Cleanup

src/cat/tricks/dyadic.tex

@@ -14,6 +14,7 @@
     \[
     \text{rk}(q) = \min\bracs{n \in \natp|x \in \mathbb{D}_n}
     \]
+
     is the \textbf{dyadic rank} of $q$.
 \end{definition}
 
@@ -32,7 +33,7 @@
 \begin{proof}
     First suppose that $\text{rk}(x) = 1$. In which case, $x \in \bracs{0, 1/2}$, and either $M(x) = \emptyset$ or $M(y) = \bracs{1}$.
 
-    Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \ref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
+    Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \autoref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
 \end{proof}
 
 \begin{proposition}
@@ -48,15 +49,18 @@
     \[
     \phi(x) + \phi(y) = g_2 + g_2 \le g_1 = \phi(x + y)
     \]
+
     In the second, $\phi(x) + \phi(y) = \phi(x + y)$.
 
-    Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \ref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
+    Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \autoref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
     \[
     \phi(x) + \phi(y) = \phi(x_0) + \phi(y) + g_{n+1} \le \phi(x_0 + y) + g_{n+1} = \phi(x + y)
     \]
+
     by the inductive assumption. Otherwise, there exists $y_0 \in \mathbb{D}_n$ such that $y = y_0 + 2^{-n-1}$, so
     \[
     \phi(x) + \phi(y) \le \phi(x_0) + \phi(y_0) + g_n = \phi(x_0) + \phi(y_0) + \phi(2^{-n}) \le \phi(x + y)
     \]
+
     by the inductive assumption.
 \end{proof}

src/cat/tricks/index.tex

@@ -1,5 +1,5 @@
 \chapter{Inequalities and Computations}
 \label{chap:tricks}
 
-\input{./src/cat/tricks/dyadic.tex}
-\input{./src/cat/tricks/product.tex}
+\input{./dyadic.tex}
+\input{./product.tex}

src/cat/tricks/product.tex

@@ -7,10 +7,12 @@
     \[
     ab \le \frac{a^p}{p} + \frac{b^q}{q}
     \]
+
     and for any $\eps > 0$,
     \[
     ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
     \]
+
 \end{lemma}
 \begin{proof}
     Since $x \mapsto \exp(x)$ is convex,
@@ -22,4 +24,5 @@
     \[
     ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
     \]
+
 \end{proof}