Thank you Sam.

src/op/example/hardy.tex

@@ -8,7 +8,7 @@
     \norm{f}_{H^p(D)} = \sup_{r \in [0, 1)} \braks{\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^p d\theta}^{1/p}
     \]
 
-    if $p < \infty$, and $\norm{f}_{H^p(D)} = \norm{f}_u$ if $p = \infty$. Then the the \textbf{Hardy space} is the space
+    if $p < \infty$, and $\norm{f}_{H^\infty(D)} = \norm{f}_u$ if $p = \infty$. Then the \textbf{Hardy space} is the space
     \[
     H^p(D) = \bracsn{f \in H(D; \complex)|\ \normn{f}_{H^p(D; \complex)} < \infty}
     \]
@@ -43,7 +43,7 @@
     \dpn{f, g}{r} = \frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})} d\theta = \frac{1}{4}\sum_{k = 0}^3 i^k [f + i^kg]_{r}^2
     \]
 
-    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\norm{f}_{H^2(D)} = \lim_{r \upto 1}[f]_r$, so
+    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\norm{f}_{H^2(D)} = \lim_{r \upto 1}[f]_r$, so the inner product
     \begin{align*}
     \dpn{f, g}{H^2(D)} &= \lim_{r \upto 1}\dpn{f, g}{r} = \lim_{r \upto 1}\frac{1}{4}\sum_{k = 0}^3 i^k [f + i^kg]_r^2 \\
     &= \frac{1}{4}\sum_{k = 0}^3 i^k \normn{f + i^kg}_{H^2(D)}^2
@@ -60,7 +60,7 @@
         \dpn{f, g}{H^2(D)} &= \lim_{r \upto 1}\sum_{n = 0}^\infty x_n\ol{y_n}r^{2n} = \sum_{n = 0}^\infty x_n\ol{y_n}
     \end{align*}
 
-    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}. Thus $\bracsn{z^n}_1^\infty$ forms an orthonormal basis for $H^2(D)$ and the given map is unitary. 
+    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}. Thus $\bracsn{z^n}_0^\infty$ forms an orthonormal basis for $H^2(D)$ and the given map is unitary. 
 \end{proof}
 
 \begin{proposition}
@@ -77,7 +77,7 @@
 
     then:
     \begin{enumerate}
-        \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. 
+        \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(x_0)$. 
         \item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$. 
     \end{enumerate}
 \end{proposition}
@@ -110,7 +110,7 @@
 
     then
     \begin{enumerate}
-        \item Let $p = \sum_{k = 0}^n a_k z^k$, $H^2_n(D) = \bracsn{q \in \complex[z]|\deg(p) \le n}$, $P \in B(H^2(D))$ be the orthogonal projection onto $H^2_n(D)$, and identify $H^2_n(D) = \complex^{n+1}$, then $Tf = Ppf$ for all $f \in H^2_n(D)$. 
+        \item Let $p = \sum_{k = 0}^n a_k z^k$, $H^2_n(D) = \bracsn{q \in \complex[z]|\deg(q) \le n}$, $P \in B(H^2(D))$ be the orthogonal projection onto $H^2_n(D)$, and identify $H^2_n(D) = \complex^{n+1}$, then $Tf = Ppf$ for all $f \in H^2_n(D)$. 
         \item For any $q \in \complex[z]$, $P(p + z^{n+1}q)f = Ppf$ for all $f \in H^2(D)$. 
         \item For any $q \in \complex[z]$, $\norm{T}_{M_{n+1}(\complex)} \le \normn{p + z^{n+1}q}_{H^\infty(D)}$. 
     \end{enumerate}

src/op/system/complete-upgrade.tex

@@ -239,7 +239,7 @@ The following is a slight generalisation of one form of the above example, but s
     \begin{align*}
     \sum_{k = 1}^n S_k (E_{ij} \otimes 1_B) S_k^* &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes b_{k, s}^*) (E_{ij} \otimes 1_B) (E_{kt} \otimes b_{k, t}^*)^* \\
     &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes b_{k, s}^*) (E_{ij} \otimes 1_B) (E_{tk} \otimes b_{k, t}) \\
-    &= \sum_{k = 1}^n E_{kk} \otimes b_{k, i}^* b_{k, j}^* = \sum_{k = 1}^n E_{kk} \otimes \phi(E_{ij})
+    &= \sum_{k = 1}^n E_{kk} \otimes b_{k, i}^* b_{k, j} = \sum_{k = 1}^n E_{kk} \otimes \phi(E_{ij})
     \end{align*}
 
     By linearity, for each $T \in M_n(\complex)$,