$L^1$ convergence to equilibrium of solutions of heat equation Let $u$ and $v$ be the weak solutions of
$$u_t - \Delta u = f$$
$$u(0)=u_0$$
and
$$-\Delta v = f$$
$$|\Omega|^{-1}\int_\Omega v =0$$
on a bounded domain $\Omega$, where $u$ and $v$ satisfy homogeneous Neumann BCs. Here we may take $f$ and $u_0$ to have spacial mean values zero.
I'm trying to prove an estimate of the form
$$\lVert{u(t)-v}\rVert_{L^1(\Omega)} \leq C(t)\lVert{u_0-v}\rVert_{L^1(\Omega)}$$
where $C(t) \to 0$ as $t \to \infty$.
I can prove this if instead of $L^1$ norms we had $L^2$, but I need $L^1$ on both sides. Does anyone know how to achieve this? The $L^2$ case can be done with testing the weak form satisfied by $u-v$ with $u-v$, Poincare's inequality and actually we will find $C(t) = e^{-Kt}$ after using it as an integrating factor.
 A: Here is an idea. Put $w=u-v$, $w_0=u_0-v$. The required inequality takes form $\lVert{w}\rVert_{L^1(\Omega)} \leq C(t)\lVert{w_0}\rVert_{L^1(\Omega)}$. Denote $G(x,y,t)$ the Green function of the Neumann problem for the heat equation in $\Omega$. There is an integral representation of the solution:
$$
w(x,t)=\int_\Omega G(x,y,t)w_0(y)\,dy.
$$
Since $w_0$ has mean value zero, denoting $G'(x,y,t)=G(x,y,t)-|\Omega|^{-1}$, it can be rewritten as
$$
w(x,t)=\int_\Omega G'(x,y,t)w_0(y)\,dy.
$$
So it is enough to prove the inequality with 
$C(t)=\max_{x,y\in \Omega}|G'(x,y,t)|$. May be it is proven somewhere that the last quantity tends to zero as $t\to+\infty$. 
If not, let $\lambda_1=0<\lambda_2<\ldots$ and $\varphi_1$, $\varphi_2,\ldots$ be eigenvalues and eigenfunctions of the Neumann problem in $\Omega$, $\|\varphi_n\|_{L_2(\Omega)}=1$. Then 
$$
G'(x,y,t)=\sum_{n=2}^\infty \varphi_n(x)\varphi_n(y)e^{-\lambda_n t}.
$$
If values $\max_{x\in \Omega}|\varphi_n(x)|$ grow not too quickly the result follows. Say, for one-dimensional case $\Omega=[0,\pi]$ eigenfunctions  $\varphi_n(x)=\cos nx$ are uniformly  bounded  and one can take the constant in the form $C(t)=Me^{-\lambda_2 t}$.
