From eb2d40258a71760328fcbb36cb66c466b12e196f Mon Sep 17 00:00:00 2001
From: Josia Pietsch <josia.pietsch@uni-bonn.de>
Date: Wed, 12 Jul 2023 18:24:36 +0200
Subject: [PATCH] typo

---
 inputs/lecture_10.tex | 9 ++++-----
 1 file changed, 4 insertions(+), 5 deletions(-)

diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex
index 0ec6a3d..31be813 100644
--- a/inputs/lecture_10.tex
+++ b/inputs/lecture_10.tex
@@ -133,7 +133,7 @@ However, Fourier analysis is not only useful for continuous probability density
         RHS &=& \lim_{T \to  \infty} \frac{1}{2 T} \int_{-T}^T e^{-\i t x} \int_{\R} e^{\i t y} \bP(\dif y) \\
             &\overset{\text{Fubini}}{=}& \lim_{T \to  \infty} \frac{1}{2 T} \int_\R \bP(dy) \int_{-T}^T \underbrace{e^{-\i t (y - x)}}_{\cos(t ( y - x)) + \i \sin(t (y-x))} \dif t\\
             &=& \lim_{T \to  \infty} \frac{1}{2T} \int_{\R} \bP(\dif y) \int_{-T}^T \cos(t(y - x)) \dif t\\
-            &=& \lim_{T \to  \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} \bP(\dif y)\\
+            &=& \lim_{T \to  \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x))}{T (y-x)} \bP(\dif y)\\
     \end{IEEEeqnarray*}
     Furthermore
     \[
@@ -141,10 +141,11 @@ However, Fourier analysis is not only useful for continuous probability density
         1, &y = x,\\
         0, &y \neq  x.
     \end{cases}
+    % TODO TODO TODO
     \]
     Hence
     \begin{IEEEeqnarray*}{rCl}
-            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x)}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right)
+            \lim_{T \to \infty} \frac{1}{2 T }\int_{\R} \frac{2 \sin(T  (y-x))}{T (y-x)} \bP(\dif y) &=& \bP\left( \{x\}\right)
     \end{IEEEeqnarray*}
     % TODO by dominated convergence?
 \end{refproof}
@@ -154,7 +155,7 @@ 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 $\phi(0) = 1$, $|\phi(1)| \le t$ and $\phi(\cdot )$ is continuous.
+        \item $\phi(0) = 1$, $|\phi(t)| \le 1$ 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
@@ -164,8 +165,6 @@ However, Fourier analysis is not only useful for continuous probability density
 \end{theorem}
 \begin{refproof}{thm:lec_10thm5}
     Part (a) is obvious.
-    % TODO
-
 
     For part (b) we have:
     \begin{IEEEeqnarray*}{rCl}