Let $f:\mathbb{R}\rightarrow [0,\infty]$ be a lower semi-continuous function and define the functional
$$
\begin{aligned}
F\_f:&\ell^1 \rightarrow [0,\infty]\\
(x\_n)\_{n=0}^{\infty} &\to \sum\_{n=0}^{N((x\_n)\_{n=0}^{\infty})} f(x\_n),
\end{aligned}
$$
where $N\left((x\_n)\_{n=0}^{\infty}\right)=\inf\left\{
N\_0\in \mathbb{N}:\, \forall n \geq N\_0,\, x\_{n} =0
\right\}$. Then is $F\_f$ lower semi-continuous?



Lower semi-continuity of length-dependent functional
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.


