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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 $x=a$

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$g_1(x)\times g_2(x) = g_1(x)$ so the product has the same properties as one of the original functions –  Henry Dec 10 '12 at 7:43
    
But at $x = a$, $0 \times -1 = 0$ no? –  sidht Dec 10 '12 at 7:51
    
It seems your statement is ambiguous about the value of $g_1(a)$. In any case $g_1(x)\times g_2(x) = -g_2(x)$ so the product still has the same continuity properties as one of the original functions –  Henry Dec 10 '12 at 7:55
    
No $g_1 = -1 \forall x$, it's a horizontal line. and $g_2 = 1$ with a hole at $x = a$ –  sidht Dec 10 '12 at 7:58
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1 Answer

up vote 1 down vote accepted

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.

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For $g_2$ what happened to the other intervals? I don't see what's wrong with my example –  sidht Dec 10 '12 at 8:11
    
Your example is fine too, in your case, $g_2 = 1_{\mathbb{R}\setminus \{a\}}$, and the same reasoning as in my example (replacing $0$ by $a$) shows that the product is not lsc. –  copper.hat Dec 10 '12 at 8:15
    
Sorry, but do you mind clarifying what Henry was saying then? –  sidht Dec 10 '12 at 8:23
    
Henry was incorrect. $g_2$ (your's or mine) is lsc., but $-g_2$ is not; they do not have the same continuity properties. –  copper.hat Dec 10 '12 at 16:13
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