Can we have $\|\nabla u \|_{L^1(B(0,1))}\leq \|\nabla u-\tilde{c} \|_{L^1(B(0,1))}$ for radially symmetric function? Suppose $B(0,1)$ is the unit ball in $\mathbb R^2$ and $u\in C^\infty(\overline{B(0,1)})$. Suppose $u$ is radially symmetric, i.e. $u(x)=u(Rx)$ for any $R\in SO(2)$. My question is, do we have 
$$\|\nabla u \|_{L^1(B(0,1))}\leq \|\nabla u-\tilde{c} \|_{L^1(B(0,1))} \tag 1$$
for arbitrary vector $\tilde{c}=(c_1,c_2)$?
In one dimension this question is true. But I meet trouble to prove it in 2 dimensions. I am actually not sure this inequality will hold or not. So if you could come up with a counterexample it would also be great. But I hope it is true...
My final goal is to have radially symmetric $u\in BV(B(0,1))$. But I think once we proved $(1)$ we would be done.
Thank you!

Yes, this is true for any even function, that is a function such that $u(-x)=u(x)$. The $L^1$ norm of a vector-valued function is the sum of $L^1$ norms of its components. So it suffices to consider one partial derivative, say $v = u_{x_1}$. Note that $v(-x)=-v(x)$, that is $v$ is odd. 
The minimum of $\|v-c\|_{L^1}$  among all $c\in\mathbb{R}$ is attained by any median of $v$, that is a number $m$ such that the sets $\{v> m\}$ and $\{v<m\}$ have equal measure. Indeed, if these sets have unequal measure, moving $m$ slightly to decrease the larger set results in the decrease of the $L^1$ norm of $v-c$. 
Since $v$ is odd, the sets $\{v> 0\}$ and $\{v<0\}$ have equal measure; thus $m$ is a median of $v$.
