# Lower semi-continuity of induced function on sequences

Let $$f:X\rightarrow [0,\infty)$$ be (resp. weakly) lower semi-continuous on the reflexive Banach space $$X$$. Let $$\ell^p(X)$$ denote the space of $$p$$-summable sequences in $$X$$, i.e.: $$\sum_{n=1}^{\infty}\|x_n\|_x^p<\infty$$; here $$1\leq p<\infty$$. Then, is the "induced" map: $$F:(x_n)\mapsto \sum_{n=1}^{\infty} f(x_n),$$ (resp. weakly) lower semi-continuous? I assume that there exists at-least one $$p$$-summable sequence for which $$F$$ is finite-valued.

Let $$x = (x_n) \in \ell^p(X)$$ and $$F_N(x) := \sum_{n=1}^N f(x_n)$$, $$N \in \mathbb{N}$$. First if each $$F_N$$ is l.s.c. (weakly or not), then $$F = \sup_{N \in \mathbb{N}} F_N$$ is l.s.c., second if each $$x \to f(x_n)$$ is l.s.c., then $$F_N$$ is l.s.c. as a sum of finite many l.s.c. functions and finally $$x \to x_n$$ is even continuous. (See f.i. Bourbaki (1989), General Topology IV.6.2). Hence your function $$F$$ is l.s.c. even without the assumption of $$F(x)$$ being finite at some point $$x$$.

• I mean to say: If f is weakly l.s.c, then is $F_N$ weakly l.s.c also? I find this confusing since if X is finite-dimensional then f is both weakly and strongly l.s.c so $F_N$ must both be weakly and strongly lsc. on $\ell^p(X)$..which I only thought happens when things are convex...
– ABIM
Jan 7, 2021 at 10:15
• Sorry, I can't follow your argument, Continuity is a topological concept, which has nothing to do with linearity, convexity, linear structure etc. per se. Of course additional structure simplifies some arguments in proving continuity. Jan 7, 2021 at 10:32
• Yes, but the weak topology is a TVS concept so it relies explicitly on linearity (therefore so does weak l.s.c-ty). Either way, my question is: is $F_N$ ever weakly lsc if $X$ is finite-dimensional and $f$ is continuous?
– ABIM
Jan 7, 2021 at 10:39
• The answer is yes: The only point is that $x \to x_n$ is continuous, since finite sums of continuous functions are continuous. Jan 7, 2021 at 11:29
• Actually, wouldn't the point be that since linear functionals are continuous then the (projection) map (in the other direction) $\pi_n:(x_n)_n\mapsto x_n$ is weakly continuous, and since $f$ is weakly cnt then $f\circ \pi_n:\ell^p\rightarrow \mathbb{R}$ is weakly cnt; so, $\sum_{n=1}^N f\circ \pi_n:\ell^p \rightarrow \mathbb{R}$ is weakly cnt and then their supremum is weakly lsc?
– ABIM
Jan 7, 2021 at 12:43