# 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
Commented 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. Commented 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
Commented 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. Commented 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
Commented Jan 7, 2021 at 12:43