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I am looking for a concrete example of a compact convex subset $K$ of $\mathbb{R}^2$ with $(0,0)$ center of symmetry, with a smooth boundary ( a $C^{\infty}$ $1$ dimensional submanifold of $\mathbb{R}^2$), such that its polar dual $K^{\circ}$ also has smooth boundary. Even better if both have analytic boundaries. Moreover, $K$ should not be an ellipse.

Comments: $K^{\circ}$, the convex dual is defined by $$K^{\circ} = \{ v \in \mathbb{R}^2\ | \ \langle v,u\rangle \le 1 \text{ for all } u \in K \}$$ (if we use $|\cdot |$ in the inequality we get the same thing).

It is easy to see that $K^{\circ}$ is compact, convex and symmetric. Moreover, it is a fact (duality theorem) that $K^{\circ \circ} = K$.

Examples: $K=\{(x,y)\ |\ \frac{x^2}{a^2} + \frac{y^2}{b^2}\le 1\}$. Then $K^{\circ} = \{(x,y)\ |\ a^2 x^2 + b^2 y^2\le 1\}$. In general, if $K=\{(x,y)\ |\ \frac{|x|^p}{a^p} + \frac{|y|^p}{b^p}\le 1\}$ then $K^{\circ} = \{(x,y)\ | \ a^q |x|^q + b^q |y|^q\le 1\}$ where $q$ is the Holder dual of $p$.

In the above example with $p$, $q$, we see that we cannot make both $K$,$K^{\circ}$ with boundary $C^{2}$, unless $p=q=2$.

max_zorn
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orangeskid
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  • @SFTP: Thanks for posting the bounty! I am really curious if we'll see an explicit example. – orangeskid Apr 12 '18 at 02:23
  • @SFTP: If we could do real elimination of quantifiers we could find the expression of the dual of $x^4 + x^2 y^2 + y^4\le 1$ as a semialgebraic set. We know already that the corresponding norm is analytic. – orangeskid Apr 16 '18 at 00:18
  • I thought one might want to involve exp somehow, made a couple of plots with wolfram alpha, they look nice but I have no idea how to compute the dual, and no reason to believe that the dual has a smooth boundary. Say, $e^x+e^{-x}+e^y+e^{-y}=e+\frac1e+2$ or $e^{x^2}+e^{y^2}=e+1$. http://www.wolframalpha.com/input/?i=plot+e%5Ex%2Be%5E(-x)%2Be%5Ey%2Be%5E(-y)%3De%2B1%2Fe%2B2,+x%5E2%2By%5E2%3D1 and http://www.wolframalpha.com/input/?source=frontpage-immediate-access&i=plot+e%5Ex%5E2%2Be%5Ey%5E2%3De%2B1 – Mirko Apr 18 '18 at 01:59
  • @Mirko: What a surprise to see the first output..! The theory in the answer below allows one to conclude in some cases that the dual is also smooth (analytic). The problem with explicit examples is that some system of equations are hard to solve explicitly. Another way would be to give a convex set by a polynomial inequality, make sure that it is smooth with the criterion below and then try to find the dual with the definition, perhaps eliminating quantifiers. It's hard to find explicitely duals. I have some examples but they are not smooth ( and not centrally symmetric). – orangeskid Apr 18 '18 at 02:33
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    the following is a side product of my experiments with wolframalpha, it is not convex, but it came to a nice shape (that may have noting to do with the problem), so here it is for amusement http://www.wolframalpha.com/input/?source=frontpage-immediate-access&i=plot+(x%5E2%2F+e%5E(x%5E2))%2B(y%5E2%2F+e%5E(y%5E2))%3D1%2Fe – Mirko Apr 18 '18 at 05:28
  • @Mirko: Wow, super nice! i took level 1/e -00.1, just to see the metamorphosis! I love this stuff! – orangeskid Apr 18 '18 at 05:41

2 Answers2

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Given a norm $\|\cdot\|$, consider the function $f(x) = \frac12 \|x\|^2$. If this function is real-analytic outside of $0$, then the unit ball of the norm has analytic boundary, being a level set of $f$ (note that $\nabla f$ vanishes only at $0$).

The Legendre-Fenchel conjugate of $f$ is $g(x) = \frac12\|x\|_*^2$ where $\|\cdot \|_*$ is the dual norm. (Reference). Thus, we want $g$ to be real-analytic as well.

A basic property of conjugate convex functions is that $\nabla g$ is the inverse of $\nabla f$. Also, the inverse of a real-analytic map with nonzero Jacobian is real-analytic.

Thus, what we need is for $f$ to have nonzero Jacobian, which is equivalent to $\|x\|^2$ having nonsingular Hessian matrix. (This is the property that fails for norms like $(x_1^4+x_2^4)^{1/4}$.)

There is an easy way to "fix" any smooth $f$ as above: just add some multiple of the squared Euclidean norm to it. This will add a positive constant to all eigenvalues of the Hessian, making it positive definite.

For example, $$ \|x\| = \sqrt{x_1^2+x_2^2+\sqrt{x_1^4+x_2^4}} $$ is a norm such that both the unit ball $\{x:\|x\|\le 1\}$ and its polar $\{x:\|x\|_*\le 1\}$ have real-analytic boundary.

  • Thank you! That is Very informative. Any chance of having a pair of dual norms like that in "explicit" form? – orangeskid Apr 09 '18 at 12:43
  • had some hopes with $|x|^2=\sqrt{x_1^4 + x_1^2 x_2^2 + x_2^4}$ but still not a nice system for the jacobian of the dual squared. – orangeskid Apr 12 '18 at 02:22
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There is a family of examples of the form $xf(x)+y\,g(y)=1$, where $f,g,f^{-1}$ and $g^{-1}$ are analytic. If $u=f(x)$ and $v=g(y)$ then the above equation becomes $xu+yv=1$, in turn $uf^{-1}(u)+v\,g^{-1}(v)=1$, which represents the dual curve.

Consider the curve $x\sinh(x) + y \sinh(y) = 1$. If we let $u=\sinh(x)$ and $v=\sinh(y)$, then $x u + y v = 1$ (matching the definition of the dual curve), and hence, using that $x=\sinh^{-1}(u)=\ln(u+\sqrt{u^2+1})$ and $y=\sinh^{-1}(v)=\ln(v+\sqrt{v^2+1})$, we obtain the equation for the dual curve $u \sinh^{-1}(u) + v \sinh^{-1}(v)=1$.

The first curve $x\sinh(x) + y \sinh(y) = 1$

first curve $x\sinh(x) + y \sinh(y) = 1$

The second curve $u \sinh^{-1}(u) + v \sinh^{-1}(v)=1$ (the dual of the first)

dual curve $u \sinh^{-1}(u) + v \sinh^{-1}(v)=1

Both curves together:

both curves together

Edit. The above is an example of the form $x f(x) + y f(y)=1$ with dual curve $u f^{-1}(u) + v f^{-1}(v)=1$. The circle is a special case of this family of examples, when $f(x)=x=f^{-1}(x)$, but an ellipse is not. A more general family of examples is $x f(x) + y\, g(y)=1$ with dual curve $u f^{-1}(u) + v\, g^{-1}(v)=1$. An ellipse is a special case of the latter family with $f(x)=\frac x{a^2},f^{-1}(x)=a^2x,g(x)=\frac x{b^2},g^{-1}(x)=b^2x$.

As an example in the latter family that is not an ellipse one may take $f(x)=\tan^{-1}(x), f^{-1}(x)=\tan(x),g(x)=\sinh(x), g^{-1}(x)=\sinh^{-1}(x)$. That is, the curve $x \tan^{-1}(x) + y \sinh(y) = 1$ with dual curve $u \tan(u) + v \sinh^{-1}(v)=1$, shown below.

a more general example

Edit. A few more examples showing how a curve and its dual may relate, on a picture.

$x\sin(x)+y\arcsin(y)=1$ with dual $u\arcsin(u)+v\sin(v)=1$, shown below

$x\sin(x)+y\arcsin(y)=1$ with dual $u\arcsin(u)+v\sin(v)=1$

$x^2+y\tan(y)=1$ with dual $u^2+v\arctan(v)=1$, shown below

$x^2+y\tan(y)=1$ with dual $u^2+v\arctan(v)=1$

$x^2+y(e^y-1)=1$ with dual $u^2+v\ln(1+v)=1$, shown below

$x^2+y(e^y-1)=1$ with dual $u^2+v\ln(1+v)=1$

Mirko
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    Yes, they are analytic since the functions $f(x, y) = x\sinh x +y\sinh y$ and $g(x, y) = x\sinh^{-1} x +y\sinh^{-1} y$ are real-analytic with nonvanishing gradient in $\mathbb{R}^2\setminus {(0, 0)}$. –  Apr 18 '18 at 04:08
  • Very nice! I am starting to understand these things better:) – orangeskid Apr 18 '18 at 04:18
  • @bro thank you, it looks like $x^2+y^2=1$ (a self-dual circle) is a special case (when $f(x)=x$, the identity function) of a family of examples of the form $xf(x)+yf(y)=1$ with dual $uf^{-1}(u)+vf^{-1}(v)=1$. I didn't think of general conditions on $f$, but it looks like in the case when $f(x)=\sinh(x)$ it helps that $x$ and $f(x)$ have the same sign (so that $x f(x)\ge0$), and that $\sinh(x)$ is monotonically increasing. An ellipse is also of this form, well not exactly, since it is not all-way symmetric, but it is of form $xf(x)+yg(y)=1$, dual $uf^{-1}(u)+vg^{-1}(v)=1,f(x)=x/a^2,g(y)=y/b^2$. – Mirko Apr 18 '18 at 04:29