Let $X$ be a metric space. A real-valued function $f : X \rightarrow \mathbb{R}$ is upper semicontinuous if it satisfies one of the followings:

$(1)$ For all $c \in \mathbb{R}$, its preimage $f^{-1}(-\infty,c)$ is open in $X$.

$(2)$ For all $x \in X$ and all $\varepsilon>0$, there exists an open neighbourhood $U$ of $x$ such that for all $y \in U,$ we have $f(y) < f(x) + \varepsilon$.

I know that finite sum of upper semicontinuous is upper semicontinuous.

Question: Suppose for each natural number $n$ , $f_n$ is a non-negative upper semicontinuous function and $(f_n)$ is decreasing. Assume that $\sum_{n=1}^{\infty}(-1)^nf_n$ converges uniformly to $g$. Is $g$ an upper semicontinuous function?

  • $\begingroup$ The sum of u.s.c. function is u.s.c., but their difference need not be u.s.c. $\endgroup$ – Rigel May 6 '17 at 13:27
  • $\begingroup$ How about uniform sum of usc? $\endgroup$ – Idonknow May 6 '17 at 15:12

Define $f_1=\chi_{[0,3]}$, $f_2=\chi_{[1,2]}$. Then $f_1-f_2$ is non-negative but not upper semicontinuous: $$ (f_1-f_2)^{-1}(-\infty,1/2)=(-\infty,0) \cup [1,2]\cup (3,\infty), $$ which is not open.

By setting $f_n=0$ for $n>2$ all the conditions in your question are satisfied, but the resulting function is not upper semicontinuous.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.