global-analysis-and-the-banach-manifold-of-class-h1-curvers

diff --git a/sections/structure.tex b/sections/structure.tex
@@ -176,31 +176,37 @@ find\dots
   \end{equation}
 
   Now given \(\xi \in H^1(E)\) fix \(t_0, t_1 \in I\) with \(\norm{\xi}_\infty
-  = \norm{\xi_{t_1}}\). Then
-  \[
+  = \norm{\xi_{t_1}}\) and \(\norm{\xi_{t_0}} \le \norm{\xi}_0\). If \(t_0 <
+  t_1\) then
+  \begin{equation}\label{eq:one-norm-le-sqrt-two-infty-norm}
     \begin{split}
       \norm{\xi}_\infty^2
       & = \norm{\xi_{t_0}}^2
-        + \int_{t_0}^{t_1} \frac{\dd}{\dd s} \norm{\xi_s}^2 \; \dd s \\
+        + \int_{t_0}^{t_1} \frac{\dd}\dt \norm{\xi_t}^2 \; \dt \\
+      & \le \norm{\xi}_0^2
+        + \int_{t_0}^{t_1} \frac{\dd}\dt \norm{\xi_t}^2 \; \dt \\
       \text{(\(\nabla\) is compatible with the metric)}
-      & = \norm{\xi_{t_0}}^2 + \int_{t_0}^{t_1}
-        2 \left\langle \xi_s, \nabla_{\frac\dd{\dd s}} \xi_s \right\rangle
-        \; \dd s \\
+      & = \norm{\xi}_0^2 + \int_{t_0}^{t_1}
+        2 \left\langle \xi_t, \nabla_{\frac\dd\dt} \xi_t \right\rangle
+        \; \dt \\
       \text{(Cauchy-Schwarz)}
-      & \le \norm{\xi_{t_0}}^2 + \int_0^1
-        2 \norm{\xi_s} \cdot \norm{\nabla_{\frac\dd{\dd s}} \xi_s} \; \dd s \\
-      & \le \norm{\xi}_\infty^2
-        + \int_0^1 \norm{\xi_s}^2 + \norm{\nabla_{\frac\dd\dt} \xi_s}^2
-        \; \dd s \\
-      & \le \norm{\xi}_\infty^2
-        + \norm{\xi}_0^2 + \norm{\nabla_{\frac\dd\dt} \xi}_0^2 \\
-      % TODO: This is actually wrong. Fix this.
-      \text{(because of equation (\ref{eq:zero-norm-le-infty-norm}))}
+      & \le \norm{\xi}_0^2 + \int_0^1
+        2 \norm{\xi_t} \cdot \norm{\nabla_{\frac\dd\dt} \xi_t} \; \dt \\
       & \le \norm{\xi}_0^2
+        + \int_0^1 \norm{\xi_t}^2 + \norm{\nabla_{\frac\dd\dt} \xi_t}^2
+        \; \dt \\
+      & = \norm{\xi}_0^2
         + \norm{\xi}_0^2 + \norm{\nabla_{\frac\dd\dt} \xi}_0^2 \\
       & \le 2 \norm{\xi}_1^2
     \end{split}
-  \]
+  \end{equation}
+
+  Similarly, if \(t_0 > t_1\) then by inverting the orientation of the curve
+  \(\gamma\) we can see that \(\norm{\xi}_\infty < \sqrt{2} \norm{\xi}_0\).
+  More precisely, if we set \(\eta(t) = \gamma(1 - t)\) and \(\xi' \in
+  H^1(\eta^* TM)\) with \(\xi'_t = \xi_{1 - t}\) then \(\norm{\xi}_\infty =
+  \norm{\xi'}_\infty \le \sqrt{2} \norm{\xi'}_1 = \sqrt{2} \norm{\xi}_1\)
+  because of equation (\ref{eq:one-norm-le-sqrt-two-infty-norm}).
 \end{proof}
 
 \begin{note}