For some $p$ in $(1, \infty)$ and any real function $f$ let:
$$L(f, a, b) = \sup \left\{ \sum_{i=1}^k \big((x_i - x_{x-1})^p + (f(x_i) - f(x_{i-1}))^p \big)^\frac{1}{p} \mid k\in\mathbb{N}, a=x_0<x_1<\ldots<x_k=b \right\}$$
be the $L^p$ length of the graph of $f$ on $[a, b]$. $L(f, a, b) \in [0, \infty]$. Let:
$$V(f, a, b) = L(f, a, b) - (b-a),$$ $$V(f) = \lim_{b \rightarrow \infty} V(f, 0, b).$$
I have two questions about $V$.
If $f_n$ is a sequence of functions such that $\lim_{m, n} V(f_m - f_n) = 0$, is there a function $f$ such that $\lim_n V(f - f_n) = 0$? In other words is the space given by $V$ complete?
If there is such a function $f$, does $\lim_n V(f_n) = V(f)$? In other words, is $V$ "continuous"?
Here is what I have tried. For each $b$, $\lim_{m, n} V(f_m - f_n, 0, b) = 0$ so $f_n$ converges to some $f$ in the supremum norm on $[0, b]$ so $\lim_n V(f - f_n, 0, b) = 0$. Since an $f$ exists on $[0, b]$ for all $b$ and all such $f$ agree, $f_n$ converges pointwise to some $f$ on $[0, \infty]$. But it is not clear to me that $f_n$ converges uniformly so it is not clear that $\lim_n V(f - f_n) = 0$. Specifically, every term of:
$$V(f - f_n) = \sum_{i=0}^\infty V(f - f_n, i, i+1)$$
goes to 0 but it does not follow that the sum does.