Let function $u\in BV(\Omega)$ be a function of bounded variation and $\Omega\subset \mathbb R^2$ be a smooth domain. I know it is possible to approximate function $u$ with polynomials, i.e.,
$$
u = \sum\_{n=1}^{+\infty} P\_n,
$$
where $P\_n$ is of polynomial of order $n$.


My question: would it be possible to design polynomial $P\_n$ so that 
$$
TV(u) = \sum\_{n=1}^{\infty}TV(P\_n),
$$
where by $TV(u)$ we mean the total variation of $u$.



Can I approximate a function of bounded variation with orthogonal polynomial?
**The answer is no.** Orthogonality is with respect to the $L^2$ norm and the $TV$ norm (of a smooth function) is the $L^1$ norm of the derivative and for this norm the $L^2$ orthogonality does not mean much.


Let $\Vert\cdot\Vert\_1$ denote the $L^1$ norm on $\Omega$.
Smooth functions are dense in BV (Theorem 2, Section 5.2.2 in [1]) so I think you can approximate u by polynomials $Q\_n$ and then you can write $P\_n=Q\_n-Q\_{n-1}$. However, for smooth functions, and in particular for polynomials, $TV(P)=|DP|(\Omega)=\Vert \nabla P\Vert\_1$.


Let $P$ and $Q$ be polynomials.
If $\nabla P$ is not parallel to $\nabla Q$ at every point of $\Omega$, then 
$\Vert \nabla P + \nabla Q\Vert\_1<\Vert \nabla P\Vert\_1 + \Vert\nabla Q\Vert\_1$.
Note that the gradients are parallel iff $\nabla P = \lambda\nabla Q$,
$\nabla (P-\lambda Q)=0$, $P=\lambda Q+c$ which is a very restrictive condition.
Therefore, unless you have this linear relation between all polynomials you have:
$$
TV(u)=|Du|(\Omega)=\lim\_{k\to\infty}\Vert\sum\_{n=1}^k \nabla P\_n\Vert\_1<\sum\_{n=1}^\infty \Vert\nabla P\_n\Vert\_1=\sum\_{n=1}^\infty TV(P\_n).
$$
Therefore you cannot get the equality that you want, but instead you get a sharp inequality.


A good introduction to BV functions is Chapter 5 in:


[1] **L. Evans, R. F. Gariepy**, *Measure theory and fine properties of functions.* Revised edition. Textbooks in Mathematics. CRC Press, Boca Raton, FL, 2015.


