Assume $a(t)\geq 0$ and $b(t)\geq 0$.
i can show the following inequality
$\mid e^{-\int_0^ta(s)ds}-e^{-\int_0^tb(s)ds} \mid\leq T\max_{0\leq t \leq T}\mid a(t)-b(t)\mid$
by writing $\mid e^{-\int_0^ta(s)ds}-e^{-\int_0^tb(s)ds}\mid =\mid -\int_{\int_0^t b(s)ds}^{\int_0^ta(s)ds} e^{-\tau}d\tau\mid \leq\mid\int_0^ta(s)ds-\int_0^tb(s)ds\mid\leq T\max_{0\leq t \leq T}\mid a(t)-b(t)\mid$
but under same assumptions i want to get a similar inequality for the following difference
$e^{\int_0^ta(s)ds}-e^{\int_0^tb(s)ds}$
unfortunately, in this case i couldn't use the above approach. My question is can you show that a similar inequality for the above difference. Please.
Thanks in advance!
For $a(s)=ns^{n-1}$ and $b(s)=0$ we have
$$e^{\int_0^ta(s)ds}-e^{\int_0^tb(s)ds}=e^{t^n}-1.$$
We have $$\max_{0\leq t \leq T}|a(t)-b(t)|=1.$$ So, the inequality
$$e^{t^n}-1\le K T \max_{0\leq t \leq T}|a(t)-b(t)|=KT$$ must holds for any $t\in[0,T].$ Thus, we must have, in particular,
$$e^{T^n}-1\le KT.$$
This is impossible, unless you allow $K$ to depend on $T.$ But I think this is not what you are looking for.
The inequality works in case you deal with negative exponents because $e^{-x}\le 1$ for $x\ge 0.$ But, if you allow positive exponents then $e^x$ goes to infinite as $x$ goes to infinite, and it is not possible to get control on $e^x-e^y$ in the way you are suggesting.