diff --git a/src/vna/vn/unlimited.tex b/src/vna/vn/unlimited.tex
index 76f2c22..cc83ac8 100644
--- a/src/vna/vn/unlimited.tex
+++ b/src/vna/vn/unlimited.tex
@@ -33,10 +33,10 @@
 \begin{proof}
     Let $\seq{\xi_n} \subset D(T^*)$ and $\xi, \eta \in H$ such that $\xi_n \to \xi$ and $T^*\xi_n \to \eta$, then for any $\theta \in D(T)$, 
     \[
-    \dpn{\eta, \theta}{H} = \limv{n}\dpn{T^*\xi_n, \theta}{H} = \limv{n}\dpn{\xi_n, T\theta}{H} = \dpn{\xi, T\theta}{H} = \dpn{T^*\xi, \theta}{H}
+    \dpn{\eta, \theta}{H} = \limv{n}\dpn{T^*\xi_n, \theta}{H} = \limv{n}\dpn{\xi_n, T\theta}{H} = \dpn{\xi, T\theta}{H}
     \]
 
-    As $D(T)$ is dense, $\eta = T^*\xi$, so the graph of $T^*$ is closed.
+    As $D(T)$ is dense, $T \circ \xi \in D(T)^*$, and $\eta = T^*\xi$. Thus the graph of $T^*$ is closed.
 \end{proof}
 
 \begin{definition}[Self-Adjoint]
@@ -71,12 +71,15 @@
 
     (2, closed range): For any $\seq{\xi_n} \subset D(T)$ and $\eta \in H$ with $(T + i)\xi_n \to \eta$, (1) implies that $\seq{\xi_n} \subset D(T)$ and $\seq{T\xi_n}$ are both Cauchy. Let $\xi = \limv{n}\xi_n$, then since $T$ is self-adjoint, it is closed, so $T\xi = \limv{n}T\xi_n \in T(H)$, and $\eta = (T + i)\xi \in (T+i)(H)$. 
 
-    (2, dense range): Let $\eta \in H$ with $\eta \perp (T + i)(H)$, then for each $\xi \in D(T)$, 
-    \[
-    0 = \dpn{(T + i)\xi, \eta}{H} = \dpn{T\xi, \eta}{H} - \dpn{\xi, i\eta}{H} = \dpn{\xi, (T - i)\eta}{H}
-    \]
-
-    As $D(T)$ is dense, $(T - i)\eta = 0$, so $\eta = 0$ by (1). 
+    (2, dense range): Let $\eta \in H$, then the following are equivalent:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\eta \perp (T + i)(H)$. 
+        \item $\dpn{T\xi, \eta}{H} = \dpn{\xi, i\eta}{H}$ for all $\xi \in D(T)$. 
+        \item $\eta \in D(T^*)$ with $T^*\eta = i\eta$. 
+        \item $\eta \in D(T)$ with $(T - i)\eta = 0$. 
+    \end{enumerate}
+    
+    By (1), $\eta \perp (T+i)(H)$ if and only if $\eta = 0$. 
 \end{proof}
 
 \begin{definition}[Cayley Transform]
@@ -120,7 +123,7 @@
 \begin{proof}
     (1): For each $n \in \natp$, $\one_{\bracsn{f \le n}}f \in B^\infty(X; \complex)$. Thus for each $\xi \in H$, by a monotone class argument, 
     \begin{align*}
-        \int_X |f|^2 dE_{E(\bracsn{f \le n}\xi, \bracsn{f \le n}\xi)} &= \int_{\bracsn{f \le n}}|f|^2 dE_{\xi, \xi} < \infty \\
+        \int_X |f|^2 dE_{E(\bracsn{f \le n})\xi, E(\bracsn{f \le n})\xi} &= \int_{\bracsn{f \le n}}|f|^2 dE_{\xi, \xi} < \infty \\
         \normn{\xi - E(\bracsn{f \le n})\xi}_H^2 &= \normn{E(\bracsn{f > n})\xi}_H^2
     \end{align*}
 
@@ -128,13 +131,15 @@
 
     It remains to show that $\int_X f dE$ is defined on the claimed domain. Let $\xi \in D(\int_X f dE)$, $\eta \in H$, and $B \in \cb_X$, then by (1) and (3) of \autoref{lemma:spectral-measure-properties}, 
     \[
-    \int_X \one_B d|E_{\xi, \eta}| \le E_{\xi, \xi}(B)^{1/2}E_{\eta, \eta}(B)^{1/2} \le E_{\xi, \xi}(B)^{1/2}\norm{\eta}_H
+    \int_X \one_B d|E_{\xi, \eta}| \le E_{\xi, \xi}(B)^{1/2}E_{\eta, \eta}(B)^{1/2}
     \]
 
     For any $\phi \in \Sigma(X; \complex)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
-    \[
-    \int_X |\phi| d|E_{\xi, \eta}| \le \norm{\eta}_H\sum_{\alpha \in \phi(X)} |\alpha| E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2} \le \norm{\eta}_H\braks{\int_X |\phi|^2 dE_{\xi, \xi}}^{1/2}
-    \]
+    \begin{align*}
+    \int_X |\phi| d|E_{\xi, \eta}| &\le \norm{\eta}_H\sum_{\alpha \in \phi(X)} |\alpha| E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2}E_{\eta, \eta}(\bracsn{\phi = \alpha})^{1/2} \\
+    &\le \norm{\eta}_H\braks{\int_X |\phi|^2 dE_{\xi, \xi}}^{1/2}
+    \end{align*}
+    
 
     By the \hyperref[Monotone Convergence Theorem]{theorem:mct}, 
     \[
@@ -297,7 +302,7 @@
 
     By (2) of \autoref{lemma:closed-unbonuded-adjoint}, $(I+T^*T): D(T^*T) \to H$ is surjective, so $\xi = 0$, and $D(T^*T)$ is dense in $D(T)$. 
 
-    (3): By (3) of \autoref{definition:unbounded-spectral-integral}, $|T|$ is self-adjoint, and in particular closed.  Since $D(T)$ is isomorphic to the graph of $|T|$, $D(T)$ is a Hilbert space. Similarly, for any $\xi \in D(T)$ with $\xi \perp D(T^*T)$ and $\eta \in D(T^*T)$, by the \hyperref[Borel functional calculus]{definition:unbounded-spectral-integral}\footnote{The fact that the spectral measure of a positive element is supported in $[0, \infty)$ is needed here.},
+    (3): By (3) of \autoref{definition:unbounded-spectral-integral}, $|T|$ is self-adjoint, and in particular closed.  Since $D(T)$ is isomorphic to the graph of $|T|$, $D(|T|)$ is a Hilbert space. Similarly, for any $\xi \in D(|T|)$ with $\xi \perp D(T^*T)$ and $\eta \in D(T^*T)$, by the \hyperref[Borel functional calculus]{definition:unbounded-spectral-integral}\footnote{The fact that the spectral measure of a positive element is supported in $[0, \infty)$ is needed here.},
     \[
     0 = \dpn{\xi, \eta}{H} + \dpn{|T|\xi, |T|\eta}{H} = \dpn{\xi, (I + T^*T)\eta}{H}
     \]
@@ -311,7 +316,7 @@
     \normn{T\xi}_H^2 = \dpn{T\xi, T\xi}{H} = \dpn{\xi, T^*T\xi}{H} = \dpn{|T|\xi, |T|\xi}{H} = \norm{|T|\xi}_H^2
     \]
 
-    By the density provided in (2) and (3), $\normn{T\xi}_H^2 = \norm{|T|\xi}_H^2$ for all $\xi \in D(T^*T)$. 
+    By the density provided in (2) and (3), $\normn{T\xi}_H^2 = \norm{|T|\xi}_H^2$ for all $\xi \in D(T)$. 
 \end{proof}
 
 
@@ -319,7 +324,6 @@
 \label{theorem:unbounded-polar-decomposition}
     Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then there exists a unique partial isometry $U \in B(H)$ with $\ker U = \ker T$ such that $T = U|T|$, where $|T|$ is the \hyperref[absolute value]{definition:unbounded-absolute-value} of $T$. 
 \end{theorem}
-% Omitted until further notice.
 \begin{proof}
     By \autoref{definition:unbounded-absolute-value}, $|T|: D(T) \to H$ is a densely defined, self-adjoint, and positive operator with $\normn{T\xi}_H = \norm{|T|\xi}_H$ for all $\xi \in D(T)$ and $\ker(T) = \ker(|T|)$. Let
     \[
@@ -331,7 +335,7 @@
     \ker(U) = |T|(D(T))^\perp = \ker(|T|) = \ker(T)
     \]
 
-    Finally, let $V: H \to H$ be a partial isometry with $T = V|T|$ and $\ker(V) = \ker(T)$, then for each $\xi \in D(T)$, $V|T|\xi = T\xi = U|T|\xi$, so $U = V$ on $|T|(D(T))$. By continuity, $U = V$ on $\ol{T(D(T))}$. As
+    Finally, let $V: H \to H$ be a partial isometry with $T = V|T|$ and $\ker(V) = \ker(T)$, then for each $\xi \in D(T)$, $V|T|\xi = T\xi = U|T|\xi$, so $U = V$ on $|T|(D(T))$. By continuity, $U = V$ on $\ol{|T|(D(T))}$. As
     \[
     \ker(V) = \ker(T) = \ker(U) = |T|(D(T))^\perp
     \]