I need help with this exercise: Prove that $\exists (f_{n})_{n=1}^{\infty}\subset BV([0,1])$ such that $f_{n}\to f$ uniformly, but $\|f_{n}-f\|_{BV}\not\to 0$.
So I proposed the sequence $(f_{n})_{n=1}^{\infty}$ where:$$f_{n}(x)=\begin{cases} 0 &\quad x=0\\ x^{1+\frac{1}{n}}\sin(\frac{1}{x}) &\quad x\in(0,1] \end{cases}$$I proved that $\|f_{n}\|_{BV}=|f_{n}(0)|+V_{0}^{1}(f_{n})=V_{0}^{1}(f_{n})<\infty$ and that $f_{n}\to f$ uniformly where $$f(x)=\begin{cases} 0 &\quad x=0\\ x\sin(\frac{1}{x}) &\quad x\in(0,1] \end{cases}$$ and such that $\|f\|_{BV}=|f(0)|+V_{0}^{1}(f)=V_{0}^{1}(f)$ is not finite.
I'm having trouble proving that $\|f_{n}-f\|_{BV}\not\to 0$. I know that $\|f_{n}-f\|_{BV}=|(f_{n}-f)(0)|+V_{0}^{1}(f_{n}-f)=V_{0}^{1}(f_{n}-f)\leq V_{0}^{1}(f_{n})+V_{0}^{1}(f)$ but this doesn't imply the norm is not zero. I also thought I could use the fact that $\|g\|_{BV}=0 \iff g\equiv0$, but I'm stucked.