Question concerning the decomposition of the Total Variation Measure

Question

Let $\Omega \subseteq \mathbb{R}^n$ be open and $u \in BV(\Omega)$. Then there exists a finite signed Radon measure $Du$, which happens to be the weak derivative of $u$, with the property: $\displaystyle \int_\Omega u div\varphi d\mathcal{L}^n = -\int_\Omega \langle\varphi,Du\rangle, ~\forall \varphi \in C_c^1(\Omega)$. Due to $Du$ being a Radon measure, we can decompose it using Lebesgues decomposition theorem: $Du = D^au + D^su$ with $D^au$ being absolutely continuous, and $D^su$ being singular with respect to the Lebesgue Measure. And as far as I understand, $D^au$ captures the smooth part of $u$, so in any neighbourhood of points in which $u$ is classically differentiable, $D^au = \nabla u \mathcal{L}^n$ and $D^su$ vanishes (please correct me if I'm wrong). On the contrary $D^su$ captures the "jump part" of $Du$. I'm not exactly sure, what the jump part of a function is. Is it just every point in which the function is not differentiable or is there any more to it? I would be very grateful about an example of a function that has a continuously differentiable part and a jump part.

Answer 1

To keep things simple you can take $n=2$, $\Omega$ to be the square of sidelength $2$ centered at the origin in $\mathbf R^2$, and let $u(x,y) = 1$ if $y > 0$ and $u(x,y) = 0$ otherwise.

The intuitive idea behind $u$ having bounded variation is that 1) the function value of $u$ only changes by $1$, and 2) the "size" of the set where the value of $u$ changes is finite. In this case, the function values of $u$ jump from $0$ to $1$ across the part of the $x$-axis that lies inside $\Omega$, whose $1$-dimensional measure equals $2$.

We can proceed a bit more formally. Suppose that $\phi \in C_c^\infty(\Omega;\mathbf R^2)$. Then $$\int_\Omega u \,\mathrm{div}\varphi \,d\mathcal{L}^2 = \int_{-1}^1 \int_{-1}^1 u\,\mathrm{div}\varphi\, dydx = \int_{-1}^1 \int_{0}^1\,\mathrm{div}\varphi\, dydx$$

Write $\varphi = (\varphi_2,\varphi_2)$. Then $\mathrm{div}\varphi = \dfrac{\partial \varphi_1}{\partial x} + \dfrac{\partial \varphi_2}{\partial y}$. Since $\varphi_1$ is supported in $\Omega$ you have $$\int_{-1}^1 \dfrac{\partial \varphi_1}{\partial x}(x,y) \, dx = 0$$ for all $y \in (-1,1)$. Thus $$\int_{-1}^1 \int_{0}^1\dfrac{\partial \varphi_1}{\partial x}(x,y)\, dydx = \int_0^1 \int_{-1}^1\,\dfrac{\partial \varphi_1}{\partial x}(x,y)\, dxdy = 0.$$ On the other hand, for each $x \in (-1,1)$ you have $$ \int_0^{1} \dfrac{\partial \varphi_2}{\partial y}(x,y)\, dy =0 - \varphi_2(x,0)$$ so that $$\int_{-1}^1\int_0^{1} \dfrac{\partial \varphi_2}{\partial y}(x,y)\, dy =0 - \varphi_2(x,0).$$ You can add these up to find that $$ \int_{-1}^1 \int_{0}^1\,\mathrm{div}\varphi\, dydx = - \int_{-1}^1 \varphi_2(x,0) \, dx.$$ At this point it is helpful to know something about Hausdorff measure - if you denote by $I$ the set $(-1,1) \times \{0\} \subset \Omega$, then the last integral may be written as $$\int_\Omega \varphi_2\, d\mathcal{H}^1|_I$$ where $\mathcal{H}^1|_I$ is the restriction of $1$-dimensional Hausdorff measure to $I$. Finally we put it all together $$\int_\Omega u \,\mathrm{div}\varphi \,d\mathcal{L}^2 = - \int_\Omega \varphi_2 \, d\mathcal{H}^1|_I = - \int_\Omega \varphi_1\, d0 - \int_\Omega \varphi_2 \, d\mathcal{H}^1|_I$$ so that $Du$ is the vector-valued measure $$Du = (0,\mathcal{H}^1|_I).$$ Since $Du$ is concentrated on a set of $2$-dimensional Lebesgue measure zero it follows that $D^a u = (0,0)$ and $Du = D^su$. Moreover note that $\|Du\| = 2$ so that $u$ has bounded variation.