I have a question regarding the proof of theorem 4.6 in https://arxiv.org/abs/1007.3899 (Hall's conjecture).

Let $S^2$ be the class of Borel subsets in $\mathbb{R}^2$ with finite and positive Lebesgue measure $|\, .\, |$. For a given set $E\in S^2$ we denote with $B_E$ the ball centered in the origin with $|E|=|B_E|$, and $P(E)$ the perimeter of $E$ (see https://en.wikipedia.org/wiki/Caccioppoli_set).

For $E\in S^2$ let $\delta P(E):=\frac{P(E)-P(B_E)}{P(B_E)}$ and $\alpha(E):=\inf\limits_{x\in\mathbb{R}^n}\frac{|E\triangle (x+B_E)|}{|B_E|}$. And for $E\in S^2$ with $\alpha(E)>0$ define $Q(E):=\frac{\delta P(E)}{\alpha(E)^2}$.

In the proof of theorem 4.6 it is claimed: If $(E_j)_j\subset S^2$ is a sequence of sets such that $$\lim\limits_{j\to\infty}Q(E_j)= Q(B) \text{ and } \lim\limits_{j\to\infty}\|H_j-1\|_{L^\infty(\partial E_j)}=0,$$ where $H_j$ stands for the mean curvature of the boundary $\partial E_j$ and $B:=B_1(0)$, then there exists a $j_0>0$ such that $E_j$ is a convex set for all $j\ge j_0$.

Does anyone know why that holds (i.e. how the convexity is concluded)? Or do you know any reference for an anything which helps to show the convexity? I appreciate any help.

**Edit:** I have found the book "Lecture Notes on Mean Curvature Flow " by Carlo Mantegazza, where in the section 4.5.1 about "an alternative proof of Grayson's theorem" there can be found a similar argument (in the last section before remark 4.5.8), namely roughly: Let $\gamma_t$ be the curvature flow of a closed, embedded, smooth curve in the plane, $\tilde{\gamma}_{q_i}$ a certain new sequence of rescaled curves with uniformly bounded curvatures in $L^2$, such that a certain subsequence converges in $C^2$ to a limit curve which is a unit circle. Assuming that the $\tilde{\gamma}_{q_i}$ has positive curvature, then they curves are convex.

Thus, I think there is a general fact which I don't know (I am not an expert in this field), giving convexity under assumptions that some sequence converges to the unit ball and an extra assumption abouts it's curverture.