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Let $A$ be a nonempty convex subset of the Euclidean plane. For each direction $\theta$, let $d_A(\theta)$ be the diameter of $A$ in that direction. That is, $d_A(\theta)$ is the distance between the supporting lines perpendicular to $\theta$.

Suppose that $B$ is another convex subset of the plane with the same diameter function as $A$, i.e., for all $\theta$, $d_B(\theta)= d_A(\theta)$. Suppose also that $B$ is symmetric about a point. (For example, WLOG think of $B\subset \mathbb{R^2}$ being symmetric about the origin, i.e. $-B=B$.) Can we conclude that $B$ has area greater than or equal to that of $A$, $|B|\geq |A|$?

In other words, I'm interested in when area is maximized for a convex region with a given diameter function, and speculating that it happens when the region is symmetric about a point. This would generalize the fact that disks have the greatest area among regions of a given constant diameter (isodiametric inequality).

This is closely related to the question How does the area of $A+A$ compare to the area of $A-A$? Namely, I think that a positive answer to my question would imply in a straightforward manner a positive answer to the linked question.

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Jonas Meyer
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    The more common name for $d_A(\theta)$ is the width function. It is expressed in terms of the support function as $d_A(\theta)=h_A(\theta)-h_A(-\theta)$. The concept makes sense in all dimensions, and your proof generalizes too. –  Jul 04 '14 at 02:19
  • @Thisismuchhealthier.: Thank you, that is useful. It explains why searching for "diameter function" wasn't very helpful. – Jonas Meyer Jul 04 '14 at 02:38

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I think I've found an answer to my question. It is yes.

In the question How does the area of $A+A$ compare to the area of $A-A$?, the affirmative answer following from the Brunn–Minkowski inequality says that $|A-A|\geq |A+A|$. Then $|\frac12(A-A)| \geq|\frac{1}{2}(A+A)|$. Because $A$ is convex, $\frac{1}{2}(A+A)=A$. On the other hand, $\frac{1}{2}(A-A)=B$, i.e., $\frac12(A-A)$ has the same diameter as $A$ in each direction, but it is symmetric about the origin. (This $B=\frac12(A-A)$ is called the central symmetrization of $A$, a phrase that helped me find a reference to the Brunn–Minkowski inequality in a paper.) Thus, $|B|\geq |A|$.

I did not realize when I posted the question that it is merely a reformulation of the linked one.

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