# fa.functional analysis – On the uniform boundedness principle and the space of functions of bounded variation

Let $$U$$ be a bounded smooth domain of $$mathbb{R}^d$$. We write $$m$$ for the Lebesgue measure on $$U$$. A function $$f in L^1(U,m)$$ has bounded variation in $$U$$ if
begin{align*} V(f,U):=sup left{int_{U}f,text{div}varphi,dm : varphi in C^1_{c}(U;mathbb{R}^d),, sup_{x in U}|varphi(x)| le 1. right}
Here, $$|varphi(x)|={varphi_1(x)^2+cdots+varphi_d(x)^2}^{1/2}$$. We denote by $$BV(U)$$ the set of functions of bounded variation in $$U$$. For $$f in BV(U)$$, we set
begin{align*} |f|_{BV(U)}=|f|_{L^1(U,m)}+V(f,U), end{align*}
where $$|f|_{L^1(U,m)}$$ is the $$L^1$$-norm of $$f$$.

The following theorem is well-known (now $$U$$ is a bounded smooth domain):

Let $${f_n}_{n=1}^infty$$ be a bounded sequence in $$BV(U)$$. Then, there exists a subsequence $${f_{n_k}}_{k=1}^infty$$ and $$f in L^1(U,m)$$ such that $$lim_{k to infty}f_{n_k}=f$$ in $$L^1(U,m)$$.

My question

Let $${f_n}_{n=1}^infty$$ be a (not necessarily bounded) sequence in $$BV(U)$$. We assume that $${f_n}_{n=1}^infty$$ is a bounded sequeene in $$L^1(U,m)$$, and for any $$varphi in C_{c}^1(U;mathbb{R}^d)$$,
begin{align*} lim_{n to infty}int_{U}f_n,text{div}varphi,dm=int_{U}f,text{div}varphi,dmquad {rm (A)} end{align*}
for some $$f in BV(U)$$. Of course, this does not imply that $${f_n}_{n=1}^infty$$ is a bounded sequence in $$BV(U)$$ (the condition (A) is not sufficient to use the uniform boundedeness principle). However, $$C_{c}^infty(U;mathbb{R}^d)$$ (with the topology induced by the seminorms) is a montel space. Then, the condition (A) and the uniform boundedness principle should show that $$lim_{n to infty}f_n =f$$ in a strong sense.

What does this convergence mean? If the convergence is metrizable, please tell me the metric.