_{n\times n}$ is isomormphic to $\ell (\Bbb{R}^n)$, that is, a space of continuous linear maps.

Then, \begin{align}\Vert e^{A}-e^{B}\Vert&=\big\Vert\int^{1}_{0}e^{(1-t)A}(A-B)e^{tB}dt\big\Vert\\ &\leq \int^{1}_{0}\Vert e^{(1-t)A}(A-B)e^{tB}\Vert dt\\ &\leq\Vert A-B\Vert \int^{1}_{0}\Vert e^{(1-t)A}\Vert\Vert e^{tB}\Vert dt\end{align} I got stuck at this point. Please, any help about the this and the first one?

Answer 1

With

$\rho(t) = e^{(1 - t)A}e^{tB}, \tag 0$

we have

$\rho(1) = e^B, \tag 1$

$\rho(0) = e^A; \tag 2$

thus,

$e^B - e^A = \rho(1) - \rho(0) = \displaystyle \int_0^1 \dot \rho(s) \; ds; \tag 3$

from (0),

$\dot \rho(t) = e^{(1 - t)A}(-A)e^{tB} + e^{(1 - t)A}Be^{tB} = e^{(1 - t)A}(B - A)e^{tB}; \tag 4$

we substitute (4) into (3):

$e^B - e^A = \displaystyle \int_0^1 e^{(1 - s)A}(B - A)e^{sB} \; ds, \tag 5$

and negate:

$e^A - e^B = \displaystyle \int_0^1 e^{(1 - s)A}(A - B)e^{sB} \; ds; \tag 6$

that's the first one.

For the second,

$\Vert e^A - e^B \Vert = \left \Vert \displaystyle \int_0^1 e^{(1 - s)A}(A - B)e^{sB} \; ds \right \Vert \le \displaystyle \int_0^1 \left \Vert e^{(1 - s)A}(A - B)e^{sB} \right \Vert \; ds$ $\le \displaystyle \int_0^1 \left \Vert e^{(1 - s)A} \right \Vert \left \Vert A - B \right \Vert \left \Vert e^{sB} \right \Vert \; ds = \Vert A - B \Vert \displaystyle \int_0^1 \left \Vert e^{(1 - s)A} \right \Vert \left \Vert e^{sB} \right \Vert \; ds$ $\le \Vert A - B \Vert \displaystyle \int_0^1 e^{(1 - s) \Vert A \Vert} e^{s \Vert B \Vert} \; ds \le \Vert A - B \Vert \displaystyle \int_0^1 e^{(1 - s)r}e^{sr} \; ds = e^r \Vert A - B \Vert. \tag 7$