Let $f(x)$ be a left continuous and non-increasing real-valued function. Can I prove that $f(x)x$ is upper semicontinuous?
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$\begingroup$ obvious: the product of two semicontinuous functions is semicontinuous and $f(x)$ decreasing and left-continuous is semicontinuous $\endgroup$– ralphJan 24, 2014 at 18:21
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$\begingroup$ What do you mean "semicontinuous".... typically we talk about semicontinuous at a point. Like in this picture: en.wikipedia.org/wiki/File:Upper_semi.svg $\endgroup$– SquirtleJan 24, 2014 at 19:17
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$\begingroup$ @Squirtle: ...And a function that is semicontinuous at every point of its domain of definition is simply said to be "semicontinuous". $\endgroup$– Alex M.Nov 6, 2018 at 21:38
2 Answers
This is not only obvious, it's also false. Let $$f(x)=\begin{cases} 1,\quad &x\le -1;\\ 0, & x>0\end{cases}$$ The product $g(x)=xf(x)$ is not upper semicontinuous at $-1$, because $$g(-1)=-1 < \limsup_{x\to -1}g(x)=0$$
According to proposition 2 on page 362 of Bourbaki's "General Topology: Chapters 1–4", the product of two positive upper-semicontinuous functions is upper-semicontinuous. It follows that the answer two your question depends on the domain of definition of your function:
- if it is contained in $[0, \infty)$, then $xf$ is upper-semicontinuous;
- if it contains negative numbers, the answer given by @you can call me AI shows that $xf$ may fail to be upper-semicontinuous.