Let $\Omega \subset \mathbb{R}^d$ a regular domain (compact boundary $C^1$) e let $u \in H^1(\Omega)$. Let $tr(u) \equiv \alpha$.
I would like to prove that the truncated function $$\bar{u}=u \chi_{\{u < \alpha \}}+\alpha \chi_{\{u \geq \alpha \}}$$ is still a Sobolev function with $\dfrac{\partial u}{\partial x_i}=\dfrac{\partial u}{\partial x_i}\chi_{\{u <1\}}$ and $tr(\bar{u})=1$.
I tried to apply the direct definition with the existence of weak derivative but I wasn't able to do that.
Yes, this is the case.
Note that $u-\alpha\in H_0^1(\Omega)$. It is a known result that $\max(v,0)\in H_0^1(\Omega)$ if $v\in H_0^1(\Omega)$, see here for example (using $G(x)=\max(x,0)$), or here.
Using $$ \bar u = \max(u-\alpha,0)+\alpha $$ it follows that $\bar u\in H^1(\Omega)$.
old answer (before an important correction was made in the original post):
No, this is not the case.
We choose $d=1$, $\Omega=(0,2\pi)$ and the function $$ u(x) = 2+\sin(x). $$ Clearly, $u\in H^1(\Omega)$ and $\alpha=2$. However, $\bar u$ is not continuous and has a jump at $x=\pi$: $$ \bar u(x) = \begin{cases} 1 & x\in (0,\pi) \\ 2+sin(x) & x\in [\pi,2\pi) \end{cases} $$ Therefore it is not a Sobolev function. (Note that for $d=1$ the functions in $H^1(\Omega)$ are continuous.)
