1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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: $$ \begin{aligned} \psi^*(x) &= x \cdot y^* - \psi(y^*) \\ &= x \cdot (y_0 + q x/|x|) + \sqrt{r^2-q^2} \\ &= x \cdot y_0 + q |x| + \frac{q}{|x|} \\ &= x \cdot y_0 + q\frac{|x|^2+1}{|x|} \\ &= x \cdot y_0 + \frac{|x| \ r}{\sqrt{1+|x|^2}}\frac{|x|^2+1}{|x|} \\ &= x \cdot y_0 + r \sqrt{1+|x|^2} \end{aligned} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.