# A lower bound for total variation in multivariable functions

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

• 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? May 26 at 5:56
• @user378654 yes, you are right, I forgot to say that the function can't be constant, that it must change values May 26 at 14:27