# Is a function $u\in \mathrm{SBV}(\Omega)$ with these additional properties essentially bounded?

Some related earlier discussion can be found here.

Let $\Omega\subset \mathbb R^N$ be open bounded with smooth boundary, $\mathcal H^{N-1}(\partial\Omega)<\infty$ and $u\in SBV(\Omega)$. Then $$Du = \nabla u\mathcal L^N\lfloor\Omega + (u^+-u^-)\nu_u\mathcal H^{N-1}\lfloor S_u,$$ where $S_u$ is the jump set of $u$, $\nu_u$ is a measurable field of normals on $S_u$, and $u^+$ and $u^-$ are the corresponding one-sided approximate limits. In addition, suppose the following three properties: $$\|T(u)\|_{L^\infty(\partial \Omega)}< C_1\tag 1$$ $$\|\nabla u\|_{L^\infty(\Omega)}<C_2 \tag 2$$ $$\int_{S_u}|u^+-u^-|\,d\mathcal H^{N-1}<C_3 \tag 3$$ Here $C_1$, $C_2$, and $C_3$ are positive constants and $T$ denotes the trace operator.

Question: Can I estimate $\|u\|_{L^\infty}$ by the constants $C_1$, $C_2$, $C_3$ and constants that depend on $N$ and $\Omega$? If yes, can I further weaken (2) to $$\int_\Omega|\nabla u|^2<C_2 ?$$

Thank you!

No, such an estimate for $\|u\|_{L^\infty}$ cannot hold.
Just consider a “wedding-cake” with infinitely many layers of height 1, i.e. $u=\sum_{k=1}^\infty \mathbf{1}_{B(0,r_k)}$ for a sequence of positive radii $(r_k)$ that decreases to $0$ suitably quickly.
Added in edit: Take $\Omega=B(0,1)$ and $0<r_{k+1}<r_k<1$ for all $k\in\mathbb{N}$ with $r_k\downarrow 0$ so quickly that $$\sum_{k=1}^\infty\mathcal{H}^{N-1}(\partial B(0,r_k))<\infty.$$ Then also $\sum_k|B(0,r_k)|<\infty$ and the partial sums of the series defining $u$ are in $\mathrm{BV}(\Omega)$ and converge to $u$ in $L^1(\Omega)$. By the usual $\mathrm{BV}$-compactness result [1, Theorem 3.23] we obtain $u\in\mathrm{BV}(\Omega)$. Observe that the points where $u$ is not approximately continuous is given by $S_u=\bigcup_k\partial B(0,r_k)$. Moreover, $u$ is locally constant away from $S_u$. Hence $\nabla u$ (i.e. the part of $Du$ absolutely continuous with respect to the $N$-dimensional Lebesgue measure) vanishes and the Cantor part of $Du$ is supported in $S_u$. By [1, Proposition 4.2] the Cantor part is trivial and $u\in\mathrm{SBV}(\Omega)$.
So $Tu=0$, $\nabla u=0$ and $$\int_{S_u}|u^+-u^-|\,\mathrm{d}\mathcal{H}^{N-1} = \int_{S_u}1\,\mathrm{d}\mathcal{H}^{N-1}=\mathcal{H}^{N-1}(S_u)<\infty.$$ However, $u$ is not in $L^\infty(\Omega)$.