First, since you are mapping a piecewise linear curve ($\partial\Omega_h$) onto a smooth one ($\partial\Omega$), a diffeomorphism $\omega: \Omega_h \rightarrow \Omega$ does not exist in the strong sense. What you want is a piecewise differentiable homeomorphism.

With that being said, as shown in this paper, Proposition 4.7, there exists a piecewise differentiable homeomorphism $\omega:\Omega_h \rightarrow \Omega$ that fulfils a second-order pointwise estimate

\begin{equation} \tag{1} \|\omega - Id\|_{L^\infty(\Omega_h)} \leq Ch^2, \end{equation}

but its Jacobian fulfils only a first-order pointwise estimate

\begin{equation} \tag{2} \label{jacobian} \|J - 1\|_{L^\infty(\Omega_h)} \leq Ch, \end{equation}

where $J$ is the weak Jacobian, because $\omega$ is piecewise differentiable only. Estimate \eqref{jacobian} is optimal, so you cannot get a second-order pointwise estimate for the Jacobian. A quick way to see this is that you are mapping the piecewise linear curve $\partial\Omega_h$ onto the smooth one $\partial\Omega$, which is interpolation in disguise. Now, piecewise linear interpolation provides first-order accuracy for the gradients, only.

If you are happy with an $L^p$, $1\leq p <+\infty$ estimate instead of $L^\infty$ in \eqref{jacobian}, then you can even get arbitrarily good accuracy, i.e. for any $\varepsilon > 0$ you can choose $\omega$ sufficiently close to the identity such that

\begin{equation} \tag{3} \|J - 1\|_{L^p(\Omega_h)} \leq \varepsilon, \end{equation}

as you correctly said in your comment.

First, since you are mapping a piecewise linear curve ($\partial\Omega_h$) onto a smooth one ($\partial\Omega$), a diffeomorphism $\omega: \Omega_h \rightarrow \Omega$ does not exist in the strong sense. What you want is a piecewise differentiable homeomorphism.

With that being said, as shown in this paper, Proposition 4.7, there exists a piecewise differentiable homeomorphism $\omega:\Omega_h \rightarrow \Omega$ that fulfils a second-order pointwise estimate

\begin{equation} \tag{1} \label{mapping} \|\omega - Id\|_{L^\infty(\Omega_h)} \leq Ch^2, \end{equation}

but its Jacobian fulfils only a first-order pointwise estimate

\begin{equation} \tag{2} \label{jacobian} \|J - 1\|_{L^\infty(\Omega_h)} \leq Ch, \end{equation}

where $J$ is the weak Jacobian, because $\omega$ is piecewise differentiable only. Estimate \eqref{jacobian} is optimal, so you cannot get a second-order pointwise estimate for the Jacobian. A quick way to see this is that you are mapping the piecewise linear curve $\partial\Omega_h$ onto the smooth one $\partial\Omega$, which is interpolation in disguise. Now, piecewise linear interpolation provides first-order accuracy for the gradients, only.

If you are happy with $L^p$, $1\leq p <+\infty$ estimates instead of $L^\infty$ in \eqref{mapping} and \eqref{jacobian}, you can improve those estimates as follows. Since the special choice of $\omega$ mentioned above is the identity everywhere on $\Omega_h$ except for a $h$-thick neighbourhood (or stripe) $S$ of the boundary $\partial \Omega_h$ (i.e. $\partial \Omega_h \subset S \subset \Omega_h$ with $\text{area}(S) \le Ch$), then it holds that

\begin{equation} \tag{3} \begin{split} \|\omega-Id\|_{L^p(\Omega_h)} &= \|\omega-Id\|_{L^p(S)} \le \text{area}(S)^{\frac{1}{p}} \|\omega-Id\|_{L^\infty(S)}\\ &= \text{area}(S)^{\frac{1}{p}}\|\omega-Id\|_{L^\infty(\Omega_h)} \le Ch^{2+\frac{1}{p}}; \end{split} \end{equation}

\begin{equation} \tag{4} \begin{split} \|J-1\|_{L^p(\Omega_h)} &= \|J-1\|_{L^p(S)} \le \text{area}(S)^{\frac{1}{p}} \|J-1\|_{L^\infty(S)}\\ &= \text{area}(S)^{\frac{1}{p}}\|J-1\|_{L^\infty(\Omega_h)} \le Ch^{1+\frac{1}{p}}. \end{split} \end{equation}

The bottom line is that $\omega$ is sufficiently close to the identity, as you correctly said in your comment.

First, since you are mapping a piecewise linear curve onto a smooth one, a diffeomorphism $\omega: \Omega_h \rightarrow \Omega$ does not exist in the strong sense. What you want is a piecewise differentiable homeomorphism.

With that being said, as shown in this paper, Proposition 4.7, there exists a piecewise differentiable homeomorphism $\omega:\Omega_h \rightarrow \Omega$ that fulfils a second-order pointwise estimate

\begin{equation} \tag{1} \|\omega - Id\|_{L^\infty(\Omega_h)} \leq Ch^2, \end{equation}

but its Jacobian fulfils only a first-order pointwise estimate

\begin{equation} \tag{2} \label{jacobian} \|J - 1\|_{L^\infty(\Omega_h)} \leq Ch, \end{equation}

where $J$ is the weak Jacobian, because $\omega$ is piecewise differentiable only. Estimate \eqref{jacobian} is optimal, so you cannot get a second-order pointwise estimate for the Jacobian. A quick way to see this is that you are mapping a piecewise linear curve onto a smooth one, which is interpolation in disguise. Now, piecewise linear interpolation provides first-order accuracy for the gradients, only.

If you are happy with an $L^p$, $1\leq p <+\infty$ estimate instead of $L^\infty$ in \eqref{jacobian}, then you can even get arbitrarily good accuracy, i.e. for any $\varepsilon > 0$ you can choose $\omega$ sufficiently close to the identity such that

\begin{equation} \tag{3} \|J - 1\|_{L^p(\Omega_h)} \leq \varepsilon, \end{equation}

as you correctly said in your comment.

First, since you are mapping a piecewise linear curve ($\partial\Omega_h$) onto a smooth one ($\partial\Omega$), a diffeomorphism $\omega: \Omega_h \rightarrow \Omega$ does not exist in the strong sense. What you want is a piecewise differentiable homeomorphism.

With that being said, as shown in this paper, Proposition 4.7, there exists a piecewise differentiable homeomorphism $\omega:\Omega_h \rightarrow \Omega$ that fulfils a second-order pointwise estimate

\begin{equation} \tag{1} \|\omega - Id\|_{L^\infty(\Omega_h)} \leq Ch^2, \end{equation}

but its Jacobian fulfils only a first-order pointwise estimate

\begin{equation} \tag{2} \label{jacobian} \|J - 1\|_{L^\infty(\Omega_h)} \leq Ch, \end{equation}

where $J$ is the weak Jacobian, because $\omega$ is piecewise differentiable only. Estimate \eqref{jacobian} is optimal, so you cannot get a second-order pointwise estimate for the Jacobian. A quick way to see this is that you are mapping the piecewise linear curve $\partial\Omega_h$ onto the smooth one $\partial\Omega$, which is interpolation in disguise. Now, piecewise linear interpolation provides first-order accuracy for the gradients, only.

If you are happy with an $L^p$, $1\leq p <+\infty$ estimate instead of $L^\infty$ in \eqref{jacobian}, then you can even get arbitrarily good accuracy, i.e. for any $\varepsilon > 0$ you can choose $\omega$ sufficiently close to the identity such that

\begin{equation} \tag{3} \|J - 1\|_{L^p(\Omega_h)} \leq \varepsilon, \end{equation}

as you correctly said in your comment.