# Calculating Legendre Transform

Let $\Omega,\Omega^*$ be bounded domain in $\mathbb{R}^n$, $B_r(y_0)$ is a small ball of radius $r$ center at $y_0\in\Omega^*$, in $\Omega^*$

define a function $\psi(y)=-\sqrt{(r^2-|y-y_0|^2)}$ on $B_r(y)$

I want to calculate the function obtained by applying the Legendre transform on $\psi$, namely $$u_0(x)=\sup\{x\cdot y-\psi(y):y\in B_r(y_0)\}$$

What is the explicit expression of $u_0$?

I tried to apply the calculus method to find the interior maximum, i.e. fix a point $y^*\in B_r(y_0)$

By differentiation, $$D_y\Big(x\cdot y+\sqrt{r^2-|y-y_0|^2}\Big)=x-\frac{y-y_0}{\sqrt{r^2-|y-y_0|^2}}$$

So I deduce $u_0(x)=x\cdot y^*+\sqrt{(r^2-|y^*-y_0|^2)}=x\cdot y^*+\frac{y^*-y_0}{x}$,

but my question is, can I have an expression of $u_0(x)$ only in terms of $x$ but without $y^*$? it semms difficult to make $y$ as subject of formula. Thanks for help!

## 1 Answer

For a fixed point $x$, the optimal $y^*$ satisfies: $$x-\frac{y^*-y_0}{\sqrt{r^2-|y^*-y_0|^2}} = 0$$ So the vectors $x$ and $y-y_0$ must be multiples of each other. Let $q = |y^*-y_0|$, so $y^* = y_0 + qx/|x|$. Then by some algebraic manipulation, we can solve for $q$ and hence $y^*$:

$$|x|=\frac{q}{\sqrt{r^2-q^2}} \Leftrightarrow |x|^2(r^2-q^2)=q^2 \Leftrightarrow q =\frac{|x| \ r}{\sqrt{1+|x|^2}}$$

$$y^*-y_0 = \frac{x \ r}{\sqrt{1+|x|^2}}$$

So plugging this into the expression for $\psi^*(x)$ gives: \begin{aligned} \psi^*(x) &= x \cdot y^* - \psi(y^*) \\ &= x \cdot (y_0 + q x/|x|) + \sqrt{r^2-q^2} \\ &= x \cdot y_0 + q |x| + \frac{q}{|x|} \\ &= x \cdot y_0 + q\frac{|x|^2+1}{|x|} \\ &= x \cdot y_0 + \frac{|x| \ r}{\sqrt{1+|x|^2}}\frac{|x|^2+1}{|x|} \\ &= x \cdot y_0 + r \sqrt{1+|x|^2} \end{aligned}