Let $v$ be a function of bounded variation with the total variation defined as $$ |Dv|(\Omega) =\sup\left\{- \int_\Omega v\, {\rm div}\, \phi\; dx:\, \phi \in C^\infty_0(\Omega, \mathbb{R}^n),\, |\phi| \leq 1 \right\}. $$

In the case of $n=1$, we can use either this definition or the usual one to prove for non-constant functions that change sign:

$$ |Dv|(\Omega) \geq \lVert v^+ \rVert_{\infty} + \lVert v^-\rVert_{\infty}. $$

Assume now that $v$ is defined in $\Omega = B_{r_{3}}$ and $v$ is radial and decreasing (i.e. $v(x) = v(y)$ iff $|x|=|y|$ and $v(x) \geq v(y)$ iff $|x| \geq |y|$), I want to know if there is a simillar bound for $|Dv|(\Omega)$ for any dimension.

For example if the function is equal to $\lVert v^+ \rVert_{\infty}$ in $B_{r_{1}}$, then is $0$ in $B_{r_{2}}\setminus B_{r_{1}}$ and then equal to -$\lVert v^- \rVert_{\infty}$ in $B_{r_{3}}\setminus B_{r_{2}}$ the total variation will be (at least by my computations):

$$ |Dv|(B_{r_{3}}) = \lVert v^+ \rVert_{\infty}|\partial B_{r_{1}}| + \lVert v^-\rVert_{\infty}|\partial B_{r_{2}}|. $$

So I think that maybe one can find that for any $v$ radial and decreasing one can maybe prove that

$$ |Dv|(\Omega) \geq a\lVert v^+ \rVert_{\infty} + b\lVert v^-\rVert_{\infty} $$ for some $a,b$ constants related to where $v$ is positive and negative.

Obs1: $|.|$ denotes the measure of a set (in the case of the spheres would be the surface measure, the Hausdorff measure)

Obs2: $B_{r}$ is the ball centered in the origin with radius $r$

Obs3: $f^\pm = \max\{\pm f,0\}$

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    $\begingroup$ I don't really get your 1D inequality (or maybe your notation): isn't this just false for nontrivial constant functions, where the left is $0$ but the right is not? $\endgroup$
    – user378654
    May 26 at 5:56
  • $\begingroup$ @user378654 yes, you are right, I forgot to say that the function can't be constant, that it must change values $\endgroup$ May 26 at 14:27


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