\begin{proposition}\label{separating-equilibrium}
A separating equilibrium where the green firm labels and the non-green firm does not label is sustainable under beliefs $\mu_{s=1}=1$ and $\mu_{s=0}\leq q$  if:
\begin{enumerate}
    \item $p > \delta$, and $\gamma_I > 0$
    \item $p=\delta$
    \item $\frac{L_0+\sqrt{L_0^2-4\gamma_I A}}{2A} < p \leq \mu_0 \delta$, $L_0 \leq \frac{2 \left( \delta \gamma_I \mu_0 - \sqrt{B} \right)}{\delta \mu_0 q}$, and $\gamma_I \geq q$, where $A=\frac{q-\mu_0 \delta(1-L_0)}{\mu_0 \delta q}$ and  $B=\delta \gamma_I \mu_0 \left( \delta \gamma_I \mu_0 - \delta \mu_0 q + q^2 \right)$
\end{enumerate}

\end{proposition}

\begin{appendixproof}[Proof of Proposition \ref{separating-equilibrium}]

Let $\mu_{s=0}\equiv \mu_0$. The purchase probabilities are:
\begin{align}
\theta_{s=0} &= \begin{cases}
    1-\frac{p}{\mu_0 \delta} \text{ if } \frac{p}{\mu_0 \delta} \in [0,1] \\
    0 \text{ if } \frac{p}{\mu_0 \delta} > 1
\end{cases}\\
\theta_{s=1} &= \begin{cases}
    1-\frac{p}{\delta} \text{ if } \frac{p}{\delta} \in [0,1] \\
    0 \text{ if } \frac{p}{\delta} > 1
\end{cases}
\end{align}
Since $\mu_0 \leq q \leq 1$, we have that $\frac{p}{q \delta} > 1$ implies $ \frac{p}{\mu_s \delta}>1$, and $\frac{p}{\mu_s \delta} \in [0,1]$ implies $\frac{p}{q \delta} \in [0,1]$. Therefore the possible combinations $(\theta_{s=0},\theta_{s=1})$ are:
\begin{enumerate}
    \item $(0,0)$
    \item $\left(0,1-\frac{p}{\delta}\right)$
    \item $\left(1-\frac{p}{\mu_0 \delta},1-\frac{p}{\delta}\right)$
\end{enumerate}

In a separating equilibrium, a green firm should prefer to signal, and the brown firm should prefer not to signal. The respective conditions are:

\begin{align}
\theta_{s=1} p(1-L_0)-C-a+\gamma_I &> \theta_{s=0} p-C-a \\
(\theta_{s=1}(1-L_0)-\theta_{s=0})p +\gamma_I &>  0 \label{separatinginequality1}
\end{align}
and 
\begin{align}
\theta_{s=0} p - C  & \geq \theta_{s=1} p(1-L_0-K_0) -C \\ 
\Rightarrow \theta_{s=1}(1-L_0-K_0)-\theta_{s=0} & \leq 0 \label{separatinginequality2}
\end{align}
i) If $\theta_{s=1}=0$ and $\theta_{s=0} =0$, then $p > \delta$, and Inequality \ref{separatinginequality2} is always satisfied, and Inequality \ref{separatinginequality1} is satisfied if $\gamma_I>0$. 

ii) In the second case, the conditions yield $\mu_0 \delta < p \leq \delta$. Substituting the probabilities into Inequality \ref{separatinginequality1}:
\begin{align}
\left(1-\frac{p}{\delta}\right)(1-L_0)p +\gamma_I &>0 \label{separatinginequality3} \\
\Rightarrow -(1-L_0)p^2+\delta(1-L_0)p+\delta\gamma_I &> 0
\end{align}
Since the leading coefficient is negative, \( p \) must be between the roots:

\begin{align}\label{separatinginequality4}
  \frac{\delta}{2} - \frac{\sqrt{\delta^2(1-L_0)^2 + 4(1-L_0)\delta\gamma_I}}{2(1-L_0)}< p < \frac{\delta}{2} + \frac{\sqrt{\delta^2(1-L_0)^2 + 4(1-L_0)\delta\gamma_I}}{2(1-L_0)}
\end{align}
The left side is always negative, so the inequality reduces to:
\begin{align}\label{separatinginequality5}
  0< p < \frac{\delta}{2} + \frac{\sqrt{\delta^2(1-L_0)^2 + 4(1-L_0)\delta\gamma_I}}{2(1-L_0)}
\end{align}
Next substituting the probabilities into Inequality \ref{separatinginequality2}
\begin{align}
    \left(1-\frac{p}{\delta}\right)(1-L_0-K_0)  &\leq 0 \label{separatinginequality6} \\
    p &\geq \delta
\end{align}
So this case only holds when $p=\delta$.

iii) In the third case, the condition is $p \leq \mu_0 \delta$. Substituting the probabilities into Inequality \ref{separatinginequality1}:
\begin{align}\label{separatinginequality7}
    \left(\left(1-\frac{p}{\delta}\right)(1-L_0)-\left(1-\frac{p}{\mu_0\delta}\right)\right)p +\gamma_I &>  0 \\
    \left(\frac{q-\mu_0 \delta(1-L_0)}{\mu_0 \delta q}\right)p^2-L_0 p+\gamma_I &> 0
\end{align}
Because $\mu_0 \leq q$ by assumption, the coefficient of the leading term is positive. The solution set of the inequality is:
\begin{equation}\
    p < \frac{L_0-\sqrt{L_0^2-4\gamma_I A}}{2A}
\end{equation}
or 
\begin{equation}
    p > \frac{L_0+\sqrt{L_0^2-4\gamma_I A}}{2A}
\end{equation}
where $A=\frac{q-\mu_0 \delta(1-L_0)}{\mu_0 \delta q}$. The first solution is always negative, so only the second solution is admissible. Therefore the valid price range is:
\begin{align}
    \frac{L_0+\sqrt{L_0^2-4\gamma_I A}}{2A} < p \leq \mu_0 \delta
\end{align}
We must also ensure the discriminant is non-negative:
\begin{align}
    L_0^2-4\gamma_I A &\geq 0 \\
    \Rightarrow \mu_0 \delta q L_0^2-4\gamma_I \mu_0 \delta L_0+4 \gamma_I(\mu_0\delta-q)&\geq 0
\end{align}
The solutions are 
\begin{equation}\label{separatinginequality8}
L_0 \leq \frac{2 \left( \delta \gamma_I \mu_0 - \sqrt{\delta \gamma_I \mu_0 \left( \delta \gamma_I \mu_0 - \delta \mu_0 q + q^2 \right)} \right)}{\delta \mu_0 q}
\end{equation}

and

\begin{equation}\label{separatinginequality9}
L_0 \geq \frac{2 \left( \delta \gamma_I \mu_0 + \sqrt{\delta \gamma_I \mu_0 \left( \delta \gamma_I \mu_0 - \delta \mu_0 q + q^2 \right)} \right)}{\delta \mu_0 q}.
\end{equation}
Finally, we must ensure the discriminant of Inequalities \ref{separatinginequality8} and \ref{separatinginequality9} is non-negative, which happens if $\mu_0(\gamma_I-q)\geq -q^2$. We can ensure this is true by assuming $\gamma_I \geq q$. If this is the case, then Inequality \ref{separatinginequality9} cannot hold, because it is greater than 2, and $L_0$ must be less than unity. 
\end{appendixproof}

can you check that the intuitive criterion holds in these three cases for the separating PBE?