Do we have a version of Grönwall lemma for $t<b$?

Question

Let $y(\cdot)$ be a continuous mappings defined on $[a,b]$ and $K\in (0,+\infty]$ a constant. An application of the Grönwall lemma is the following simple case: if we have for all $t\in[a,b]$
$$y(t)\leq K + \int_a^ty(s) ds ,$$ then we will have for all $t\in [a,b]$ $$y(t)\leq K e^{\int_a^tds}=Ke^{(t-a)}. $$ My question is that do we still have such result if we change the order of the integral $\int_a^t ds$ to $\int_t^b ds$ in other words do we have the following implication: $$\forall t\in [a,b]\; \;\;\; y(t)\leq K + \int_t^by(s) ds \implies \forall t\in [a,b]\; \;\;\; y(t)\leq K e^{\int_t^bds}=Ke^{(b-t)}.$$

Answer 1

By taking : $z(t)=y(-t)$ for all $t\in [-b,-a]$, and changing the measure $ds=-d\tau$ we obtain $$z(t)\leq K+\int_{-b}^t z(\tau)d\tau $$ and then by applying the classic Gronwall lemma we get that : $$z(t)\leq Ke^{t+b} \;\; \forall t\in [-b,-a],$$ in other words $$y(t)\leq Ke^{b-t} \;\; \forall t\in [a,b]$$