The projection onto the null space of total variation operator

Let $TV$ denote the total variation semi-norm over domain $\Omega\subset \mathbb R^2$ which is open bounded with smooth boundary.

Let $\mathcal N$ denote the null space of $TV$. That is, a function belongs to $\mathcal N$ should be a constant. Let $P$ denote the projection operator onto $\mathcal N$.

My question: let a function $u\in L^\infty$ be given. Then do we have $$P(u) = \frac1{|\Omega|}\int_\Omega u\,dx$$ hold? i.e., the projection gives the average of $u$?

thank you!

• "The" projection operator? A subspace by itself is not sufficient to define a projection onto it. If you have an inner product, then you can use that to define a particular projection, but a semi-norm is not enough. – Paul Sinclair Oct 17 '16 at 23:22
• @PaulSinclair this is what I thought... but I'm reading a paper in which they use the exact thing I wrote here, and it troubles me a lot. But I don't have a way to fix this since bounded variation does not have an inner product ... – spatially Oct 17 '16 at 23:29
• @PaulSinclair I'll link the paper I'm reading. That might be helpful – spatially Oct 17 '16 at 23:32
• Either they define the projection in some way elsewhere in the paper, or they meant "a" projection operator and failed to notice the error. Certainly $P$ defines a projection onto $\mathcal N$, but so does any multiple of $P$, and one can easily build other projections. – Paul Sinclair Oct 18 '16 at 14:00

The projection operator you wrote down is the projection onto $\mathcal N$ with respect to the $L^2$-scalar product: Let $u\in L^2(\Omega)$ and $v\in \mathcal N$ be given, that is, $v$ is constant. Then $$\int_\Omega( u-P(u))v dx = v \cdot |\Omega| (P(u)-P(u))=0.$$