Given $\Omega\subset \mathbb R^N$ open bounded smooth boundary.
We define $$ TV(u,\alpha):=\sup\left\{\int_\Omega u\operatorname{div}v\,dx,\, v\in C_c^\infty(\Omega;\,\mathbb R^2),\,\|v\|_{L^\infty}\leq \alpha\right\} $$ for any $\alpha>0$, and $$ BV(\Omega,\alpha):=\{u\in L^1(\Omega),\,TV(u,\alpha)<\infty\} $$ Clearly, as $\alpha=1$, we have $BV(\Omega,1)=BV(\Omega)$ in usual definition of bounded variation.
My question is: what is the space $$ BV(\Omega,\infty):=\bigcap_{\alpha>0}BV(\Omega,\alpha) $$ I feel the space $BV(\Omega,\infty)$ only contains the constant function but I am not sure since I think for different value of $\alpha_1$ and $\alpha_2$, $BV(\Omega,\alpha_1)$ is equivalent to $BV(\Omega.\alpha_2)$ since $$\alpha_2 TV(u,\alpha_1)=\alpha_1 TV(u,\alpha_2).$$
Any help is really welcome!
Yes, a very simple scaling argument shows $$TV(u,\alpha) = \alpha \, TV(u,1).$$ Hence, $BV(\Omega, \alpha) = BV(\Omega)$ for all $\alpha > 0$.