From 2b34e3b5fbfaa3db283a060dbb7a772b327e1416 Mon Sep 17 00:00:00 2001
From: Josia Pietsch <josia.pietsch@uni-bonn.de>
Date: Wed, 12 Jul 2023 16:39:58 +0200
Subject: [PATCH] fixed typo

---
 inputs/lecture_10.tex | 25 +++++++++++++------------
 1 file changed, 13 insertions(+), 12 deletions(-)

diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex
index 1ffaadd..0ec6a3d 100644
--- a/inputs/lecture_10.tex
+++ b/inputs/lecture_10.tex
@@ -61,24 +61,24 @@ where $\mu = \bP X^{-1}$.
 
 \begin{theorem} % Theorem 3
     \label{thm:lec10_3}
-    Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
+    Let $\bP \in M_1(\R)$ such that $\phi_\bP \in L^1(\lambda)$.
     Then $\bP$ has a continuous probability density given by
     \[
-    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) \dif t.
+    f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\bP}(t) \dif t.
     \]
 \end{theorem}
 
 \begin{example}
     \begin{itemize}
-        \item Let $\bP = \delta_{\{0\}}$.
+        \item Let $\bP = \delta_{0}$.
           Then
           \[
-          \phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
+          \phi_{\bP}(t) = \int e^{\i t x} \dif \delta_0(x) = e^{\i t 0 } = 1
           \]
         \item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
             Then
             \[
-            \phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
+            \phi_{\bP}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
             \]
     \end{itemize}
 \end{example}
@@ -88,20 +88,21 @@ where $\mu = \bP X^{-1}$.
         If $x_n \to  x$, then $f(x_n) \to  f(x)$.
     \end{claim}
     \begin{subproof}
-        If $e^{-\i t x_n} \phi(t) \xrightarrow{n \to  \infty}  e^{-\i t x } \phi(t) $ for all $t$.
-
-        Then
+        Suppose that
+        $e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x} \phi(t)$
+        for all $t$.
+        Since
         \[
         |e^{-\i t x} \phi(t)| \le  |\phi(t)|
         \]
-        and $\phi \in L^1$, hence $f(x_n) \to  f(x)$
+        and $\phi \in L^1$,
+        we get $f(x_n) \to  f(x)$
         by the dominated convergence theorem.
-
     \end{subproof}
 
     We'll show that for all  $a < b$ we have
     \[
-    \bP\left( (a,b] \right)  = \int_a^b (x) \dif x.\label{thm10_3eq1}
+    \bP\left( (a,b] \right)  = \int_a^b f(x) \dif x.\label{thm10_3eq1}
     \]
     Let $F$ be the distribution function of $\bP$.
     It is enough to prove \autoref{thm10_3eq1}
@@ -153,7 +154,7 @@ However, Fourier analysis is not only useful for continuous probability density
     Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
     Then
     \begin{enumerate}[(a)]
-        \item $\phi(0) = 1$, $|\phi(t)| \le t$ and $\phi(\cdot )$ is continuous.
+        \item $\phi(0) = 1$, $|\phi(1)| \le t$ and $\phi(\cdot )$ is continuous.
         \item $\phi$ is a \vocab{positive definite function},
             i.e.~
             \[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge  0