Let $\Omega,\Omega^*$ be bounded domain in $\mathbb{R}^n$ and $u_0$ be a uniformly convex function define on $\Omega$. Suppose the gradient of $u_0$ maps $\Omega$ into a subdomain of $\Omega^*$, i.e. $\omega^*=Du_0(\Omega)\subset\Omega^*$
Apply the Legendre transform of $u_0(x)$ to give a function $$v_0(y)=\sup_{x\in\Omega}\{x\cdot y-u_0(x)\}:y\in \omega^*$$
Now on $\Omega^\delta=\{x\in\mathbb{R}^n|\text{ dist}(x,\Omega)<\delta\}$, define a function $u_1$ which is also convex, and extends $u_0$ to $\Omega^\delta$ (i.e. $u_1=u_0$ in $\Omega$)
Consider the Legendre transform of $u_1$ to give a function $v$, $$v(y)=\sup_{x\in\Omega^\delta}\{x\cdot y-u_1(x)\},y\in\Omega^*$$. I want to ask, is $v$ the same as $v_0$ in $\omega^*$?
Yes, $v(y)=v_0(y)$ for every $y\in \omega^*$. Indeed, let $\hat x$ be such that $Du_0(\hat x)=y$. The function $w(x) = x\cdot y-u_1(x)$ has zero gradient at $\hat x$. Since $w$ is convex, its value at $\hat x$ is equal to its minimum.
The proof does not use the assumption that $u_0$ is uniformly convex.