Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. For $m \in \mathbb{N}$ and $M > 0$ we denote by $T_M$ the truncation of vectors in $\mathbb{R}^m$ to length $M$, i.e., $$T_M(x) = \begin{cases} x & \text{if } \|x\| \le M, \\ M \, \frac{x}{\|x\|} & \text{if } \|x\| > M.\end{cases}$$ (We use the Euclidean norm in $\mathbb{R}^m$)
Then $T_M$ is globally Lipschitz with modulus $1$ and thus, $x \mapsto T_M (u(x))$ belongs to $H_0^1(\Omega)^m$ for all $u \in H_0^1(\Omega)^m$, see, e.g., here. We denote this function by $T_M u$.
Do we have $T_M u \to u$ in $H_0^1(\Omega)^m$ as $M \to \infty$?
Some thoughts:
- Since $T_M u$ is bounded in $H_0^1(\Omega)^m$ and converges pointwise to $u$, we get weak convergence in $H_0^1(\Omega)$.
- The case $m = 1$ is classical. Here, one can use a cain rule and $\|\nabla(T_M u)(x)\| \le \|\nabla u(x)\|$ f.a.a. $x \in \Omega$. Then one can use dominated convergence for the gradient to get strong convergence in $H_0^1(\Omega)$. Such a chain rule seems not exist in the vectorial case above.
The following is inspired by the usual proof that for $u\in H^1(\Omega)$ also $u^+\in H^1(\Omega)$.
We have $$ T_M u = u \min\left(1, \frac M{\|u\|} \right) = u\left (1+ \min\left(0, \frac M{\|u\|} -1\right) \right) $$
Define the following approximation of $f(x)=\min(0,x)$ $$ f_\sigma(t):= \begin{cases}\sigma - \sqrt{x^2 +\sigma^2} & x<0\\0& x\ge 0\end{cases}, $$ with the properties $$ \ |f_\sigma(t)|\le |\min(0,t)|, \quad |f_\sigma'(t)|\le1 . $$ Then define the approximation of $T_M$ $$ T_M^\sigma u = u \cdot \left(1 + f_\sigma\left( \frac M{\sqrt{\|u\|^2+\sigma}}-1 \right )\right) $$
$$ \begin{aligned} \nabla (T_M^\sigma u)_i& = \nabla u_i \cdot \left(1 + f_\sigma\left( \frac M{\sqrt{\|u\|^2+\sigma}}-1 \right )\right)\\ &\qquad+ u_i f_\sigma'\left( \frac M{\sqrt{\|u\|^2+\sigma}}-1 \right) \frac{-M}{(\sqrt{\|u\|^2+\sigma})^3}(\sum_j u_j \nabla u_j) \end{aligned}$$
Here we have the pointwise majorants $$ \left|f_\sigma\left( \frac M{\sqrt{\|u\|^2+\sigma}}-1 \right )\right| \le \max\left(0, 1- \frac M{\sqrt{\|u\|^2+\sigma}}\right) \le 1 $$ and $$ \|T_M^\sigma u\| \le 2 \|u\| . $$ In the next estimate we only consider points, where $M\le \sqrt{\|u\|^2+\sigma}$, as otherwise $f_\sigma'(\dots)=0$ and the estimate is trivial. In this case $$ \begin{aligned} \|(\nabla T_M^\sigma u)_i\| &\le 2 \|\nabla u_i\| + M^{-2}\|u_i\| \|\sum_j u_j \nabla u_j\| \le 2 \|\nabla u_i\| + M^{-2}\|u_i\| \|u\| \cdot (\sum_j \|\nabla u_j\|^2)^{1/2}\\ &\le 2 \|\nabla u_i\| + \sum_j \|\nabla u_j\|. \end{aligned}$$ Both bounds are integrable and independent of $\sigma$ (and of $M$). Hence $T_M^\sigma u \to T_Mu$ in $L^2(\Omega)$, and one can also pass to the limit $\sigma\searrow0$ in the equation $$ \int_\Omega T_M^\sigma u_i D_j\phi =-\int_\Omega D_jT_M^\sigma u_i \phi \quad \phi \in C_0^\infty(\Omega) $$ to obtain a characterization of $\nabla T_Mu$ as pointwise limit of $\nabla T_M^\sigma u_i$. The complicated part is passing to the limit here: $$ u_i f_\sigma'\left( \frac M{\sqrt{\|u\|^2+\sigma}}-1 \right ) \frac{-M} {(\sqrt{\|u\|^2+\sigma})^3}. $$ We have $f_\sigma'(x) = - \frac x{x^2+\sigma^2}$ for $x<0$. If $M\le \|u\|$ then $$ \begin{split} f_\sigma'\left( \frac {M -\sqrt{\|u\|^2+\sigma}}{\sqrt{\|u\|^2+\sigma}} \right ) & = - \frac {M -\sqrt{\|u\|^2+\sigma}}{\sqrt{\|u\|^2+\sigma}} \frac1{ \sqrt{\frac {(M -\sqrt{\|u\|^2+\sigma})^2}{\|u\|^2+\sigma} + \sigma^2}}\\ &= - \frac {M -\sqrt{\|u\|^2+\sigma}}{ \sqrt{(M -\sqrt{\|u\|^2+\sigma})^2 + (\|u\|^2+\sigma)\sigma^2}} \to 1 \text{ for } \sigma \searrow0 \end{split} $$ Hence if $M\le \|u\|$ then $$ \nabla T_M^\sigma u_i \to \nabla u_i \min\left(1, \frac M{\|u\|} \right) - u_i \cdot 1\cdot \frac{M}{\|u\|^3}\sum_j u_j \nabla u_j. $$ This yields the following expression for $\nabla (T_M u)_i$: $$ \nabla (T_M u)_i = \nabla u_i \min\left(1, \frac M{\|u\|} \right) - \begin{cases} u_i \cdot \frac{M}{\|u\|^3}\sum_j u_j \nabla u_j & \text{ if } M\le \|u\|\\ 0 & \text{ if } M> \|u\|\\ \end{cases} $$ Passing to the limit $M\to\infty$ in the first term is trivial, the second one can be bounded using $M\le \|u\|$ by $$ \left|u_i \cdot \frac{M}{\|u\|^3}\sum_j u_j \nabla u_j\right| \le \sum_j \|\nabla u_j\|, $$ hence $\nabla(T_M u)_i\to 0$ in $L^2(\Omega)$ for $M\to\infty$ by Lebesgue dominated convergence.