Let $h:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ be a continuous function such that $\lim_{x\to \pm\infty}h(x,y)=0$ for all $y$. Then we know that if $L(y)=\min_{x\in \mathbb{R}}h(x,y)$ exists, it must be $L(y)\leq 0$. Now, is it true that if $L(y)$ is well defined in a interval $(a,b]$ with $a<b$ then it is also well defined at the point $a$?
Intuitively this seems to be right, but I couldn't find a way to prove it.
Any hint or suggestion would be appreciated.
The answer is no. Let $f(t)$ be a positive-valued function with $f(0)>1$ and $f(-1)<1$ and $\lim_{t\to\pm\infty} f(t)=0$, such as $f(t)=2/(1+2t^2)$ or $f(t)=2e^{-t^2}$ (its exact value is unimportant). Then define $$ h(x,y) = f(x)\big(1-f(xy-1)\big). $$ Note that $h(x,0) = f(x)\big(1-f(-1)\big) > 0$ for all $x\in\Bbb R$. However, for $y\ne0$, $$ h\big(\tfrac1y,y\big) = f(y)\big(1-f(0)\big) < 0. $$ Once the limits as $x\to\pm\infty$ are verified, these inequalities show that $L(y)$ exists for all $y\ne0$ but does not exist for $y=0$.
(At the core, this example exploits the useful fact that the hyperbola $xy=1$ is a closed set in $\Bbb R^2$ whose projection onto either coordinate is not closed.)