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}<infty.

end{align*}

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.**