$\min$ rather than $\inf$ since $\tilde F_0$ is compact.) However, $\min_{F_0} I(g)$ is at least as big as $\inf_{U_\delta}
I(g)$, since $F_0 \subset U_\delta$. Thus $\min_{F_0} I(g) =
\lim_{\delta \to 0} \inf_{U_\delta} I(g)$.

Minimizing the rate function on the boundary
============================================

&gt;From now on we will work exclusively with graphons rather than with graphs. From Theorem \[thm1\], all questions boil down to “minimize the rate function over such-and-such region”. The first region we study is the lower boundary of $(e,t)$-space, beginning with the first (flat) scallop:

\[flat\] If $e \le 1/2$ and $t=0$, then $\min_{F_0}
  I(g) = I_0(2e)/2$, and this minimum is achieved at the graphon $$\label{2fold} g_0(x,y) = \begin{cases}2e &\hbox{if }x<\frac{1}{2} < y \hbox{  or } y<\frac{1}{2}<x; \cr 0 & \hbox{otherwise.}\end{cases}$$ Furthermore, any other minimizer is equivalent to $g_0$, corresponding to the same reduced graphon.

Since $t(g)$ is identically zero, the measure of the set $\{(x,y) \in [0,1]^2 | g(x,y) =0\}$ is at least 1/2. Otherwise, the graphon $\bar g(x,y) = \begin{cases}1 & \hbox{if } g(x,y)>0; \cr
    0& \hbox{otherwise,}\end{cases}$ would have no triangles and would have edge density greater than 1/2, which is impossible. So we restrict attention to graphons that are zero on a set of measure at 