and we obtain $$\delta I \ge \frac{1}{4} \iint I_0''(g) \delta g^2 \ge \frac{1}{2} 
\iint \delta g(x,y)^2,$$ where we used the fact that $I_0''(u) \ge 2$ for all $u$.

Any global minimizer must be $O(\epsilon)$ close to $g_0$, and hence $O(\epsilon)$ close to our specified perturbative minimizer. This means that the only way for them to differ is through a complicated bifurcation of minimizers at $g_0$, despite the uniform bounds on $\delta I$ as we approach the boundary. The difference between these hypothetical minimizers and $g_0$ would not be pointwise small, but would merely be small in an $L^1$ sense.

For instance, consider graphons of the form $$g(x,y) = \begin{cases} p & x<c<y \hbox{ or } y<c<x; \cr \alpha & x,y < c; \cr \beta & x,y > c, \end{cases}$$ where $c$ is a parameter that we will vary and $p$, $\alpha$ and $\beta$ are constants that depend on $c$. For each $c$ sufficiently close to $1/2$, it is possible to find a graphon of this form such that $\iiint g(x,y) g(y,z) g(x,z) dx \, dy \, dz = t$ and $\iint g(x,y) dx\, dy =e$, and such that the Euler-Lagrange equations (\[ELeq\]) are satisifed. Call this graphon $g_c(x,y)$. A lengthly 