Let $g:\Omega\to \mathbb{R}$, $\Omega\subseteq \mathbb{R}^2$ open, a polynomial functions in the variables $x$ and $y$, i.e. $$ g(x,y)= \sum_{\ell=0}^{m}\sum_{\mu+\nu=\ell} b_{\mu\nu}x^{\mu}y^{\nu}. $$ Suppose $[-1,+1]\times [-1,+1]\subset \Omega$ and that the function $g$ satisfy the conditions: $$ \begin{array}{ccc} g(x,-1)<0 \mbox{ for all } x\in[-1,1] &\quad \mbox{ and }\quad & g(x,1)>0 \mbox{ for all } x\in[-1,1] \\ \end{array} $$ For each $x\in [-1,+1]$ fix the set $$ \Lambda(x)=\left\{y\in [-1,1] \left|\begin{array}{rl} g(x,y)&=0\\ (\partial/\partial y)g(x,y)&>0 \end{array}\right.\right\} $$ Set the function $\varphi:[-1,+1]\to [-1,+1]$ by $$ \varphi(x)= \min\overline{\Lambda(x)}, $$ that is, $\varphi(x)$ is the minimum of the set of adherent points of set $\Lambda(x)$.
Question. Is the function $ \varphi:[-1,+1]\to [-1,+1] $ continuous?
My attempt. In fact, let $x_n\to \tilde{x}\in [-1,+1]$. I'm trying to prove that $\lim_{n\to\infty}\varphi(x_n)=\varphi(\tilde{x})$.For each $x_n\in [-1,+1]$ there is a $y_n\in\overline{\Lambda(x_n)}$ such that $$ y_n=\min \overline{\Lambda(x_n)} $$ I'm trying to use the fact that both $\Lambda(x_n)$ and the range $[-1,+1]$ are sequentially compact. But I'm not getting any success. The problem is that when $ y_n $ varies for $ y_{n + 1} $ the set $ \Lambda (x_n)$ varies for $ \Lambda (x_{n + 1}) $.
First of all, $\varphi(x)$ may not even be defined, since $\Lambda(x)$ can be empty. For instance, if $g(x,y)=y^3$ then $\Lambda(x)$ is empty for all $x$.
Even when $\varphi$ is defined, it may not be continuous. For instance, let $g(x,y)=(y^2-x/4)(y-2/3)$. Then $2/3\in\Lambda(x)$ for all $x$ so $\varphi(x)$ is always defined. Moreover, $2/3$ is the only point in $\Lambda(x)$ if $x\leq 0$, so $\varphi(x)=2/3$ for $x\leq 0$. But when $x>0$, $\varphi(x)=-\sqrt{x}/2$, and in particular $\varphi(x)$ approaches $0$ as $x\to 0$ from above.