For every convex compact set $K$ of area $1$ in $\mathbb{R}^2$, among all ellipses of area $1$ there exists an ellipse $E$ such that the area of the symmetric difference between $K$ and $E$ is smallest possible.



> 
> **Questions.**
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> 
> (a) Is $E$ unique?
> 
> 
> (b) If the answer is "yes", does the same hold for the analogous question in higher dimensions?
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> 



Approximating a convex disk by an ellipse
Question (a) has a positive answer in the centrally symmetric case. The proof is involved and I will only summarize the strategy here. Full details can be found in [this ArXiv paper](https://arxiv.org/abs/1702.03808). Comments, suggestions, corrections etc are welcome.


Let $K \subset \mathbb{R}^2$ be a compact convex set of area $\pi$. To prove uniqueness of best ellipse approximation, it is sufficient to consider ellipses in the family $E\_t := \{(x,y) \in \mathbb{R}^2 ; e^t x^2 + e^{-t} y^2 \le 1 \}$ (because any pair of ellipses can be put inside the family by applying a suitable area-preserving linear map). I actually show that the function $A(t) := \mathrm{area} (K \vartriangle E\_t)$ has a unique minimum, and is strictly decreasing (resp. increasing) to the left (resp. right) of it. Equivalently, the function is "strictly quasiconvex".


First, I consider where $K$ is **regular**: this means that that $K$ has a smooth boundary which is transverse to all ellipses $\partial E\_t$, except for a finite number of non-degenerate (i.e. quadratic) tangencies, all of which occur outside the envelope hyperbola $xy = \pm 1/2$. In this regular case, I prove the following more precise "uniform quasiconvexity" properties:


* The function $A$ is $C^1$ everywhere and $C^2$ except at the tangency parameters.
* If $t$ is not a tangency parameter, then:
$$
\max\{|A'(t)|, A''(t)\} > \delta(A(t)) > 0, \tag{$\star$}
$$
where $\delta$ is some positive continuous function on the interval $(0,2\pi)$ that **does not depend on $K$**.


It's not difficult to convince oneself that these properties imply the strict quasiconvexity of the function $A$. Actually any uniform limit of functions with the properties above is also strictly quasiconvex. 


Finally, an arbitrary $K$ can always be perturbed with respect to the symmetric difference metric so that it becomes regular. So it follows that the function $A$ is always strictly quasi-convex.


How are the quasiconvexity properties proved (in the regular case)?


By applying a suitable linear map, it is sufficient to consider $t=0$. Suppose this is not a tangency parameter (as the case of tangency is actually easier). Similarly to my older "answer", one can deduce formulas for the derivatives $A'(0)$ and $A''(0)$. Actually $A'(0)$ depends only on the points of intersection between $\partial K$ and the unit circle $\partial E\_0$, while $A''(0)$ depends also on the angles of intersection. An inspection of the second formula shows that $A''(0)$ has a strong "tendency" for being positive; for example if no two consecutive intersection points are separated by an angle $>\pi/2$ then $A''(0)>0$. What we need to show in order to obtain the main inequality $(\star)$ is that if $A''(0) \leq 0$ (and $A(0)$ is not too close to $0$ or $2\pi$) then $A'(0)$ cannot be too close to zero. This is done by a case-by-case analysis, combining analytic and geometric arguments -- I can only refer to the paper for the details.


**UPDATE** (May 2018): The assumption that the ellipses have exactly the same area as $K$ is never used; actually if we consider ellipses whose area is a number $\lambda$ between the areas of the John and the Loewner ellipses of $K$ then the analogous uniqueness theorem holds. So I ultimately obtain a positive answer in dimension $2$ for a question of [Artstein-Avidan and Katzin](https://arxiv.org/abs/1612.01128) about uniqueness of "maximal intersection ellipsoids". 


