# Lower semi-continuity of length-dependent functional

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?

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.