Uniqueness for second order PDE $-\Delta u = c(u)$ I'm trying to solve the following problem:
Let $\Omega$ be an open set in $\mathbb R^n$ and consider the equation
\begin{cases}
-\Delta u = c(u) & \text{ on } \Omega \\
 u = 0 & \text{ on } \partial\Omega.
\end{cases}
The function $c:\mathbb R \to \mathbb R$ satisfies 
$$\sup_{t\in \mathbb R} |c'(t)| \leq \alpha < \lambda_1 (\Omega),$$
where $\lambda_1 (\Omega)$ representes the first eigenvalue of the laplacian in $\Omega$, and $\alpha$ is a constant.
I want to show that this PDE has a unique solution.
 A: I can show it in the case that $\Omega$ is bounded.
Let $(\lambda_i)$ be the eigenvalues of the Dirichlet laplacian $L:=-\Delta$ and $(\phi_i)$ its corresponding eigenfunctions $(-\Delta \phi_i=\lambda_i\phi_i)$ which constitute an orthonormal basis of $L^2(\Omega)$. It is known by the spectral theorem for compact Hermitian operators that the eigenvalues are positive and have no accumulation points. Thus, we can suppose $0<\lambda_1(\Omega)=:\lambda_1\leq\lambda_2 \leq\dots$.
Suppose $u,v\in H^2(\Omega)$ are solutions of the boundary value problem and let $w:=u-v$. Then $w$ is a solution of
\begin{cases}
L w = c(u)-c(v) & \text{ in } \Omega \\
 w = 0 & \text{ on } \partial\Omega.
\end{cases}
Now write $ c(u)-c(v)$ in terms of the derivative of $c$ using the mean value theorem to get $L w (x)= c'\circ\xi(x) \cdot (u(x)-v(x)) = c'\circ\xi(x) \cdot w(x)$ for some $\xi(x)$ lying between $u(x)$ and $v(x).$
By Parseval's identity,
$$\begin{align}
\Vert w \Vert^2 &=\sum |\langle w,\phi_i\rangle |^2 \\
&=\sum \frac 1 {\lambda_i^2}|\langle w,\lambda_i\phi_i\rangle |^2 \\
&=\sum \frac 1 {\lambda_i^2}|\langle w,L \phi_i\rangle |^2 \\
&=\sum \frac 1 {\lambda_i^2}|\langle L w,\phi_i\rangle |^2 \quad \text{since }L\text{ is self-adjoint,}\\
&=\sum \frac 1 {\lambda_i^2}|\langle (c'\circ\xi ) \cdot w,\phi_i\rangle |^2 \\
&\leq \sum \frac 1 {\lambda_1^2}|\langle (c'\circ\xi) \cdot w,\phi_i\rangle |^2=\left\Vert \frac{c'\circ\xi}{\lambda_1}w\right\Vert^2. 
\end {align}$$
The last norm is finite because $|c'\circ \xi /\lambda_1|$ is bounded above by $ \alpha/\lambda_1<1$ and, thus, doesn't affect the convergence of the integral.
However, if $w$ were not identically $0$ we could say that $$\int \left|\frac{c'\circ\xi}{\lambda_1}w\right|^2\le \int \left|\frac{\alpha}{\lambda_1}w\right|^2 < \int |w|^2,$$
so that  $\Vert w \Vert^2 \leq \left\Vert \frac{c'\circ\xi}{\lambda_1}w\right\Vert^2 < \Vert w \Vert^2$, a contradiction.
