Proof of Proposition 2.2

From the analysis in Subsection 2.4, it suffices to consider the case $% \kappa =\emptyset$, and Proposition 2.1 reduces our task to veryfying: for each $(\emptyset ,g,G)\in Std((0,1))$, there exists a sequence of simple deformations satisfying $(f_m,\nabla f_m)\rightarrow (g,G)$ and

$$\lim\limits_{m\rightarrow \infty }\sum\limits_{z\in \Gamma (f_m)}h([f_m](z))=L\int\limits_0^1(\nabla g(x)-G(x))dx.$$ (9)

Our first step in constructing such a sequence is to consider the case $% \nabla g(x)=G(x)$ for all $x\in (0,1)$. In this case, $(\emptyset ,g,\nabla g)=$ $(\emptyset ,g,G)\in Std((0,1))$, and we may put $\kappa _m:=\emptyset$and f_m:=g for every positive integer m. Thus, $\Gamma (f_m)=\emptyset$and $\sum\limits_{z\in \Gamma (f_m)}h([f_m](z))=0=\int_0^1(\nabla g(x)-G(x))\,dx$ for all m, so that (2.9) is evident.

The remaining case to consider is when there exists $x_0\in (0,1)$ for which $\nabla g(x_0)$ $\neq G(x_0).$ The continuity of $\nabla g$ and of G and the relations $\nabla g(x)=\det \nabla g(x)\geq \det G(x)=G(x)$ for all $x\in (0,1)$ imply that

$$I:=\int\limits_0^1(\nabla g(x)-G(x))dx>0.$$ (10)

By the definition $L:=\liminf_{\zeta \rightarrow 0+}\frac{h(\zeta )}\zeta$, we may choose $m\mapsto \zeta _m>0$ such that $\lim_{m\rightarrow \infty }\zeta _m=0$ and $\lim_{m\rightarrow \infty }\frac{h(\,\zeta _m\,)}{\zeta _m}% =L.$ Moreover, we may assume that $m\mapsto \zeta _m$ is strictly decreasing and, because I>0, that $\zeta _m<I$ for all m. Let m now be given, and note that, because the function $y\mapsto$ $\int_0^y(\nabla g(x)-G(x))\,dx$ is continuous, we may choose a number x_q1 $\in (0,1)$ such that

$$\int_0^{x_{q_{_1}}}(\nabla g(x)-G(x))\,dx=\zeta _m.$$ (11)

For x in the interval (0,x_q1), put

$$f_m(x):=g(0+)+\int\limits_0^xG(\xi )d\xi$$ (12)

and note that $\nabla f_m(x)=G(x)$ and

$$\begin{eqnarray*}0 &\leq &g(x)-f_m(x)=\left( g(0+)+\int\limits_0^x\nabla g(\xi )... ...q \int\limits_0^{x_{q_1}}(\nabla g(\xi )-G(\xi ))d\xi =\zeta _m. \end{eqnarray*}$$

To define f_m on (x_q1,1), we first consider the case

$$\int_{x_{q_{_1}}}^1(\nabla g(x)-G(x))\,dx\leq \zeta _m.$$

For $x\in (x_{q_1},1)$ we define

$$f_m(x):=g(x_{q_1})+\int\limits_{x_{q_1}}^xG(\xi )d\xi$$ (13)

and we note as above that $\nabla f_m(x)=G(x)$ and $0\leq g(x)-f_m(x)\leq \zeta _m.$ Relations (2.12) and (2.13) tell us that

$$\begin{eqnarray*}f_m(x_{q_1}+)-f_m(x_{q_1}-) &=&g(x_{q_1})-\left( g(0+)+\int\lim... ...\int\limits_0^{x_{q_1}}(\nabla g(\xi )-G(\xi ))d\xi =\zeta _m>0, \end{eqnarray*}$$

and it follows that $\Gamma (f_m)=\{x_{q_1}\}$ and $[f_m](x_{q_1})=\zeta _m>0.$ Consequently,
$(\{x_{q_1}\},f_m)\in Sid((0,1))$, the condition (ii) in Proposition 2.2 is satisfied, and

$$\left\vert f_m(x)-g(x)\right\vert \leq \zeta _m\ {\rm and }\ \left\vert \nabla f_m(x)-G(x)\right\vert =0\qquad$$ (14)

for all $x\in (0,1)\,\backslash \,\{x_{q_1}\}.$

In the remaining case

$$\int_{x_{q_{_1}}}^1(\nabla g(x)-G(x))\,dx>\zeta _m,$$

we may choose x_q2 $\in (x_{q_1},1)$ such that

$$\int_{x_{q_{_1}}}^{x_{q_2}}(\nabla g(x)-G(x))\,dx=\zeta _m,$$

and for x in the interval (x_q1,x_q2) we define f_m(x) again using the formula (2.13), so that the relations (2.14) are satisfied for all $x\in (x_{q_1},x_{q_2})$ and $[f_m](x_{q_1})=\zeta _m>0$. If we now put $M_m:=\lfloor \frac I{\zeta _m}$, i.e., the greatest integer less than or equal to the number $\frac I{\zeta _m}>1$, then we may choose for each $k\in \left\{ 1,2,\ldots ,M_m\right\}$ a point $x_{q_k}\in (0,1)$such that

$$x_{q_1}<\cdots <x_{q_k}<x_{q_{k+1}}<\cdots <x_{q_{M_m}},$$ (15)

 

$$\int_0^{x_{q_1}}(\nabla g(x)-G(x))\,dx \int_{x_{q_{_1}}}^{x_{q_2}}(\nabla g(x)-G(x))\,dx$$ =

$$\cdots =\int_{x_{q_{M_m-1}}}^{x_{q_{M_m}}}(\nabla g(x)-G(x))\,dx=\zeta _m,$$ = (16)

and

$$\int_{x_{q_{M_m}}}^1(\nabla g(x)-G(x))\,dx\leq \zeta _m.$$ (17)

We now define $\kappa _m:=\left\{ x_{q_1},\cdots ,x_{q_k},x_{q_{k+1}},\cdots ,x_{q_{M_m}}\right\}$ and, setting x_q0:=0 and x_qMm+1:=1,we put

$$f_m(x):=g(x_{q_k})+\int_{x_{q_k}}^{x_{q_{k+1}}}G(\xi )\,d\xi ... ...{k+1}})\ { \rm and }\ k\in \left\{ 0,1,2,\ldots ,M_m\right\} .$$ (18)

This definition yields a simple deformation $(\kappa _m,f_m)\in Sid((0,1))\$with $\Gamma (f_m)=\kappa _m$, $[f_m](z)=\zeta _m$ for all $% z\in \Gamma (f_m),$ and satisfying (2.14) for all $x\in (0,1)\backslash \kappa _m$. We now may write

$$\begin{eqnarray*}\sum\limits_{z\in \Gamma (f_m)}h([f_m](z)) &=&\sum\limits_{z\in... ...ta _m)}{\zeta _m},(\frac{I\,}{% \zeta _m}+1)h(\zeta _m)\right) . \end{eqnarray*}$$

In other words, for all positive integers m,

$$\begin{array}{ll} \frac{\,h(\zeta _m)}{\zeta _m}\int_0^1(\nab... ...}{\zeta _m}(\int_0^1(\nabla g(x)-G(x))dx+\zeta _m) \end{array}$$

and, letting $m\rightarrow \infty$, we obtain the desired relation(2.9).