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Let be $(X, \{ p_i \}_{i \in I} )$ a locally convex space, $M_0\subset X$ a bounded and nonempty set and $f = l + I_{M_0}$ where l is a continuous function and \begin{equation*} I_{M_0}(x)= \begin{cases} 1 & \text{if} \hspace{.1cm} x \in M_0 \\ 0 & \text{if} \hspace{.1cm} x \notin M_0 \end{cases} \end{equation*} Then f is a sequentially lower semicontinuous function? I guess the problem comes down to prove that $I_{M_0}$ is sequentially lower semicontinuous since $l$ is a continuos function. Update: $M_0$ is sequentially closed.

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  • $\begingroup$ The prototypical lsc. functions are indicator functions of open sets. Take $l=0$, $M_0 = [0,1]$ where $X$ is the reals. $\endgroup$ – copper.hat Jun 18 at 23:55
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The question has been updated in a comment.

False even on the real line. Can you think of a bounded non-empty set $M_0$ such that $I_{M_0}$ is not l.s.c.? Hint: take $M_0=(0,1]$.

Answer to the edited question: again consider the real line. if $M_0$ is closed then $I_{M_0}$ is upper semicontinuous, not lower semicontinuous. For example take $M_0=[0,1]$ and consider the sequence $x_n =1+\frac 1 n$. This sequence tends to $x=1$, $I_{M_0}(x_n)=0$ for all $n$, $I_{M_0}(x)=1$. Hence $I_{M_0}$ is not lower semicontinuous.

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  • $\begingroup$ What about this math.stackexchange.com/questions/706671/… ? $\endgroup$ – The Student Jun 18 at 23:40
  • $\begingroup$ Your set is just bounded and non-empty. There is no assumption on whether it is open or closed. $\endgroup$ – Kabo Murphy Jun 18 at 23:44
  • $\begingroup$ Sorry for omitting it, $M_0$ is sequentially closed. $\endgroup$ – The Student Jun 18 at 23:50

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