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Fix a real number $\alpha>1$ and an integer $n \geq 1$. Let $T_\alpha$ be the mapping defined on the set $\mathcal{E}$ of closed convex subsets of $\mathbb{R}^n$ by \begin{equation*} T_\alpha(E) = \{\alpha x + (1-\alpha) y, x \in E, y \in E\}. \end{equation*} It is clear that $T_\alpha(E) \in \mathcal{E}$ and $E \subseteq T_\alpha(E)$ for all $E \in \mathcal{E}$.

I am interested in the properties of the operator $T_{\alpha}$. My questions are the following:

  1. is there a standard name for this operator?
  2. if $F \in \mathcal{E}$, under which conditions does there exist $E \in \mathcal{E}$ such that $F=T_\alpha(E)$?
  3. does $T_\alpha(E)=T_\alpha(E')$ imply $E=E'$?

Thank you.

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  • $\begingroup$ what have you done so far? I think this is a difficult question (three of them) and question 2 in particular need not have an easily described answer. I have considered a similar problem, never published (yet) and don't quite recall it, but for example take a half-disc $H$, is $H=T_\alpha(E)$, some $\alpha>1$, some $E$. Take a square $S$, same question. In particular, how do you represent the corner points? Do you know any $F$ as in your question 2 that cannot be represented as $T_\alpha(E)$. Why not also consider $\alpha\in(0,1)$? $\endgroup$ – Mirko Jun 16 at 22:53
  • $\begingroup$ Given that $\alpha>1$ and $x,y$ run through $E$ independent of each other, I would write $T_\alpha(E)$ as $\alpha E-(\alpha-1)E$. Yes, a square $S$ could be represented as $T_\alpha(E)$ (and $E$ is unique for a given $\alpha$, itself a square of a suitable size...at least when $S$ has the origin as the intersection point of its diagonals, and likely in general(?)). A half-disc cannot be represented as $T_\alpha(E)$ for any $E$ (given $\alpha>1$). Note that $-E$ is a reflection of $E$, so $\alpha E$ and $-(\alpha-1)E$ have opposite orientation. $\endgroup$ – Mirko Jun 16 at 23:25
  • $\begingroup$ The answer to question 3 is no, at least if you allow unbounded sets $E$ (that are closed and convex). For example, $\Bbb R=T_\alpha(E)$ for any $\alpha>1$ and any half-ray$E$, so $\Bbb R=T_\alpha(\Bbb R)=T_\alpha(E)=T_\alpha(E')$, any half-rays $E,E'$ of the form $[a,\infty)$ or $(-\infty,b]$. $\endgroup$ – Mirko Jun 16 at 23:45
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This is more of an extended comment for now (but gives a good start I hope). The answer to question 3 is negative, at least of we consider unbounded convex closed sets. I need to think of the bounded case, might be then the answer to question 3 is positive? Re question 1, I doubt there is a name for this specific case, namely with coefficients $\alpha$ and $-(\alpha-1)$. In general of course one talks about Minkowski sum of two sets $A$ and $B$ defined as $A+B=\{a+b:a\in A, b\in B\}$. I personally also use the term Minkowski difference $A-B=\{a-b:a\in A, b\in B\}$ though this is the same as the Minkowski sum $A+(-B)$ where $-B$ is the opposite of $B$, or minus $B$, i.e. $-B=\{-b:b\in B\}$. I have studied (but never got around to submit a paper) the set $A-A$, but for the operator in question $$T_\alpha(E) = \{\alpha x + (1-\alpha) y\ ;\ x \in E, y \in E\}$$ I prefer to think of it as the Minkowski difference $$T_\alpha(E) = \alpha E - (\alpha-1)E$$ (where $\gamma E=\{\gamma x\ ;\ x\in E\}$). Note that, since $\alpha>1$, we have that $\alpha-1>0$ so $\alpha E$ and $(\alpha-1)E$ are similar copies of $E$ having the "same direction" (with $(\alpha-1)E$ smaller in size), and $-(\alpha-1)E$ having "opposite direction".

Note also one "graphical" way to think of Minkowski addition $A+B$ (or subtraction $A-B$, in a similar way), works in the special case when the origin $0$ belongs to $B$. (A picture would help but I won't draw it for now.) For each $a\in A$ translate $B$ so that the origin goes to $a$ (formally the translated copy is $B+a$) and take the union of all such translated copies. The result is $A+B$. It is instructive to do this with sets in the plane, sliding $B$ around, guided by the points $a$ in $A$, and seeing what is the area swept by these copies of $B$, the result being $A+B$. For example if $A$ is a triangle (containing the origin) and $B=-A$ (the opposite triangle), then $A+B=A-A$ is a hexagon (with pairs of parallel sides). Note how (1) $A-A$ is clearly bigger, thicker and "more rounded" than $A$, and (2) say $A$ is a triangle having one horizontal side (to be specific) then the horizontal line at the opposite vertex is a supporting line intersecting the triangle in at only one point, whereas there is no horizontal line (supporting or not) intersecting the hexagon $A-A$ in only one point. Note also that $A-A$ is symmetric, $A-A=-(A-A)$.

So, I think question 2 may not have an easily described answer, and I would just list a couple of examples, to have some basis for a better understanding hopefully to come.

If $S$ is any closed convex set that is symmetric about the origin, i.e. $S=-S$ then there is a similar set $P=\gamma S$ (for a suitable $\gamma$ to be chosen below) with $S=T_\alpha(P)=\alpha P - (\alpha-1)P$. Note that if $P=-P$ then $\alpha P-(\alpha-1)P=\alpha P+(\alpha-1)P$, so we just need $P=-P$ with $S=\alpha P + (\alpha-1)P$. Note also that:
Lemma. For any convex $E$ and any positive $\gamma$ and $\alpha$ we have $\alpha E+\gamma E=(\alpha +\gamma)E$. Proof. Indeed, if $z=\alpha x+\gamma y$ for some $x,y\in E$ then $v=\frac{\alpha}{\alpha+\gamma}x+ \frac{\gamma}{\alpha+\gamma}y\in E$ and $\alpha v+\gamma v=(\alpha +\gamma)v=z$. In particular the Lemma implies that $\alpha P+(\alpha-1)P=(2\alpha-1)P$. So we could take $P=\frac1{2\alpha-1}S$, then $P=-P$ and
$\alpha P-(\alpha-1)P=\alpha P+(\alpha-1)P=$
$=(2\alpha-1)P=(2\alpha-1)\frac1{2\alpha-1}S=S$.

The above generalizes to the case when $R=t+S$ where $S=-S$, i.e. $R$ is a translation of a symmetric closed convex set $S$ (by the vector $t\in\Bbb R^n$). In this case let $Q=t+P$ where (as above) $P=\frac1{2\alpha-1}S$. We have:
$T_\alpha(Q)=\alpha Q-(\alpha-1)Q=$
$\alpha(t+P)-(\alpha-1)(t+P)=$
$t+\alpha P-(\alpha-1)P=$
$t+\alpha P+(\alpha-1)P=$
$t+(2\alpha-1)P=$
$t+(2\alpha-1)\frac1{2\alpha-1}S=t+S=R$.

Note that in a similar way one could show that if $F=T_\alpha(E)$ for some closed convex $F,E$, and if $t+F$ is the translation of $F$ by a vector $t$, then $t+F=T_\alpha(t+E)$. Indeed $T_\alpha(t+E)=\alpha(t+E)-(\alpha-1)(t+E)=$
$t+\alpha E-(\alpha-1)E=t+T_\alpha(E)=t+F$.

Also, if $Y$ is a triangle in $\Bbb R^2$ then there is no closed convex $E$ with $Y=T_\alpha(E)$. More generally suppose $Y\subset\Bbb R^2$ is a bounded closed convex set such that there are two parallel supporting lines $l$ and $k$ such that $l\cap Y$ is a singleton, and $k\cap Y$ contains a non-degenerate line segment. (Recall that a supporting line for a bounded closed convex planar set is a line that intersects the set, and such that the set is contained in one of the two closed half-planes determined by the line. There ought to be a generalization for $\Bbb R^n,n\ge3$, using supporting hyperplanes, but I will stick with the planar case, as described above.)
Without loss of generality (and for ease of expression) suppose that $l$ and $k$ are horizontal. (This could be achieved using rotation, which is clearly ok to be used). Suppose, towards a contradiction, that there were a closed and convex (and necessarily bounded) set $E$ with $Y=T_\alpha(E)$.
Case 1. The top horizontal supporting line of $E$ intersects $E$ in a singleton, and the bottom horizontal supporting line of $E$ also intersects $E$ in a singleton. Then it is easy to see that the same applies to $\alpha E-(\alpha-1)E$ (i.e both the top and bottom horizontal supporting lines of the latter set intersect it in singletons), a contradiction, since one of the horizontal supporting lines of $Y$ intersects $Y$ in a non-degenerate line segment.
Case 2. At least one of the two horizontal supporting lines of $E$ intersects $E$ in a non-degenerate line segment. In this case (exercise for the reader) both the top and the bottom horizontal supporting lines of $\alpha E-(\alpha-1)E$ intersect the latter set in non-degenerate line segments, a contradiction, since one of the horizontal supporting lines of $Y$ intersects $Y$ in a singleton.

I tend to believe that the answer to question 3 is yes, for bounded closed convex sets $E,E'$ (but I didn't think of the proof and cannot rule out that there may be some examples...but, update, Oliv posted an answer that indeed the answer to question 3 is positive in the bounded case).
For the unbounded case there are easy examples even for $n=1$. Call a subset of the real line of the form $[a,\infty)$ or $(−\infty,b]$ a (closed) ray. If $E$ and $E'$ are any two rays, then $\Bbb R=T_\alpha(\Bbb R)=T_\alpha(E)=T_\alpha(E')$, so $T_\alpha(E)=T_\alpha(E')$ does not imply $E=E'$, and $T_\alpha(\Bbb R)=T_\alpha(E)$ does not imply $\Bbb R=E$.

The following is not an answer to question 2, yet it does give the feeling that one gets a little bit better understanding as to what one might expect from an answer. As noted earlier, one could rewrite $T_\alpha(E)$ as $T_\alpha(E) = \alpha E - (\alpha-1)E$.
Then using the $\alpha\gamma$ Lemma, one could further rewrite $T_\alpha(E)$ as
$T_\alpha(E) = (1+\alpha-1)E - (\alpha-1)E=$
$1E+(\alpha-1)E - (\alpha-1)E=E+H$, where $H=(\alpha-1)E - (\alpha-1)E$, and $H$ is symmetric, $H=-H$. If $\alpha$ is very close to $1$ (so $\alpha-1$ is positive but very close to $0$) then $H$ is a very small symmetric set, and $F=E+H$ has generally "the shape of" $E$, though a bit more rounded. One the other hand, if $\alpha$ is very big, then, intuitively, $E$ is very small, almost negligible, compared to the big set $H$, so in this case $F=E+H$ would tend to look almost like a symmetric set. This is a bit imprecise of course, but hopefully contributes to a better understanding of what $F=E+H$ might look like. A precise characterization, if possible, might be based upon the study of sets $E+H$, where $E$ is closed and convex, and $H=(\alpha-1)E - (\alpha-1)E$ (with the hope that studying such sets might be easier than working directly with the definition of $T_\alpha(E)$, or at least to complement it).

Suppose that $F$ is a convex closed set that is not a translate of a symmetric convex closed set (i.e. not of the form $R=t+S$ where $S=-S$). Then a necessary condition for $F$ to be of the form $T_\alpha(E)$ for some convex closed $E$ is that $F$ be decomposable, that is $F=K+L$ where at least one of $K$ and $L$ is not of the form $\lambda F+t$ for some $\lambda\ge0$ and $t\in\Bbb R^n$. Indeed $F=E+H$ (where $H=(\alpha-1)E - (\alpha-1)E$), and $H$ is symmetric, while $F$ is not a translate of a symmetric set, thus $H$ cannot be of the form $\lambda F+t$. (It might be that $F$ has to be decomposable regardless of whether it is a translate of a symmetric set or not.)

A triangle in the plane is indecomposable. Most (in Baire category sense) closed convex sets are indecomposable (so they cannot be of the form $T_\alpha(E)$ for any closed convex $E$). That seems to suggest that very few sets $F$ are of the form $T_\alpha(E)$. There are some $F=T_\alpha(E)$ that are not a translate of a symmetric set (e.g. in the plane when $E$ is an equilateral triangle with side $1$ and $\alpha=2$, for a specific example, then $E-E$ is a regular hexagon with side length $1$, and $F=E+E-E$ is a hexagon with opposite sides of length $1$ and $2$.

There seem to be plenty of literature to study on (in)decomposable closed convex sets (convex bodies), I found some references though I am looking for better ones (or at least more of them).

G. T. Sallee,
Minkowski decomposition of convex sets
Israel Journal of Mathematics
September 1972, Volume 12, Issue 3, pp 266–276. https://link.springer.com/article/10.1007%2FBF02790753

Walter Meyer
Indecomposable Polytopes
Transactions of the American Mathematical Society
Vol. 190 (Apr., 1974), pp. 77-86 https://www.jstor.org/stable/1996951?seq=1#page_scan_tab_contents
https://pdfs.semanticscholar.org/828e/4a4fd2a0696bf31be94091402195dee6dae1.pdf

A Course on Convex Geometry
Daniel Hug, Wolfgang Weil
University of Karlsruhe
revised version 2009/2010
January 24, 2011
https://www.fmf.uni-lj.si/~lavric/hug&weil.pdf
(exercises 8,9 sect.3.1, p.71)

It is not clear to me if there will be a good characterization, since the answer to question 3 in yes, for bounded $E$. That is, each bounded convex closed $E$ produces its own $F=T_\alpha(E)$. There are as many $F$'s as $E$'s. It might be interesting to describe a procedure that, given any closed convex $F$ would produce closed convex $E$ with the property that either $F=T_\alpha(E)$ or else $F\not=T_\alpha(E')$ for any closed convex $E'$. If $F$ is a translate of a symmetric closed convex set, such a procedure was described above (and always produces) $E$ with $F=T_\alpha(E)$. But perhaps there is some procedure that always produces some $E$, and if $F\not=T_\alpha(E)$ then we would know that there is no $E'$ with $F=T_\alpha(E')$. But all this is becoming too contemplative, so I will just post this edit for now.

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  • $\begingroup$ If $\alpha$ is very big then $T_\alpha(E)=\alpha E-(\alpha-1)E$ is very "close" to a set of the form $G-G$ (with $\alpha E\approx G\approx(\alpha-1)E$) and the set $G-G$ is symmetric (i.e. $G-G=-(G-G)$). So, intuitively, the answer to question 2 (for large and perhaps not so large $\alpha$) is yes, only for sets $F$ that are very close to being symmetric. I need to think what exactly this means and how to formalize it (but that would be another day). $\endgroup$ – Mirko Jun 17 at 6:26
  • $\begingroup$ If $F=T_\alpha(E)$ then $F=\alpha E-(\alpha-1)E=$ $(1+\alpha-1)E-(\alpha-1)E=$ $E+(\alpha-1)E-(\alpha-1)E=E+H$ where $H$ is the symmetric set $(\alpha-1)E-(\alpha-1)E$. Given a closed convex $F$ could we find closed convex $C$ and $K$, with $K$ also symmetric, such that $F=C+K$? If so, could we find closed convex $E$ with $F=T_\alpha(E)=E+H$ (where $H=(\alpha-1)E-(\alpha-1)E$)? $\endgroup$ – Mirko Jun 17 at 7:19
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    $\begingroup$ Thank you for all this. I think I figured out the answer to Question 3 in the bounded case. I share your intuition about 2 and the "symmetry" requirement but I am having a hard time formalizing it. Anyway, thanks for your help! $\endgroup$ – Oliv Jun 17 at 17:33
  • $\begingroup$ @Oliv If $F$ is not symmetric,but $F=T_\alpha(E)=E+H$ where $H=(\alpha-1)E-(\alpha-1)E$, then (since $F$ is not symmetric but has a symmetric summand $H$) we have that $F$ is not indecomposable. Most convex bodies (Baire category sense) are indecomposable. Google Minkowski sum indecomposable, in particular see encyclopediaofmath.org/index.php/Minkowski_addition (and some difference between $n=2$ and $n\ge3$ cases) I wish they had more specific references. P.S. What is your preferred reference for convex sets (e.g. basic facts about supporting functions, enough examples)? Thank you $\endgroup$ – Mirko Jun 18 at 5:24
  • $\begingroup$ Your use of the word "indecomposable" helped me find an interesting literature that might be helpful for this problem, thank you very much. My main reference is simply Rockafellar's textbook ("Convex analysis"). $\endgroup$ – Oliv Jun 18 at 17:42
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This is a proof that the answer to 3 is positive if $E$ and $E'$ are bounded. Combined with Mirko's counterexamples if $E$ and $E'$ are unbounded, this provides the answer to Question 3. I am still interested in the other two questions.

For any closed, convex and bounded set $A$, let $h_A: \mathbb{R}^n \rightarrow \mathbb{R}$ be the support function of $A$, defined by $h_A(x)=\max_{a \in A}{a \cdot x}$.

Take any closed, convex and bounded set $E$. Note that \begin{equation*} h_{T_\alpha(E)}(x) =\alpha h_E(x) + (\alpha-1) h_{-E}(x) = \alpha h_E(x) + (\alpha-1) h_{E}(-x). \end{equation*}

Similarly, \begin{equation*} h_{T_\alpha(E)}(-x) =\alpha h_E(-x) + (\alpha-1) h_{E}(x). \end{equation*}

Solving this system to eliminate $h_E(-x)$ yields \begin{equation*} \alpha h_{T_\alpha(E)}(-x) - (\alpha-1) h_{T_\alpha(E)}(-x) =(2\alpha-1) h_E(x). \end{equation*}

Thus, $T_\alpha(E)=T_\alpha(E')$ implies $h_E(x)=h_{E'}(x)$ for all $x$. By uniqueness of the convex set associated with a given support function $h$, this implies $E=E'$.

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