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


*

*Gronwall's Lemma

*Mean Value Theorem

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