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Define a norm on functions $f : G^r \to \Rbb$ by
\[
  \|f\|
  = (\|f\|_{\square, r})
  = \sup_{\varphi \in \mathcal{F}} |\langle f, \varphi \rangle|
  = \sup_{\varphi \in \mathcal{F} \cup (-\mathcal{F})} \langle f, \varphi
  \rangle
\]
This has a dual:
\[
  \|\varphi\|^*
  = \sup_{\|f\| \le 1} \langle f, \varphi \rangle
.\]
By definition, $|\langle f, \varphi \rangle| \le \|f\| \|\varphi\|^*$.
Note
\[
  \|f\|
  = \sup_{\varphi \in \mathcal{F}} |\langle f, \varphi \rangle|
  = \sup_{\varphi \in \mathcal{F}} |\Ebb_x f(x) \varphi(x)|
  \le \|f\|_1
,\]
and by duality
\[
  \|\varphi\|^*
  \ge \|\varphi\|_\infty
.\]

\begin{fclemma}[]
  % Lemma 3.9
  \label{lemma:3.9}
  The unit ball of $\|\bullet\|^*$ is just $\conv(\mathcal{F} \cup
  (-\mathcal{F}))$.
\end{fclemma}

We will use Hahn--Banach to prove this.

\begin{fcthm}[Hahn--Banach]
  % Theorem 3.10
  \label{thm:3.10}
  Let $K$ be a closed convex body in $\Rbb^n$, and suppose $\theta \in \Rbb^n
  \setminus K$.
  Then there exists $f \in \Rbb^n$ such that $\langle f, \theta \rangle > 1$
  while $\langle f, \eta \rangle \le 1 ~\forall \eta \in K$.
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/624802b916784aab.png}
\end{center}

\begin{proof}
  Suppose $\varphi \in \mathcal{F} \cup (-\mathcal{F})$.
  Then
  \[
    \|\varphi\|^*
    = \sup_{\|f\| \le 1} \langle f, \varphi \rangle
    \le 1
  .\]
  So $\varphi$ is in the unit ball of $\|\bullet\|^*$.
  Convex combinations of elements in $\mathcal{F} \cup (-\mathcal{F})$ are also
  in the unit ball by the triangle inequality.

  For the converse, we need Hahn--Banach.
  Suppose $\varphi \notin \conv(\mathcal{F} \cup (-\mathcal{F}))$.
  By \cref{thm:3.10}, there exists $f$ such that $\langle f, \varphi \rangle >
  1$ but $\langle f, \eta \rangle \le 1 ~\forall \eta \in \conv(\mathcal{F} \cup
  (-\mathcal{F}))$.
  The latter implies $\|f\| \le 1$.
  But
  \[
    1
    < \langle f, \varphi \rangle
    \le \|f\| \|\varphi\|^*
    \le \|\varphi\|^*
  .\]
  Hence $\varphi$ does not lie in the unit ball of $\|\bullet\|^*$.
\end{proof}

This implies that the unit ball of $\|\bullet\|^*$ is closed under
multiplication.
Hence, $\forall \varphi, \psi : G \to \Rbb$, then
\[
  \left\| \frac{\varphi}{\|\varphi\|^*} \frac{\psi}{\|\psi\|^*} \right\|^*
  \le 1
  \quad \implies \quad
  \|\varphi \psi\|^*
  \le \|\varphi\|^* \|\psi\|^*
.\]
(dual norm is submultiplicative).

\begin{fcthm}[Dense Model (Theorem 3.7')]
  % Theorem 3.7'
  \label{thm:3.7p}
  For all $\eps > 0$, if $\nu : X \to [0, \infty)$ satisfying $\|\nu - 1\| \le
  \exp(-\eps^{-O(1)})$ and $f : X \to [0, \infty)$ satisfies $0 \le f \le \nu$,
  then there exists $\tilde{f} : X \to [0, 1]$ such that $\|f - \tilde{f}\| \le
  \eps$.
\end{fcthm}

\begin{proof}
  Given $f : X \to [0, \infty)$, $0 \le f \le \nu$, it suffices to show that
  there exists $\tilde{f} : X \to \left[0, 1 + \frac{\eps}{2}\right]$ with $\|f
  - \tilde{f}\| \le \frac{\eps}{2}$, assuming that $\|\nu - 1\| \le \eps'$ for
  some sufficiently small $\eps'$.

  Suppose that there is no such $\tilde{f}$, i.e. $f$ cannot be written as $f =
  f_1 + f_2$, with
  \[
    f_1
    \in K_1 = \{g : X \to [ 0, 1 + \eps / 2 ] \}
    \qquad \text{and} \qquad
    f_2
    \in K_2 = \{h : X \to \Rbb : \|h\| \le \eps / 2\}
  .\]
  In other words: we are supposing that $f \notin K_1 + K_2$.
  Note that $K_1$ and $K_2$ are both convex bodies, and both contain $0$.
  Thus $K_1 + K_2$ is convex and contains $K_1$ and $K_2$.

  By \nameref{thm:3.10}, there exists $\psi : X \to \Rbb$ such that $\langle f,
  \psi\rangle > 1$ and $\langle g, \psi \rangle \le 1 ~\forall g \in K_1 +
  K_2$.
  Taking $g = \left( 1 + \frac{\eps}{2} \right) \indicator{\psi > 0} \in K_1$,
  we note that
  \[
    \langle (1 + \eps / 2)\indicator{\psi > 0}, \psi \rangle
    = \langle (1 + \eps / 2), \psi_+ \rangle
    \le 1
  \]
  where $\psi_+(x) = \max \{0, \psi(x)\}$.
  Thus $\langle 1, \psi_+ \rangle \le (1 + \eps / 2)^{-1}$.

  On the other hand, if $\langle g, \psi \rangle \le 1 ~\forall g \in K_2$, so
  $\langle g', \psi \rangle \le \frac{2}{\eps}$ for all $g' : X \to \Rbb$ with
  $\|g'\| \le 1$.
  So $\|\psi\|^* \le \frac{2}{\eps}$.

  So far, we have
  \[
    1
    < \langle f, \psi \rangle
    \le \langle f, \psi_+ \rangle
    \le \langle \nu, \psi_+ \rangle
    = \langle \nu - 1, \psi_+ \rangle + \ub{\langle 1, \psi_+ \rangle}_{\le (1 +
    \eps / 2)^{-1}}
  .\]
  So if we had $\|\psi_+\|^*$ bounded above, we would be done.
  Indeed, by Weierstrass approximation, there exists a polynomial $P$ such that
  \[
    |P(x) - \max \{0, x\}|
    \le \frac{\eps}{8}
    \qquad \forall x \in \left[ -\frac{2}{\eps}, \frac{2}{\eps} \right]
  .\]
  Recall that earlier we used duality to show $\|\psi\|_\infty \le \|\psi\|^*
  \le \frac{2}{\eps}$, which is why we only need the polynomial approximation to
  hold in the interval $\left[ -\frac{2}{\eps}, \frac{2}{\eps} \right]$.

  Let $P(x) = a_d x^d + \cdots + a_1 x + a_0$.
  We can do this with $R \defeq \sum_{i = 0}^{d} |a_i| \left( \frac{2}{\eps}
  \right)^i = \exp(-\eps^{-O(1)})$.
  Now
  \[
    \|P \psi\|^*
    \le \sum_{i = 1}^{d} |a_i| \|\psi^i\|^*
    \le \sum_{i = 1}^{d} |a_i| (\|\psi\|^*)^i
    \le R
  \]
  and
  \[
    \langle \nu - 1, \psi_+ \rangle
    = \langle \nu - 1, P \psi \rangle + \langle \nu - 1, \psi_+ - P \psi \rangle
    \le \ub{\|\nu - 1\| \|P \psi\|^*}_{\le \eps' \cdot R}
    + \ub{\|\nu - 1\|_1}_{\le 1} \ub{\|\psi_+ - P\psi\|_\infty}_{\le
    \frac{\eps}{8}}
  .\]
  Choosing $\eps'$ such that $\eps' \cdot R \le \frac{\eps}{8}$, we arrive at a
  contradiction.
\end{proof}