Is this set of function belongs to $L^\infty$?

Let $\Omega\subset \mathbb R^N$ be open bounded with smooth boundary. Let $u\in SBV\cap L^\infty(\Omega)$ be given. We write $$Du = \nabla u\lfloor \mathcal L^N + (u^+-u^-)\otimes \nu_u\mathcal H^{N-1}\lfloor S_u$$ where $S_u$ is the jump set of $u$.

Then define set $\mathcal F$ such that $$\mathcal F:=\{v\in SBV(\Omega)\cap L^\infty(\Omega),\|\nabla v\|_{L^\infty}\leq C, T[v]=T[u],S_v\subset S_u\}$$ where $C>0$ is a constant.

Then, do I have that $v\in\mathcal F$ is $L^\infty$ uniformly bounded?

If not, how may I modify $\mathcal F$ so that my requirement is true? Maybe require more on jump set? Since I feel my conjecture may not be true since I give no requirement on how much it can jump for a function $v\in\mathcal F$.

Thank you!

Let $\Omega=[-1,1]$ and $u(x)=u(-x)=0$ for $x\in[2^{-2k+1},2^{-2k+2})$ and $u(x)=u(-x)=2^{-k}$ for $x\in[2^{-2k},2^{-2k+1})$, $k=1,\ldots,n,\ldots$.
Doesn't this $u$ belong to your space $SBV\cap L^\infty$ ? Because if no condition is imposed on the jumps, the set $\mathcal F$ clearly contains piecewise constant functions with the same jump set $S_u=\{\pm 2^{-n}:\ n=1,\ldots\}$ and arbitrarily large values.
• I would try including (in the definition of $\mathcal F$) a condition that the singular (jump) part of $Dv$ has its (pointwise) norm dominated by that of the jump part of $Du$, or the weaker condition that $\int_{S_u} |v^+-v^-|\ d\mathcal{H}^{N-1}$ is bounded. – Jean Duchon Aug 11 '15 at 11:40
• I think it does imply $L^\infty$-boundedness of $\mathcal F$. This needs a proof in general, but in 1 dimension it's obvious. The lower semicontinuity is another question. – Jean Duchon Aug 11 '15 at 12:23