g(x,z) dx\right)^2 = 2 f_1(z)^2 \\
\int_{1/2}^1 (\delta g(x,z))^2 dx & \ge & 2 \left ( \int_{1/2}^1
\delta g(x,z) dx\right)^2 = 2 f_2(z)^2. \end{aligned}$$ Since $I_0''(\epsilon)$ and $I_0''(2e-\epsilon)$ are positive, $\delta I$ is bounded from below by $$\begin{aligned}
 &&  \frac{1}{4} \iint \limits_{[0,1]^2} I_0''(g(x,y)) \delta g(x,y)^2 dx \, dy 
+\frac{ I''(\epsilon)}{2}\left [\int_0^{1/2} f_1(z)^2 dz
+ \int_{1/2}^1 f_2(z)^2 dz \right ] \cr 
&+& \frac{I_0''(2e-\epsilon)}{2} \left [ \int_0^{1/2}f_2(z)^2 dz + 
\int_{1/2}^1 f_1(z)^2 dz \right ]  \cr 
&+& 3\lambda_2 \int_0^1 dz \left [\epsilon(f_1(z)^2 + f_2(z)^2) 
+ 2(2e-\epsilon)f_1(z)f_2(z) \right ] \end{aligned}$$

Collecting terms and applying equation (\[lambda2\]), this bound becomes $$\begin{aligned}
 \frac{1}{4} \iint \limits_{[0,1]^2} I_0''(g(x,y)) \delta g(x,y)^2 dx \, dy &+&  \int_0^{1/2} \!\!dz [c_1 f_1(z)^2 + c_2 f_2(z)^2 + 2 c_3 f_1(z)f_2(z)] \cr 
&+& \int_{1/2}^1 \!\!dz [c_1 f_2(z)^2 + c_2 f_1(z)^2 + 2 c_3 f_1(z)f_2(z)], \end{aligned}$$ where $$\begin{aligned}
c_1 &=& \frac{I_0''(\epsilon)}{2} + \frac{\epsilon (I_0'(2e-\epsilon)-I_0'(\epsilon))}
{2(e-\epsilon)^2} \\
c_2 &=& \frac{I_0''(2e-\epsilon)}{2} + \frac{\epsilon (I_0'(2e-\epsilon)-I_0'(\epsilon))}
{2(e-\epsilon)^2} \\
c_3 &=& \frac{(2e-\epsilon)(I_0'(2e-\epsilon)-I_0'(\epsilon))}{2(e-\epsilon)^2}.\end{aligned}$$ Note that all coefficients are positive, and that $c_2>1$. As $\epsilon \to 0$, $c_1$ goes to $+\infty$ as $1/\epsilon$, while $c_3$ only diverges as $-\ln(\epsilon)$. Since $c_1 c_2 > c_3^2$ for small $\epsilon$, the integrand for each $z$ is positive semi-definite, so the integral over $z$ is non-negative, 