(a\int_{P_+\cap\{y-\lambda=0\}}x^2 \,d\sigma_0\right).
\end{align*}
Thus $f'(\lambda)\neq 0$ unless $\lambda=-v_{P}((0,-1)),0$ or $v_P((0,1))$. Since $f(-v_{P}((0,-1)))=f(v_{P}((0,1)))=0$ we get $\mathcal L (\bar u')=f(\lambda_0)>0$, a contradiction.


\emph{Case-2.} The set $\mathcal S_0$ is a polytope which has an edge $E$ totally lies in $P_+$. From the proof of Proposition \ref{ker-thm}, there is a dominate convex, simple piecewise linear function
\begin{align}\label{u-test-func}
\bar u'=\max\{l_{\mathbf u},0\}\not\equiv0
\end{align}
where $l_{\mathbf u}(x,y)=u_1x+u_2y-\lambda_0$ with $\mathbf u=(u_1,u_2),~u_1\geq0$ so that
\begin{align}\label{L-u=0}
\mathcal L(\bar u')=\int_{P_+\cap\{u_1x+u_2y-\lambda_0\geq0\}}(u_1(x-a)+u_2y)ax^2\,dx\wedge dy=0
\end{align}
and the crease of $\bar u'$ equals to $E$. Then $\lambda_0>0$. Consider
$$f(\lambda):=\int_{P_+\cap\{u_1x+u_2y-\lambda\geq0\}}(u_1(x-a)+u_2y)ax^2\,dx\wedge dy,~0\leq \lambda\leq v_{P}(\mathbf u).$$
Here, $v_P(\cdot)$ is the support function of $P$. Since $\alpha\in\text{Int}(P_+)$, we have $v_P(\mathbf u)>u_1a$.

By direct computation,
\begin{align*}
f'(\lambda)&=-|\mathbf u|\int_{P_+\cap\{u_1x+u_2y-\lambda=0\}}(u_1(x-a)+u_2y)ax^2 \,d\sigma_0\\
&=-(\lambda-u_1a)\left(a|\mathbf u|\int_{P_+\cap\{u_1x+u_2y-\lambda=0\}}x^2 \,d\sigma_0\right).
\end{align*}
Obviously, we have
\begin{align}\label{f'}
&f(0),f'(0)\geq0,~f'(au_1)=f'(v_P(\mathbf{u}))=0,\notag\\
&f'(\lambda)>0~\text{when}~0<\lambda<u_1a~\text{and}~f'(\lambda)<0~\text{when}~u_1a<\lambda< v_P(\mathbf{u}).
\end{align}
Also note that $f(v_P(\mathbf{u}))=0$. By our assumption, we have $\lambda_0\in(0,v_{P}(\mathbf{u}))$ such that $f(\lambda_0)=0$. However, by \eqref{f'}, $f(\lambda)>0$ for any $\lambda \in(0, v_P(\mathbf{u}))$. We obtain a contradiction again.


\emph{Case-3.}
The set $\mathcal S_0=P\cap W_\alpha$. Thus $\bar u|_{P\cap W_\alpha}=0$. From \emph{Case-2.2} in the proof of Proposition \ref{ker-thm} we see that $\mathcal L(\ell_{\mathbf{u_0}})=0$ for $\mathbf{u_0}=(1,0)$. Consider a point $z_0$ in the interior of $P_+$. Let $\phi_0$ be a support function of $\bar u$ at $z_0$ and $\mathcal T_0$ be the contact set $\{\bar u=\phi_0\}$. Then $\nabla\phi_0\in\mathcal V_+$ and $\mathcal T_0\subsetneq P_+$. Otherwise $\bar u=\phi_0$ and one directly gets \eqref{ker-L=eq}.


We have two subcases:


\begin{figure}[h]
\begin{tikzpicture
\fill[color=gray!20]  (0.846,-0.846-0.8) -- (0,-2) -- (0,2) -- (0.564,1.2+0.564);
\draw[semithick] (-1.41,1.41) -- (0,2) -- (1.41,1.41) -- (2,0) -- (1.41,-1.41) -- (0,-2) -- (-1.41,-1.41) -- (-2,0) -- (-1.41,1.41);
\draw[very thick]  (0.564,1.2+0.564)  -- (0.846,-0.846-0.8);
\draw [very thick, -latex] (0,0) -- (1,0);
\draw (1,0) node{$\bullet$};
\draw (0,0) node{$\bullet$};
\draw (-0.2,-0.2) node {$O$};
\draw (1.3,0.2) node {$\alpha$};
\draw (0.3,-1.73/2) node {$\mathcal T_0$};
\draw (0.85,+1.73/2) node {$E$};
\draw (0.4,-2.25) node {$W_{\alpha}$};
\draw[dashed] (-2.4,0) -- (2.4,0);
\draw[dashed] (0,-2.4) -- (0,2.4);
\draw[very thick]  (0.846,-0.846-0.8) -- (0,-2) -- (0,2) -- (0.564,1.2+0.564);
\end{tikzpicture}
\caption{\emph{Case-3.1}}
\end{figure}

\vskip 10pt

\emph{Case-3.1.} There is a point $z_0$ so that $\mathcal T_0$ intersects an interior point of $P\cap W_\alpha$. Then $\phi_0$ is affine and nonpositive along $P\cap W_\alpha$, attaining its maximum $0$ at an interior point. Hence $\phi_0|_{P\cap W_\alpha}=0$ and $\nabla \phi_0=c\alpha$ for some constant $c>0$. Since $z_0\not\in W_\alpha$, we conclude that $\mathcal T_0$ is the intersection of some half-space with $P_+$, which has an edge $E\subset P_+$ with vertices on $\partial P$. In particular, $E$ does not intersect $W_\alpha$. Suppose that
\begin{align}\label{edge-eq}
E\subset\{(x,y)|xv_1+yv_2-\lambda_0=0\}
\end{align}
for some $\mathbf v=(v_1,v_2)\in\mathcal V_+$ and $\lambda_0>0$. Using the arguments of Propostion \ref{ker-thm}, we conclude that
$\mathcal L(\bar v)=0$  for
$$\bar v=\max\{xv_1+yv_2-\lambda_0,0\}.$$
This is impossible as showed in \emph{Case-2}.


\begin{figure}[h]
\begin{tikzpicture
\fill[color=gray!20] (0.564,1.2+0.564) -- (1.41,1.41) -- (2,0) -- (1.41,-1.41) -- (0.141,-0.141-1.8);
\draw[semithick] (-1.41,1.41) -- (0,2) -- (1.41,1.41) -- (2,0) -- (1.41,-1.41) -- (0,-2) -- (-1.41,-1.41) -- (-2,0) -- (-1.41,1.41);
\draw[very thick]  (0.564,1.2+0.564)  -- (0.141,-0.141-1.8);
\draw [very thick, -latex] (0,0) -- (1,0);
\draw (1,0) node{$\bullet$};
\draw (0,0) node{$\bullet$};
\draw (-0.2,-0.2) node {$O$};
\draw (1.3,0.2) node {$\alpha$};
\draw (1.