3
$\begingroup$

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 $u$.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I see sir. So I can surely have orthogonality in the sense of $L^2$ norm but not $TV$ seminorm. Also, could you expand a bit about how you get the contradiction? $\endgroup$
    – Covepe
    Apr 28, 2018 at 8:58
  • $\begingroup$ @Covepe I restructured by answer so I hope the contradiction is more transparent now. $\endgroup$ Apr 28, 2018 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.