# Prove that a function is sequentially lower semicontinuous

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.

• The prototypical lsc. functions are indicator functions of open sets. Take $l=0$, $M_0 = [0,1]$ where $X$ is the reals. – copper.hat Jun 18 at 23:55

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.
• Sorry for omitting it, $M_0$ is sequentially closed. – The Student Jun 18 at 23:50