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

diff --git a/sections/structure.tex b/sections/structure.tex
@@ -178,7 +178,7 @@ find\dots
   Now given \(\xi \in H^1(E)\) fix \(t_0, t_1 \in I\) with \(\norm{\xi}_\infty
   = \norm{\xi_{t_1}}\) 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
@@ -199,15 +199,16 @@ find\dots
         + \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 \(\norm{\xi}_\infty \le \sqrt{2} \norm{\xi}_1\).
-  More precisely, if we set \(\eta(t) = \gamma(1 - t)\) and \(\zeta \in
-  H^1(E)\) with \(\zeta_t = \xi_{1 - t}\) -- a section of \(E\) along \(\eta\)
-  -- then \(\norm{\xi}_\infty = \norm{\zeta}_\infty \le \sqrt{2} \norm{\zeta}_1
-  = \sqrt{2} \norm{\xi}_1\) because of the inequality
-  (\ref{eq:one-norm-le-sqrt-two-infty-norm}).
+  \]
+
+  Similarly, if \(t_0 > t_1\) then
+  \[
+    \norm{\xi}_\infty^2
+    = \norm{\xi_{t_0}}^2 + \int_{t_0}^{t_1} \frac{\dd}\dt \norm{\xi_t}^2 \; \dt
+    = \norm{\xi_{t_0}}^2 + \int_{1 - t_0}^{1 - t_1}
+      \frac{\dd}\dt \norm{\xi_{1 - t}}^2 \; \dt
+    \le 2 \norm{\xi}_1^2
+  \]
 \end{proof}
 
 \begin{note}