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\begin{fclemma}[]
  \label{lemma:5.1}
  Assuming:
  - $c > 0$ is such that
  \[
    \h(xy) \ge c(x\h(y) + y\h(x))
  \]
  for every $x, y \in [0, 1]$
  - $\mathcal{A}$ is a family of sets such that
  every element (of $\bigcup \mathcal{A}$) belongs
  to fewer than $p|\mathcal{A}|$ members of $\mathcal{A}$
  Then:
  $\ent{A \cup B} > c(1 - p)(\ent A + \ent B)$.
\end{fclemma}

\begin{proof}
  Think of $A, B$ as characteristic functions.
  Write $A_{< k}$ for $(A_1, \ldots, A_{k - 1})$
  etc. By the \nameref{lemma:1.3} it is enough to
  prove for every $k$ that
  \[
    \cent{(A \cup B)_k}{(A \cup B)_{<k}}
    > c(1 - p)(\cent{A_k}{A_{<k}} +
    \cent{B_k}{B_{<k}})
  .\]
  By \nameref{submod},
  \[
    \cent{(A \cup B)_k}{(A \cup B)_{< k}}
    \ge \cent{(A \cup B)_k}{A_{<k}, B_{<k}}
  .\]
  For each $u, v \in \{0, 1\}^{k - 1}$ write $p(u)
  = \Pbb(A_k = 0 \mid A_{< k} = u)$, $q(v) =
  \Pbb(B_k = 0 \mid B_{< k} = v)$.

  Then
  \[
    \cent{(A \cup B)_k}{A_{<k} = u, B_{< k} = v}
    = \h(p(u)q(v))
  \]
  which by hypothesis is at least
  \[
    c( p(u) \h(q(v)) + q(v) \h(p(u)) )
  .\]
  So
  \[
    \cent{(A \cup B)_k}{(A \cup B)_{<k}}
    \ge c \sum_{u, v} \Pbb(A_{< k} = u)
    \Pbb(B_{<k} = v) \left( p(u) \h(q(v)) + q(v)
    \h(p(u)) \right)
  .\]
  But
  \[
    \sum_u \Pbb(A_{< k} = u) \Pbb(A_k = 0 \mid
    A_{< k} = u)
    = \Pbb(A_k = 0)
  \]
  and
  \[
    \sum_v \Pbb(B_{<k} = v) \h(q(v))
    = \sum_v \Pbb(B_{<k} = v) \cent{B_k}{B_{< k} =
    v}
    = \cent{B_k}{B_{<k}}
  .\]
  Similarly for the other term, so the RHS equals
  \[
    c(\Pbb(A_k = 0) \cent{B_k}{B_{<k}} + \Pbb(B_k =
    0) \cent{A_k}{A_{<k}})
  ,\]
  which by hypothesis is greater than
  \[
    c(1 - p)(\cent{A_k}{A_{<k}} + \cent{B_k}{B_{<k}})
  \]
  as required.
\end{proof}

This shows that if $\mathcal{A}$ is \gls{uclosed},
then $c(1 - p) \le \half$, so $p \ge 1 -
\frac{1}{2c}$. Non-trivial as long as $c > \half$.

We shall obtain $\frac{1}{\sqrt{5} - 1}$. We start
by proving the diagonal case -- i.e. when $x =
y$.

\begin{fclemma}[Boppana]
  \label{lemma:5.2}
  For every $x \in [0, 1]$,
  \[
    \h(x^2) \ge \phi x \h(x)
  .\]
\end{fclemma}

\begin{proof}
  Write $\psi$ for $\phi^{-1} = \frac{\sqrt{5} -
  1}{2}$. Then $\psi^2 = 1 - \psi$, so $\h(\psi^2)
  = \h(1 - \psi) = \h(\psi)$ and $\phi \psi = 1$,
  so $\h(\psi^2) = \phi\psi \h(\psi)$. Equality also
  when $x = 0, 1$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e5dfb88e0fc54db3.png}
  \end{center}
  Toolkit:
  \begin{align*}
    \ln 2 \h(x)
    &= -x \ln x - (1 - x) \ln (1 - x) \\
    \ln 2 \h'(x)
    &= -\ln x - 1 + \ln(1 - x) + 1 \\
    &= \ln(1 - x) - \ln x \\
    \ln 2 \h''(x)
    &= -\frac{1}{x} - \frac{1}{1 - x} \\
    \ln 2 \h'''(x)
    &= \frac{1}{x^2} - \frac{1}{(1 - x)^2}
  \end{align*}
  Let $f(x) = \h(x^2) - \phi x \h(x)$. Then
  \begin{align*}
    f'(x)
    &= 2x\h'(x^2) - \phi \h(x) - \phi x \h'(x) \\
    f''(x)
    &= 2\h'(x^2) + 4x^2 \h''(x^2) - 2\phi\h'(x) -
    \phi x \h''(x) \\
    f'''(x)
    &= 4x \h''(x^2) + 8x \h''(x^2) + 8x^3\h'''(x^2)
    - 3\phi\h''(x) - \phi x\h'''(x) \\
    &= 12x \h''(x^2) + 8x^3 \h'''(x^2) - 3\phi
    \h''(x) - \phi x \h'''(x) \\
  \end{align*}
  So
  \begin{align*}
    \ln 2 f'''(x)
    &= \frac{-12x}{x^2(1 - x^2)}
    + \frac{8x^3(1 - 2x^2)}{x^4(1 - x^2)^2}
    + \frac{3\phi}{x(1 - x)}
    - \frac{\phi x(1 - 2x)}{x^2(1 - x)^2} \\
    &= \frac{-12}{x(1 - x^2)}
    + \frac{8(1 - 2x^2)}{x(1 - x^2)^2}
    + \frac{3\phi}{x(1 - x)}
    - \frac{\phi(1 - 2x)}{x(1 - x)^2} \\
    &= \frac{-12(1 - x^2) + 8(1 - 2x^2) + 3\phi(1
    - x)(1+  x)^2 - \phi(1 - 2x)(1 + x)^2}
    {x(1 - x)^2(1 + x)^2}
  \end{align*}
  This is zero if and only if
  \[
    -12 + 12x^2 + 8 - 16x^2 + 3\phi(1 + x - x^2
    -x^3) - \phi(1 - 3x^2 - 2x^3)
    = 0
  \]
  which simplifies to
  \[
    -\phi x^3 - 4x^2 + 3\phi x - 4 + 2\phi = 0
  .\]