Let us define $$E(t) = \int_{-T}^{t - \lambda(t)^2} \frac{ds}{|\ln(T - s)|^2(t - s)} \quad \text{and} \quad E^{\varepsilon}(t) = \int_{-T}^{t - \lambda(t)^2 - \varepsilon}\frac{ds}{|\ln(T - s)|^2(t - s)},$$ for $t \in [-T/4, T]$, $\lambda(t)$ defined by $$\lambda(t) = \frac{T - t}{|\ln(T - t)|^2},$$ and $0 < \varepsilon < T$ taken arbitrarily small (I aim to take $\varepsilon \to 0$ eventually). I aim to prove that, $\exists C > 0$ independent of $\varepsilon$ such that $$|E^\varepsilon(t) - E^\varepsilon (T)| \le C|E(t) - E(T)|.$$ Using the answer given in this post we can show that we actually don't need that absolute values, so that I aim to prove $$E^\varepsilon(t) - E^\varepsilon (T) \le C(E(t) - E(T)) \quad \Leftrightarrow \quad 0 \le (CE(t) - E^\varepsilon(t)) - (CE(T) - E^\varepsilon(T)).$$ Now note that $$E(T) = \frac{1}{|\ln(2T)|} \quad \text{and} \quad E^\varepsilon(T) = \frac{1}{|\ln(2T)|} - \frac{1}{|\log \varepsilon|},$$ which in the end leads to $$0 \le (C - 1) \left(\int_{-T}^{t - \lambda(t)^2} \frac{1}{|\ln(T -s)|^2(t - s)}ds - \frac{1}{|\ln(2T)|}\right) + \int_{t - \lambda(t)^2 - \varepsilon}^{t - \lambda(t)^2} \frac{1}{|\log(T -s)|^2(t - s)}ds - \frac{1}{|\log(\varepsilon)|}.$$ Basically, if I denote $G^C(t) = CE(t) - E^\varepsilon(t)$, I just want to show that $G^C(T) \le G^C(t)$ for every $t \in [-T/4, T]$. The annoying thing is that this $E(t)$ is hard to manipulate as we cannot explicitly compute the integral. I computed $G^C(t) - G^C(T)$ using Python for $T=4.10^{-5}$, $C = 4$ and $\varepsilon = T.10^{-5}$ (see picture below).
We see that the green curve is always "far" above zero in the interval we consider, so that the desired inequality seems reasonable (I did it for various other epsilon and the shape remain roughly the same).
However, I really struggle to prove it. I thought about computing the first derivative of $G^C(t)$ and show that it is decreasing in a certain range but the explicite form of the derivative is quite ugly so that I really struggle to prove the monotonicity. I also thought about showing that this function is concave, but this would involve computing the second derivative whose form is even worse to manipulate. Any idea to help would be greatly appreciated.
Some precision: It is actually fine if I can have an inequality of the type $$|E^\varepsilon(t) - E^{\varepsilon}(T)| \le C\left(|E(t) - E(T)| + \frac{1}{|\ln(\varepsilon)|^2}\right),$$ so that I allow $|E^\varepsilon(t) - E^{\varepsilon}(T)|$ to be slightly bigger than $C|E(t) - E(T)|$ (with a difference of order $1/|\ln(\varepsilon)|^2$). I manage to show that $$|E^\varepsilon(t) - E^{\varepsilon}(T)| \le C\frac{1}{|\ln(\varepsilon)|^2}$$ when $t \in [T - \varepsilon, T]$. If I could have $$|E^\varepsilon(t) - E^{\varepsilon}(T)| \le C|E(t) - E(T)| $$ in the range $[-T/4, T-\varepsilon]$ I would be delighted. None of my attempt have been successful so far.