I am trying to find a counterexample to the products of two functions which are lower semicontinous at $a$. So let those functions be $g_1,g_2$
Does $g_1 = -1$ and $g_2 = 1$ for $x\neq a$ and $0$ when $x=a$
I am trying to find a counterexample to the products of two functions which are lower semicontinous at $a$. So let those functions be $g_1,g_2$
Does $g_1 = -1$ and $g_2 = 1$ for $x\neq a$ and $0$ when $x=a$
Take $g_1(x) = -1$, $g_2 = 1_{(0,\infty)}$. Both, are lsc., but the product $f(x) = -1_{(0,\infty)}(x)$ is not lsc.
$f(\frac{1}{n})=-1$ for all $n\in \mathbb{N}$, but $f(0) = 0$, so $\liminf_{x \to 0}f(x) = -1 < 0$, or alternatively, the set $\{x| f(x) > -1\} = (-\infty,0]$ is not open.