Effective Uniformization Theorem for the Torus The famous Uniformization Theorem asserts that for any Riemannian metric $g$ on the 2-torus $T^2$ there exists a function $\rho: T^2 \to \mathbb{R}$ and a flat metric $\bar{g}$ on $T^2$ such that
\begin{equation*}
g=e^{\rho}\bar{g}.
\end{equation*}
Philosophically my question is the following:

If $g$ is "almost" a flat metric, can we say that $\rho$ is small?

More precisely, does there for every (small) $\delta >0$ exist a constant $c=c(\delta)$ with the following property and so that $c(\delta) \xrightarrow{\delta \to 0} 0$: 
Let $g$ be a metric on $T^2$ so that
\begin{equation}
\mathrm{diam}(T^2,g) \leq 1 \quad \text{and} \quad |K_g| \leq \delta,
\end{equation}
where $K_g$ is the Gauss curvature w.r.t. the metric $g$. Then there exists a function $\rho:T^2 \to \mathbb{R}$ with
\begin{equation*}
|\rho(x)| \leq c(\delta) \quad \text{for all } \, x \in T^2
\end{equation*}
so that $\bar{g}:=e^{-\rho}g$ is a flat metric.
Equivalently, can we say that the function $\rho$ whose existence is guaranteed by the Uniformization Theorem satsifies $|\rho(x)| \leq c(\delta)$ for all $x \in T^2$?
Of course the flat metric in the conformal class of $g$ is only defined up to a multiplicative constant, and so $\rho$ is also only defined up to an additive constant. Since the bound $|\rho(x)| \leq c(\delta)$ can not hold if we change $\rho$ by arbitrary constants, there should be a normalizing condition on $\rho$. In my opinion any of the following is a natural condition: 
Choose the flat metric $\bar{g}$ in the conformal class of $g$ so that the corresponding $\rho$ satisfies one of the following conditions:
\begin{equation}
\int_{T^2}\rho \, d\mathrm{vol}_{\bar{g}}=0, \quad \frac{1}{\mathrm{vol}_{\bar{g}}(T^2)}\int_{T^2}e^{\rho} \, d\mathrm{vol}_{\bar{g}}=1 \quad \text{or} \quad \frac{1}{\mathrm{vol}_{\bar{g}}(T^2)}\int_{T^2}e^{-\rho} \, d\mathrm{vol}_{\bar{g}}=1.
\end{equation}
I tried proving this by obtaining PDE estimates. For this I needed the fact that for any flat metric $\bar{g}$ on $T^2$ and any $\rho:T^2 \to \mathbb{R}$ it holds
\begin{equation}
e^{-\rho}\Delta\rho=-2K_g,
\end{equation}
where $K_g$ is the Gauss curvature of the metric $g:=e^{\rho}\bar{g}$ and $\Delta$ is the Laplace operator of the flat metric $\bar{g}$. Assuming the last of the three normalizing conditions above  I was able to prove an estimate
\begin{equation}
e^{-\rho}-1 \leq c(\delta).
\end{equation}
However, I was unable to prove a lower estimate $e^{-\rho}-1 \geq -c(\delta)$.
I would greatly appreciate an outline of how to prove it (or maybe have an idea what might work) or a reference in the literature stating this. Thanks in advance!
 A: I found an answer to the above question myself, and since it seems to be of interest to some other people, I will post the answer.
The question can be answered positively in the following form.
There exist constants $\delta_0 >0$ and $C>0$ with the following property. Let $g$ be a metric on $T^2$ such that
\begin{equation}
{\rm diam}(T^2,g)\leq 1 \quad \textit{and} \quad |\sec(g)| \leq \delta
\end{equation}
for some $\delta \leq \delta_0$. Then there exists $\rho:T^2 \to \mathbb{R}$ such that $\bar{g}:=e^{-\rho}g$ is a flat metric, and
\begin{equation}
 |\rho(x)| \leq C \delta \quad \textit{for all }\, x \in T^2.
\end{equation}
This is precisely Lemma 8.4 in the article Stability of Einstein metrics and effective hyperbolization (https://arxiv.org/abs/2206.10438) by Ursula Hamenstädt and myself. It is not possible to give the complete proof here. I try to explain the most important ingredients (though I have to sweep some technical details under the rug).
First, let's start with the right choice of $\rho$. As stated in the question, by the classic uniformization theorem there exists some function $\rho$ so that $\bar{g}:=e^{-\rho}g$ is flat. But $\rho$ is only defined up to an additive constant. We pick $\rho$ so that
\begin{equation}
\int_{T^2} \rho \,  d{\rm vol}_g = 0.
\end{equation}
It will become apparent later why this is the right normalizing condition. The key point is that $\rho$ solves the PDE
\begin{equation}
\Delta_g \rho=2K,
\end{equation}
where $\Delta_g$ is the Laplace operator of the metric $g$, and $K$ is the Gauss curvature of $g$ (here I take the sign convention $\Delta_gu=-{\rm tr}_g(\nabla^2 u)$). There are two main steps to obtain the $C^0$-estimate. First, one proves an $L^2$-estimate. This is done very classically using integration by parts and the Poincaré inequality (which holds on general manifolds with lower curvature bounds and an upper bound on the diameter). Second, one uses the de Giorgi-Nash-Moser estimates (see Theorem 8.17 in Elliptic Partial Differential Equations of Second Order by Gilbarg & Trudinger). These allow to deduce $C^0$-bounds from $L^2$-bounds. More precisely, there exists a constant $C$ so that (here I am hiding some technical details)
\begin{equation}
||\rho||_{C^0(T^2,g)} \leq C \big(||\rho||_{L^2(T^2,g)}+||\Delta_g\rho||_{L^2(T^2,g)} \big).
\end{equation}
As, by assumption, $\Delta_g \rho=2K$ is bounded by $2\delta$, this shows that it really suffices to obtain the $L^2$-estimate $||\rho||_{L^2(T^2,g)} \leq C \delta$.
So let me show how to obtain the $L^2$-estimate. By a result of Gromov, any $n$-manifold $M$ with ${\rm Ric}_M \geq -(n-1)$ and diameter bounded above ${\rm diam}(M) \leq D$ has a spectral gap, that is, there exists $C_P=C_P(n,D)$ so that for all smooth functions $u:M \to \mathbb{R}$ with $\int_M u \, d{\rm vol}_M=0$ it holds
\begin{equation}
\int_M u^2 \, d{\rm vol}_M \leq C_P \int_M |\nabla u|^2 \, d{\rm vol}_M.
\end{equation}
We will be able to apply this to $\rho$ because we chose $\rho$ in such a way that $\int_{T^2} \rho \, d{\rm vol}_g=0$. Note that, due to the Gromov-Bishop volume comparison theorem, there is a constant $V$ (not depending on anything) so that ${\rm vol}_g(T^2) \leq V$ because ${\rm diam}(T^2,g) \leq 1$ and because the curvature is uniformly bounded. Multiplying $\Delta_g \rho=2K$ with $\rho$ and invoking integration by parts we obtain
\begin{align*}
\int_{T^2} |\nabla \rho|^2 \, d{\rm vol}_g & = 2\int_{T^2} \rho \, K \, d{\rm vol}_g \\
 & \leq 4\delta \int_{T^2} 1\cdot \rho \, d{\rm vol}_g \\
& \leq 4\delta  \left(\int_{T^2} 1 \, d{\rm vol}_g\right)^{\frac{1}{2}}\left(\int_{T^2} \rho^2 \, d{\rm vol}_g\right)^{\frac{1}{2}} \\
&\leq 4 \delta V^{\frac{1}{2}} C_P^{\frac{1}{2}}\left(\int_{T^2} | \nabla \rho|^2 \, d{\rm vol}_g\right)^{\frac{1}{2}},
\end{align*}
and hence
\begin{equation}
\left(\int_{T^2} | \nabla \rho|^2 \, d{\rm vol}_g\right)^{\frac{1}{2}} \leq 4 \delta V^{\frac{1}{2}} C_P^{\frac{1}{2}}
\end{equation}
after dividing by $\left(\int_{T^2} | \nabla \rho|^2 \, d{\rm vol}_g\right)^{\frac{1}{2}}$. Therefore,
\begin{equation}
\left(\int_{T^2}  \rho^2\, d{\rm vol}_g\right)^{\frac{1}{2}} 
 \leq C_P^{\frac{1}{2}} \left(\int_{T^2} | \nabla \rho|^2 \, d{\rm vol}_g\right)^{\frac{1}{2}} \leq 4 \delta V^{\frac{1}{2}} C_P.
\end{equation}
This finishes the proof of the $L^2$-estimate, and hence the proof of the above statement (modulo details).
