A problem about $C^1$-convergence! (Elliptic theory) Let a function $u:\overline\Omega\subset\mathbb{R}^n\rightarrow\mathbb{R}$ that satisfies
$$\Delta u+f(u)=0 \ \ \ \mbox{in} \ \ \Omega,$$
and consider
$$w(x_1,...,x_{n-1})=\lim_{x_n\rightarrow+\infty}u(x_1,...,x_n).$$
Why the function $u(x_1,...,x_n)$ converges uniformly in the $C^1$ sense in compact sets of $\mathbb{R}^{n-1}$ to the function $w$, and $w$ satisfies
$$\Delta w+f(w)=0.$$
This argument is used in the paper: Further qualitative properties for elliptic equations in unbounded domains, by Berestycki, Caffarelli and Nirenberg. I didn't undesrtand this point. Someone can help me? Thanks.
 A: This is not a general statement; it is a consequence of certain assumptions in the proposition where it is used (I assume you are on page 82 of the paper like I am). 
In particular, $u$ is assumed to be a bounded monotone solution, in the sense that $\frac{\partial u}{\partial x_n} \geq 0$, and $f$ is assumed to be $C^1$ (this can be weakened to Lipschitz, probably, but since the paper says $C^1$, let's work with that). 
Let $\Omega = \omega \times \mathbb{R}_+$, where $\omega$ is some compact subset with smooth boundary of $\mathbb{R}^{n-1}$.
Now since $u$ is uniformly bounded, consider $u_k(x) = u(x', x_n + k)$ where $x' \in \mathbb{R}^{n-1}$, $x \in \mathbb{R}_+$. In any compact subset of $\Omega$, this is uniformly bounded in the $C_0$ norm as $k \rightarrow \infty$, hence $f(u_k)$ is uniformly bounded as $f$ is $C^1$.  Since $$ \Delta u_k + f(u_k) = 0$$, $u_k$ is uniformly bounded in the $C^1$ norm by elliptic estimates (see, e.g., Lemma 4.2 of Gilbarg-Trudinger), hence there is some subsequence that converges in the $C^\alpha$ norm for any $\alpha < 1$ (this is by the compact embedding of Holder spaces). Since we already know that they converge pointwise to $w$, that means that the whole sequence converges in $C^\alpha$. 
Consider now the function $u_k - u_{k'}$ as $k,k'$ go to infinity. We consider the equation
$$ -\Delta (u_k - u_{k'}) = f(u_k) - f(u_{k'}) $$
Since $f$ is Lipschitz, we have 
$$| f(u_k) - f(u_{k'}) | \leq \|f\|_{Lip} |u_k - u_{k'}| \leq \|f\|_{Lip} \|u_k - u_{k'}\|_{C^\alpha}$$
Applying elliptic estimates on the Poisson equation again, this tells us that the $u_k$ are a $C^1$ Cauchy sequence and hence their convergence to $w$ is in the $C^1$ norm. 
Now let's examine how we know $w$ satisfies the equation. Morally speaking, $w$ satisfies 
$$\Delta w + f(w) = 0$$
because in the limit, the $x_n$ derivative of the $u$ disappears - it is constant in one direction, and hence does not contribute to the Laplacian. Making this rigorous requires a detour since we only have that the convergence is in $C^1$, and here I will invoke the theory of weak solutions. We have 
$$ \int \nabla u_k \nabla \phi dx = \int f(u_k) \phi dx$$
for test functions $\phi$. Over compact sets, we have that $\nabla u_k \rightarrow \nabla w$ and $f(u_k) \rightarrow f(w)$, and hence $w$ satisfies the equation in the weak sense. From here, you can invoke elliptic estimates  to see that $w$ solves the equation in the classical sense, if you desire. 
(Expanded update of the weak solution to classical, in response to question in comment). On second thought, it may be easier to use straightforward Schauder theory rather than estimates for weak solutions. Since $w$ is a $C^1$ function and so is $f$, that means $f(w)$ is $C^1$, in particular it is at least $C^\alpha$ (on any compact subset, that is). Hence the Schauder theory means that 
$$\Delta v = - f(w)$$
has a unique $C^{2,\alpha}$ solution $v$. Since $w$ is a weak solution of the same equation, the two coincide. 
