(\ell-1)c, \end{cases}$$ with $$\iiint \limits_{[(\ell-1)c,1]^3} g(x,y) g(y,z) g(z,x) \, dx\, dy \, dz = 0,$$ and with $\iint\limits_{[0,1]^2} g(x,y) \, dx \, dy = e$. Minimizing $I(g)$ on such graphons is easy, since all but the upper right corner of the graphon is fixed. Applying Theorem \[flat\] to that corner, we get

\[scallop\] If $e>1/2$ and $t$ is the smallest value possible, then the minimum of $I(g)$ on $F_0$ is achieved by the graphon $$g_0(x,y) = \begin{cases} 1 & x< kc < y \hbox{ or } y<kc < x \hbox{ for an integer }k<\ell; \cr 
  p & (\ell-1)c < x < [1+(\ell-1)c]/2 < y \hbox{ or } (\ell-1)c < y < [1+
  (\ell-1)c]/2 < x; \cr 0 & \hbox{otherwise,} \end{cases}$$ where $$p= \frac{4c(1-\ell c)}{(1-(\ell-1)c)^2}$$ is a number chosen to make $\int\!\!\int_{[0,1]^2} g(x,y) \; dx\;dy =
e$. Furthermore, any other minimizer is equivalent to $g_0$. The minimum value of $I(g)$ is $$I(g_0) = \frac{(1-(\ell-1)c)^2}{2}I_0(p).$$

Minimizing near the first scallop
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Now that we know the minimizer [*at*]{} the (lower) boundary, we perturb it to get a minimizer [*near*]{} the boundary.

\[near\] Pick $e<1/2$ and $\epsilon$ sufficiently small. Then the graphon $$g(x,y) = \begin{cases} 2e-\epsilon & x<\frac{1}{2} < y \hbox{ or } 