Let $A_n := \{x \in \ell^1 \colon x_1 \not= 0, \ldots, x_{n-1} \not= 0\}$, $g_n \colon \ell^1 \to [0,\infty]$ be defined by $g_n(x) := f(x_n)$ if $x \in A_n$ and $g_n(x) := 0$ if $x \not\in A_n$. Then $F_f(x) = \sum_{n=1}^\infty g_n(x)$. Since the sum of two l.s.c. functions and the supremum of a sequence of (nonnegative) l.s.c. functions is l.s.c. it is sufficient to show that each $g_n$ is l.s.c. But this follows from 1. $A_n$ being open in $\ell^1$, thus $g_n$ is l.s.c. at each $x \in A_n$, since $f$ is l.s.c., and 2. that $g_n \geq 0$ and $g_n(x) = 0$ if $x \not\in A_n$.

Let $A_n := \{x \in \ell^1 \colon x_1 \not= 0, \ldots, x_{n-1} \not= 0\}$, $g_n \colon \ell^1 \to [0,\infty]$ be defined by $g_n(x) := f(x_n)$ if $x \in A_n$ and $g_n(x) := 0$ if $x \not\in A_n$. Then $F_f(x) = \sum_{n=1}^\infty g_n(x)$. Since the sum of two l.s.c. functions and the supremum of a sequence of (nonnegative) l.s.c. functions is l.s.c. it is sufficient to show that each $g_n$ is l.s.c. But this follows from 1. $A_n$ being open in $\ell^1$, thus $g_n$ is l.s.c. at each $x \in A_n$, since $f$ is l.s.c., and 2. that $g_n \geq 0$ and $g_n(x) = 0$ if $x \not\in A_n$. Thus $F_f$ is l.s.c.