Convergence of vector-valued truncation in $H_0^1(\Omega)^m$ 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.

 A: 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.
