The right truncated exponential distribution would be defined as
$$ f(x=t \mid x<T) = \frac{\lambda e^{-\lambda t}}{1-e^{-\lambda T}} $$
right?
I would like to know if this distribution preserves the memoryless property. I think that the answer is no, because the memoryless property characterizes completely the exponential distribution, so no other distribution (including truncated exponential) have it. But this is counterintuitive to me, and I cannot derive a mathematical prove for that, any help is appreciated
In order for a distribution to be memoryless, you need $$P(X>a+b|X>b)=P(X>a)$$ to hold for all $a,b>0$. Now, take $a=b=T/2$. Then $P(X>T/2)$ is nonzero, but $P(X>T|X>T/2)=0$ because $X$ is never greater than $T$.