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.
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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 $x$.
answered Dec 9, 2020 at 19:30
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$\begingroup$ 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... $\endgroup$– ABIMCommented Jan 7, 2021 at 10:15
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$\begingroup$ 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. $\endgroup$ Commented Jan 7, 2021 at 10:32
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$\begingroup$ 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? $\endgroup$– ABIMCommented Jan 7, 2021 at 10:39
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$\begingroup$ The answer is yes: The only point is that $x \to x_n$ is continuous, since finite sums of continuous functions are continuous. $\endgroup$ Commented Jan 7, 2021 at 11:29
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$\begingroup$ 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? $\endgroup$– ABIMCommented Jan 7, 2021 at 12:43