Example of a plane region with smooth boundary whose polar dual also has a smooth boundary

Question

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 $p=q=2$.

Answer 1

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.

Answer 2

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

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

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.

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

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

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