0$ such that for $\e$ sufficiently small \begin{equation} c\e^{2/3}L^{1/3}\leq \min I^{\e} \leq C\e^{2/3}L^{1/3}. \end{equation} \label{scal} \end{theorem} We show that if we restrict the admissibility class to the set of Lipschitz functions \begin{eqnarray*} {\cal B}_0&=&{\cal A}_0 \cap W^{1,\infty}(R_L)\\ &=&\{u\in W^{1,\infty}(R_L):\,|u_y|=1 \mbox{ a.e., }u_{yy} \mbox{ is a Radon measure on }R_L \\ &&\mbox{ with finite mass, } u=0 \mbox { for } x=0\} \end{eqnarray*} this statement is no longer true. In fact, we prove the following \begin{theorem} If $\Omega=R_L$ then for all functions $u\in{\cal B}_0$ $I^{\e}(u)=\infty$. \label{lav} \end{theorem} \begin{remark} It is easy to see that for all $p\in [1,\infty)$ the class \begin{eqnarray*} &\{u\in W^{1,p}(R_L):\,|u_y|=1 \mbox{ a.e., }u_{yy} \mbox{ is a Radon measure on }R_L& \\ &\mbox{ with finite mass, } u=0 \mbox { for } x=0\}& \end{eqnarray*} contains a function $u$ such that $I^{\e}(u)<\infty.$ An example for this is obtained by modifying {\bf Example 3.1} below such that \[\begin{array}{ll} \theta\in \left(\frac{1}{4},\frac{1}{2}\right) & \mbox{if }1\leq p\leq 2, \\ \\ \theta\in \left(2^{p/(1-p)},\frac{1}{2}\right) & \mbox{if }22^{p/(1-p)}.\] Furthermore, note that \begin{eqnarray*}\int_{R_L}\e |u_{yy}|\,dy\,dx &=&\e\sum_{i=0}^{\infty}\int_{x_1}^L\int_0^{2^i} 2^{i}|u_{yy}|2^{-i}\theta^i\,dy\,dx\\ &=&\e\sum_{i=0}^{\infty} (2\theta)^{i}\int_{x_1}^L\int_0^1|u_{yy}|\,dy\,dx\end{eqnarray*} and the series is convergent if and only if \[\theta<\frac{1}{2}.\] \end{remark} \paragraph{Proof of Theorem \ref{lav}.} Assume that there is a constant $K>0$ such that \[ |\nabla u|\leq K \hspace{1cm}\mbox{ for a.e. }x\in R_{L}. \] Then we have by the Cauchy-Schwarz inequality \[u^2(l,y)=\left(\int_0^l 1\cdot u_x(x,y)\,dx\right)^2 \leq \int_0^l 1^2 \,dx \cdot\int_0^l u_x^2(x,y)\,dx. \] This implies the following Poincar\'e inequality \begin{equation} \int_0^1 u^2(l,y)dy\leq l \int_0^l\int_0^1 |\nabla u(x,y)|^2 \,dx \,dy\leq CK^2l^2. \hspace{1cm} \label{poin} \end{equation} for all $l\in (0,L]$. Next we use a ``zig-zag'' inequality which was proved by Kohn and M\"uller \cite{KM94}. \begin{lemma} Let $f\in W^{1,\infty}(0,1).$ Assume that $|f'|=1$ a.e. and that $f'$ changes sign $N$ times. Then \[ \int_0^1 f^2 \,dx \geq \frac{1}{12}(N+1)^{-2}= \frac{1}{12}\left(\frac{1}{2}\int_0^1|f''|dx+1\right)^{-2}. \] \label{zigzag} \end{lemma} Lemma \ref{zigzag} implies \begin{equation} \frac{1}{12}\left(\frac{1}{2}\int_0^1|u_{yy}(l,y)|dy +1\right)^{-2}\leq \int_0^1 u^2(l,y)dy. \label{zagzig} \end{equation} Combining (\ref{poin}) and (\ref{zagzig}) we get \[ \int_0^1 |u_{yy}(l,y)|dy\geq C K^{-1}l^{-1}-2 \] where $C$ is independent of $K$ and $l$. After integration we have \[ \int_0^L\int_0^1 \e |u_{yy}(l,y)|\, dy\, dl\geq C \int_0^L l^{-1}\, dl-2\e L=\infty. \] This implies Theorem \ref{lav}. \epr We now assume that the domain is a parallelogram. To simplify the presentation assume that the parallelogram has interior angles of $\pi/4$ and $3\pi/4$. But note that our method also works for other angles (except for $\pi/2$, of course). Set $\Omega=\{(x,y):\,y