On the space $BV:=BV(\mathbb R, [0,1])$ of $[0,1]$-valued functions on $\mathbb R$ that are of bounded variation equipped with the norm $$\Vert f-g\Vert_{BV} := \Vert f-g\Vert_{L_1}+V(f-g),$$ where $$V(h) = \sup\left\{\int h(x)\phi'(x)\operatorname dx : \text{$\phi\in\mathcal C_c^1(\mathbb R)$ with $\Vert\phi\Vert_{L_\infty}\leq 1$}\right\}.$$ I am given the nonlinear functional $$T(f) = \int (h\circ f)\operatorname df,$$ where $h$ is some continuous function. I have to show continuity of $T$:
Let $(f_n)_{n\in\mathbb N}$ be a sequence in $BV$ converging to some $f$ in $BV$. Then
\begin{align*}
\vert T(f_n) – T(f)\vert &= \left\vert \int (h\circ f_n)\operatorname df_n – \int (h\circ f)\operatorname df_n + \int (h\circ f)\operatorname df_n – \int (h\circ f)\operatorname df \right\vert \\
&\leq \left\vert \int (h\circ f_n – h\circ f)\operatorname df_n\right\vert + \left\vert \int h\circ f\operatorname d(f_n – f)\right\vert.
\end{align*}
By assumption, $\Vert f_n – f \Vert_{BV}\rightarrow 0$. Due to the definition of the norm, I believe that this implies $\Vert f_n – f\Vert_{L_1}\rightarrow 0$ as well as $V(f_n – f)\rightarrow 0$. So for the first integral, \begin{align*}\left\vert \int (h\circ f)\operatorname d(f_n-f)\right \vert &\leq \int \vert h\circ f\vert \operatorname dV(f_n – f) \\&\leq \sup_{x\in\mathbb R}\vert (h\circ f)(x)\vert V(f_n-f) \\&\leq K\cdot V(f_n-f)\\&\rightarrow 0\end{align*} for some $K\geq 0$ since $f\leq 1$ and $h$ is a continuous function on a compact domain. For the second integral, $$\left\vert \int(h\circ f_n – h\circ f) \operatorname df_n\right\vert \leq \int \vert h\circ f_n – h\circ f\vert\operatorname df_n.$$ But I can't conclude the desired result from this inequalty: the integrator is $\operatorname df_n$ (instead of $\operatorname dx$), so I can't apply $\Vert f_n – f\Vert_{L_1}\rightarrow 0$.
I am also open for other approaches. If it's possible, I would like to work on $BV(\mathbb R,\mathbb R)$, i.e., no restriction of the codomain to a compact set, but I don't think it's possible.
Best Answer
The answer is that if we consider the particularization of the $n$-dimensional BV norm for $n=1$, i.e. $$\DeclareMathOperator{\Dm}{d\!} \Vert f\Vert_{BV} \triangleq \Vert f\Vert_{L_1}+\operatorname{TV}(f)\label{1}\tag{1} $$ where $\operatorname{TV}(f)$ is the Tonelli-Cesari total variation as given in the OP and in my comment, the operator $T(f)$ is discontinuous.
On the other hand, if we consider the classical $1$-dimensional BV norm $$ \Vert f\Vert_\mathscr{BV} \triangleq \Vert f\Vert_{L_1}+\operatorname{\mathscr{V}}(f)\label{2}\tag{2} $$ where $$ \operatorname{\mathscr{V}}(f)=\sup_{P \in \mathscr{P}} \sum_{i=0}^{n_{P}-1} | f(x_{i+1})-f(x_i) | $$ is the classical (or we may say Jordan) variation of $f$ and $\mathscr{P}$ is the set of finite partitions of $\Bbb R$, then $T(f)$ is continuous.
Proof of the discontinuity of $T(f)$ respect to $\Vert \cdot\Vert_{BV}$. Consider $f$ as the characteristic function of the open interval $]0, 1[$, i.e. $$ f(x) \triangleq\chi_{]0,1[}(x)= \begin{cases} 1 & x\in ]0,1[,\\ 0 & x\notin ]0,1[. \end{cases} $$ As we can easily prove, $f\in BV(\Bbb R, [0,1])$: now let's define a sequence $\{f_n\}_{n\in\Bbb N_{>0}}\subsetneq BV(\Bbb R, [0,1])$ converging to $f$ as $$ f_n(x)\triangleq \begin{cases} \varphi_n\ast f(x) = \displaystyle\int\limits_{\Bbb R}\varphi_n(x-y)f(y)\Dm y = \int\limits_0^1\varphi_n(x-y)\Dm y &x\neq 0,\\ 1-\frac{1}{2}\sin\frac{1}{n} & x=0. \end{cases} $$ where $\varphi_n(x)$ is any sequence of $C^\infty_0$ mollifiers converging to the Dirac measure as $n\to\infty$: note that also $f_n(x)\to f(x)$ pointwise for all $x\in\Bbb R\setminus\{0\}$, while for $x=0$ the pointwise limit does not exist due to the rapid oscillations of the function sequence at that point.
Now it's easy to see that $$ \Dm f = \delta(0)-\delta(1) = \lim_{n\to\infty} \Dm f_n\;\text{ in the space of Radon measures }\;\mathcal{M}(\Bbb R).\label{3}\tag{3} $$ and while this implies that $$ \int (h\circ f) \Dm f = h\circ f(0) - h\circ f(1) =h(0)-h(0)=0 $$ it also implies that $T(f_n)\nrightarrow 0$ as $n\to\infty$ since for a sufficiently large $n$ we have that $$ \begin{split} T(f_n) &= \int (h\circ f_n) \Dm f_n = \int\limits_{\Bbb R} (h\circ f_n) \Dm f_n \\ & = \int\limits_{\Bbb R} (h\circ f_n) \Dm f_n + h \circ f_n(0) - h \circ f_n(1) - h \circ f_n(0) + h \circ f_n(1)\\ & = h \circ f_n(0) - h \circ f_n(1) \\ & \qquad\qquad+ \int\limits_{-\infty}^{+\frac 12} \big(h\circ f_n- h \circ f_n|_{x=0}\big) \Dm f_n + \int\limits_{1\over 2}^{+\infty} \big(h \circ f_n - h\circ f_n|_{x=1}\big) \Dm f_n\\ & = h \left(1-\frac{1}{2}\sin\frac{1}{n} \right) - h \circ f_n(1)\\ & \qquad\qquad + \int\limits_{-\infty}^{+\frac 12} \big(h\circ f_n- h \circ f_n|_{x=0}\big) \Dm f_n + \int\limits_{1\over 2}^{+\infty} \big(h \circ f_n - h\circ f_n|_{x=1}\big) \Dm f_n \end{split}\label{4}\tag{4} $$ Now while the last three terms on the right side converge to a limit for $n\to\infty$, precisely $$ \begin{align} \lim_{n\to\infty} & h \circ f_n(1) = h(0), \\ \lim_{n\to\infty} & \int\limits_{-\infty}^{+\frac 12} \big(h\circ f_n- h \circ f_n|_{x=0}\big) \Dm f_n = 0, \\ \lim_{n\to\infty} & \int\limits_{1\over 2}^{+\infty} \big(h \circ f_n - h\circ f_n|_{x=1} \big) \Dm f_n = 0 \end{align} $$ the first term, i.e. $h \left(1-\frac{1}{2}\sin\frac{1}{n} \right)$ does not, thus $\vert T(f_n) - T(f)\vert \nrightarrow 0$ for $n\to\infty$ and the functional $T:{BV}(\Bbb R, [0,1]) \to \Bbb R$ is not continuous respect to the topology of ${BV}(\Bbb R, [0,1])$.
Proof of the continuity of $T(f)$ respect to $\Vert \cdot \Vert_\mathscr{BV}$. This is simply a consequence of the fact that $\Vert \cdot \Vert_\mathscr{BV}$ forces any converging sequence in the corresponding Banach space $\mathscr{BV}(\Bbb R, [0,1])$ to converge also pointwisely.
Final notes.