The piece-wise average of $W^{1,1}$ function Let $Q=(0,1)\times(0,1)$ be given, and $u\in W^{1,1}(Q)$. Define
$$
\bar u(x) = \int_{0}^1 u(x_1,t)\,dt
$$
for $x=(x_1,x_2)\in Q$. That is, $\bar u$ is sort of a piece-wise average of $u$.
Since $u\in W^{1,1}$, so $u$ is $\mathcal H^1$ a.e. defined and hence $\bar u(x)$ is well-defined.
My question: would it be possible to show that $\bar u\in W^{1,1}(Q)$, and, if yes, do we have 
$$
\|\nabla u\|_{L^1}\geq \|\nabla \bar u\|_{L^1}
$$ 
as well?

I am thinking of using smooth approximation to do it. It looks to me that if $u\in C^\infty(Q)\cap W^{1,1}$ then the result hold.

PS: the reason I ask this question is I wish to construct the following approximation:
Define $Q_k:=(0,1)\times(k/n,(k+1)/n)$ for $1\leq k<n\in\mathbb N$. Then I wish to define an approximation sequence $u_n$ of $u$ such that $u_n\in W^{1,1}(Q_k)$ for each $k$ and
$$
\partial_{x_2}u_n=0
$$
for $x\in Q_k$. I wish the sequence $u_n$ has following properties:
$$
u_n\to u\text{ in }L^2\tag 1
$$
$$
\|\nabla u\|_{L^1(Q)}\geq \limsup_{n\to\infty}\sum_{k=1}^n \|\nabla u_n\|_{L^1(Q_k)}\tag2
$$
and 
$$
\|Tu\|_{L^1(\{1\}\times(0,1))} = \lim_{n\to\infty}\sum_{k=1}^n\|Tu_n\|_{L^1(\{1\}\times(k/n,(k+1)/n))}
$$
where $T$ stands for the truce operator.
I am planning to build such $u_n$ by using above construction in each $Q_k$. If the above statement is true, then I am done. 
 A: Like what you said, we consider the mapping $\bar{(\cdot)}$ which sends $u$ to $\bar u$. I want to show that this is a bounded linear mapping $W^{1,1}(Q)\to W^{1,1}(Q)$ and $\|\nabla \bar u \|_1\le \| \nabla u\|$. It suffice to check $ u$ smooth. So 
$$\begin{split} \bar u (x_1+ h, x_2) - \bar u(x_1,x_2) &= \int_0^1 u(x_1+h, t) dt - \int_0^1 u(x_1, t)dt \\
& =\int_0^1 \big(u(x_1+h, t) - u(x_1,t) \big)dt \\
&= \int_0^1 \int_0^h \partial_1 u(x_1+s, t)ds dt
\end{split}$$
Thus $\partial_1 u$ exists and is $\partial_1 \overline u = \overline{\partial_1u} $. Thus (since $\partial_2 \bar u  =0$)
$$\begin{split}
\|\nabla \overline u\|_{L^1} &= \int_0^1\int_0^1 | \partial_1 \bar u(x_1, x_2)| dx_1dx_2 \\
&= \int_0^1\int_0^1 \left| \int_0^1 \partial_1u (x_1, t)dt\right|dx_1dx_2 \\
&= \int_0^1 \left| \int_0^1 \partial_1u (x_1, t)dt\right|dx_1 \\
&\le \int_0^1  \int_0^1 \left|\partial_1 u(s, t)\right| dtds \le \|\nabla u\|_{L^1}.
\end{split}$$
Since similarly one has $\| \overline  u\|_{L^1}\le \| u\|_{L^1}$, the mapping extends to a bounded linear operator $W^{1,1}(Q) \to W^{1,1}(Q)$ with the required inequality. 
