Q: How to find a tight lower bound for $\frac{1}{2t}\int_0^t \int_0^{x+\epsilon} E(x, y) d y d x$ where $E$ is defined in a specific way? Suppose that $E:[0,1] \to \mathbb{R}$ and $E(x, y)=e^{\frac{-c^2(x-y)^2}{\sigma^2}}\left(e^{\frac{h(x, y)(2 c|x-y|-h(x, y))}{\sigma^2}}-1\right)$ where $c > 0$ and $\sigma>0$ are some constants, and $h:[0,1] \to \mathbb{R}$ is such that $\lim_{x \to y} h(x,y)=0$.
I need to find a lower bound on $\frac{1}{2t}\int_0^t \int_0^{x+\epsilon} E(x, y) d y d x$, that doesn't depend on $t$ when $t \to 0$. $t$ is always positive.
One of my attempts was to use the mean value theorem, saying that:
$$
\begin{aligned}
\frac{1}{2 t} \int_0^t \int_0^{x+\epsilon} E(x, y) d y d x & \geq \frac{1}{2 t} \int_0^t \int_{\frac{\epsilon}{2}}^\epsilon E(x, y) d y d x \\
&=\frac{1}{2 t} \int_0^t \frac{\epsilon}{2} E\left(x, y^*\right) d x \\
&=\frac{\epsilon}{4} E\left(x^*, y^*\right)
\end{aligned}
$$
Where $0 \leq x^* \leq t$ and $\frac{\epsilon}{2} \leq y^* \leq\epsilon$. So when $t \rightarrow 0 , x^* \rightarrow 0$ and we have
$$
\frac{1}{2 t} \int_0^t \int_0^{x+\epsilon} E(x, y) d y d x \geq \frac{\epsilon}{4} E\left(0, y^*\right)
$$
But then we wouldn't know what $y^*$ is. I could make the bounds in the inner integral be something other than $[\frac{\epsilon}{2},\epsilon]$, something like $[\epsilon-t, \epsilon]$ so that $y^* \to \epsilon$ when $t \to 0$, but then the lower bound would depend on $t$ and also it wouldn't be that tight.
I would appreciate any help!
