estimates to each piece.

Since $h(x,y)$ and $I''(g)$ are piecewise constant, all of our integrals break down into integrals over different quadrants. Let $R_1$ and $R_2$ be the following subsets of $[0,1]^2$: $$R_1 = \{x,y < 1/2\} \cup \{x,y > 1/2\}, \qquad 
R_2 = \{x<1/2<y\} \cup \{y < 1/2 < x\}.$$ For each $z$, we define the functions $f_1(z) = \int_0^{1/2} \delta g(x,z) dx$ and $f_2(z) = \int_{1/2}^1 \delta g(x,z) dx$. The second variation in $I$ is then $$\begin{aligned}
 &&  \frac{1}{4} \iint \limits_{[0,1]^2} I_0''(g) \delta g(x,y)^2 dx\,dy   
+ \frac{I_0''(\epsilon)}{4}\iint \limits_{R_1} \delta g(x,y)^2
dx \, dy + \frac{I_0''(2e-\epsilon)}{4} \iint \limits_{R_2} \delta g(x,y)^2 dx \, dy \cr
&+&  3 \lambda_2  \int_0^1 dz \left [
\epsilon \iint \limits_{R_1}  \delta g(x,z) \delta g(y,z) dx \, dy
 +  (2e-\epsilon) \iint \limits_{R_2} \delta g(x,z) 
\delta g(y,z) dx \, dy \right ] \cr
&=& \frac{1}{4} \iint \limits_{[0,1]^2} I_0''(g(x,y)) \delta g(x,y)^2 dx \, dy 
+ \frac{I_0''(\epsilon)}{4}\iint \limits_{R_1} \delta g(x,z)^2
dx \, dz  + \frac{I_0''(2e-\epsilon)}{4} \iint\limits_{R_2} \delta g(x,z)^2 dx \, dz  \cr 
&& \qquad \qquad +  3 \lambda_2 \int_0^1 \epsilon \left [f_1(z)^2
+ f_2(z)^2)\right ] + 2(2e-\epsilon ) f_1(z) f_2(z)\,   dz \end{aligned}$$

Note that by Cauchy-Schwarz, $$\begin{aligned}
\int_0^{1/2} (\delta g(x,z))^2 dx & \ge & 2 \left ( \int_0^{1/2}
\delta 