Added a section on non-unital C^*-algebras.
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Bokuan Li
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\input{./order.tex} \input{./order.tex}
\input{./positive.tex} \input{./positive.tex}
\input{./state.tex} \input{./state.tex}
\input{./gns.tex} \input{./gns.tex}
\input{./non-unital.tex}

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\section{Non-Unital $C^*$-Algebras}
\label{section:non-unital-cstar}
\begin{proposition}
\label{proposition:c-star-unitisation}
Let $A$ be a non-unital $C^*$-algebra and $\td A$ be its unitisation, then there exists a unique norm $\norm{\cdot}_{\td A}: \td A \to [0, \infty)$ such that:
\begin{enumerate}
\item For each $x \in A$, $\norm{x}_{\td A} = \norm{x}_A$.
\item $(\td A, \norm{\cdot}_{\td A})$ is a unital $C^*$-algebra.
\end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Theorem 15.1]{Zhu}}}. ]
For each $x \in A$ and $\lambda \in \complex$, let
\[
\norm{x + \lambda}_{\td A} = \sup_{\substack{y \in A \\ \norm{y}_A \le 1}} \norm{xy + \lambda y}_A
\]
be the operator seminorm corresponding to $\td A$ acting on $A$. If $\norm{xy + \lambda y}_A = 0$ for all $y \in A$, then $xy = -\lambda y$ for all $y \in A$. Given that $A$ is non-unital, $\lambda = 0$. Since $A$ is a $C^*$-algebra, $\norm{x}_A^2 = \norm{xx^*}_A = 0$, and $x = 0$ as well. Thus $\norm{\cdot}_{\td A}$ is indeed a norm on $\td A$.
(1): Let $x \in A \setminus \bracs{0}$, then since $A$ is a Banach algebra, $\norm{x}_{\td A} \le \norm{x}_A$. On the other hand, as $A$ is a $C^*$-algebra,
\[
\norm{x}_{\td A} \ge \norm{x \cdot \frac{x^*}{\norm{x}_A}}_A = \frac{\norm{x}_A^2}{\norm{x}_A} = \norm{x}_A
\]
(2): As $\norm{\cdot}_{\td A}$ is the operator norm corresponding to $\td A$ acting on $A$, $(\td A, \norm{\cdot}_{\td A})$ is a Banach algebra. Moreover, for any $x \in A$ and $\lambda \in \complex$,
\begin{align*}
\norm{x + \lambda}_{\td A}^2 &= \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{(x + \lambda)y}_A^2
= \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{y^*(x + \lambda)^*(x + \lambda)y}_A \\
&\le \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{(x + \lambda)^*(x + \lambda)y}_A = \norm{(x + \lambda)^*(x+\lambda)}_{\td A}
\end{align*}
so $(\td A, \norm{\cdot}_{\td A})$ is a $C^*$-algebra.
Finally, \autoref{corollary:c-star-unique-norm} implies that there can be at most one norm on $\td A$ making it a unital $C^*$-algebra, so the constructed norm is unique.
\end{proof}
\begin{definition}[Approximate Identity]
\label{definition:banach-approximate-identity}
Let $A$ be a Banach algebra and $\angles{e_\beta}_{\beta \in B} \subset A$ be a net, then $\angles{e_\beta}_{\beta \in B}$ is an \textbf{approximate identity} of $A$ if:
\begin{enumerate}
\item For each $\beta \in B$, $\norm{e_\beta}_{A} \le 1$.
\item For every $x \in A$, $e_\beta x \to x$ and $x e_\beta \to x$.
\end{enumerate}
\end{definition}
\begin{definition}[Increasing Approximate Identity]
\label{definition:increasing-approximate-identity}
Let $A$ be a $C^*$-algebra and $\angles{e_\beta}_{\beta \in B} \subset A$ be an approximate identity, then $\angles{e_\beta}_{\beta \in B}$ is \textbf{increasing} if:
\begin{enumerate}
\item For each $\beta \in B$, $e_\beta \ge 0$.
\item For each $\beta, \gamma \in B$ with $\beta \le \gamma$, $e_\beta \le e_\gamma$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma:cstar-approximate-identity-existence}
For each $m, n \in \natp$ with $m \le n$,
\begin{enumerate}
\item $\sup_{t \in [0, \infty)} \paren{1/n + t}^{-2}t \le n/4$.
\item For every $t \in [0, \infty)$, $m^{-1}(m^{-1} + t)^{-1} \ge n^{-1}(n^{-1} + t)^{-1}$.
\item For every $t \in [0, \infty)$, $(1/n + t)^{-1}t = 1 - (1/n + t)^{-1}/n$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1): For each $t \in [0, \infty)$,
\begin{align*}
0 \le \paren{t - \frac{1}{n}}^2 &= t^2 - \frac{2t}{n} + \frac{1}{n^2} = t^2 + \frac{2t}{n} + \frac{1}{n^2} - \frac{4t}{n} \\
0 &\le \paren{t + \frac{1}{n}}^2 - \frac{4t}{n}
\end{align*}
so $\frac{4t}{n} \le \paren{t + \frac{1}{n}}^2$ and $\frac{n}{4} \ge \paren{\frac{1}{n} + t}^{-2}t$.
(2): For each $t \in [0, \infty)$,
\[
\frac{1}{m}\paren{\frac{1}{m} + t}^{-1} = \frac{1}{1 + mt} \ge \frac{1}{1 + nt} = \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}
\]
(3): For every $t \in [0, \infty)$,
\begin{align*}
\paren{\frac{1}{n} + t}^{-1}t &= \paren{\frac{1}{n} + t}^{-1}\braks{\paren{\frac{1}{n} + t} - \frac{1}{n}} \\
&= \paren{\frac{1}{n} + t}^{-1}\paren{\frac{1}{n} + t} - \frac{1}{n}\paren{\frac{1}{n} + t}^{-1} \\
&= 1 - \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}
\end{align*}
\end{proof}
\begin{theorem}
\label{theorem:cstar-approximate-identity-existence}
Let $A$ be a unital $C^*$-algebra, $I \subset A$ be a left ideal, and $\cf \subset 2^I$ be the collection of all finite subsets of $I$, directed under inclusion. For each $F \in \cf$, let
\[
p_F = \sum_{x \in F}x^*x \quad e_F = \paren{\frac{1}{|F|} + p_F}^{-1}p_F
\]
then $\angles{e_F}_{F \in \cf} \subset I \cap \ol{B_A(0, 1)}$ is an increasing net of positive elements such that $xe_F \to x$ for all $x \in I$.
\end{theorem}
\begin{proof}
Since $p_F$ is positive, $1/|F| + p_F$ is invertible by \autoref{proposition:positive-spectrum}, and the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} implies that $0 \le e_F \le 1_A$. As $I \subset A$ is a left ideal, $e_F \in I \cap \ol{B_A(0, 1)}$. Now,
\begin{align*}
&\sum_{x \in F}[x(e_F - 1_A)]^*[x(e_F - 1_A)] = \sum_{x \in F}(e_F - 1_A)x^*x(e_F - 1_A) \\
&= (e_F - 1_A)p_F(e_F - 1_A) = e_F^2p_F - 2e_Fp_F + p_F \\
&= (1/|F| + p_F)^{-2}p_F \cdot \braks{p_F^2 - 2\paren{\frac{1}{|F|} + p_F}p_F + \paren{\frac{1}{|F|} + p_F}^2}
\end{align*}
where
\begin{align*}
&p_F^2 - 2\paren{\frac{1}{|F|} + p_F}p_F + \paren{\frac{1}{|F|} + p_F}^2 \\
&= - p_F^2 -\frac{2p_F}{|F|} + \frac{1}{|F|^2} + \frac{2p_F}{|F|} + p_F^2 = \frac{1}{|F|^2}
\end{align*}
so
\[
\sum_{x \in F}[x(e_F - 1_A)]^*[x(e_F - 1_A)] = \frac{1}{|F|^2}(1/|F| + p_F)^{-2}p_F
\]
By (1) of \autoref{lemma:cstar-approximate-identity-existence}, $\sup_{t \in [0, \infty)} (1/|F| + t)^{-2}t \le |F|/4$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that
\[
0 \le \sum_{x \in F}[x(e_F - 1_A)]^*[x(e_F - 1_A)] \le \frac{1}{|F|^2} \cdot \frac{|F|}{4} = \frac{1}{4|F|}
\]
Thus for each $x \in F$, $0 \le [x(e_F - 1_A)]^*[x(e_F - 1_A)] \le 1/(4|F|)$. In particular, $\norm{xe_F - x}_A \le \sqrt{1/4|F|}$.
To see that $\angles{e_F}_{F \in \cf}$ is increasing, let $F, G \in \cf$ with $F \subset G$, then $p_F \le p_G$, and $(1/|F| + p_F)^{-1} \ge (1/|F| + p_G)^{-1}$ by \autoref{lemma:cstar-inversion-order-reversing}. By (2) of \autoref{lemma:cstar-approximate-identity-existence},
\[
\frac{1}{|F|}\paren{\frac{1}{|F|} + t}^{-1} \ge \frac{1}{|G|} \paren{\frac{1}{|G|} + t}^{-1}
\]
for all $t \in [0, \infty)$. For each $t \in [0, \infty)$, rewrite
\begin{align*}
\paren{\frac{1}{|F|} + t}^{-1}t &= 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + t}^{-1} \\
\paren{\frac{1}{|G|} + t}^{-1}t &= 1 - \frac{1}{|G|}\paren{\frac{1}{|G|} + t}^{-1}
\end{align*}
Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus},
\begin{align*}
e_F &= 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + p_F}^{-1} \le 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + p_G}^{-1} \\
&\le 1 - \frac{1}{|G|}\paren{\frac{1}{|G|} + p_G}^{-1} = e_G
\end{align*}
\end{proof}
\begin{corollary}
\label{corollary:cstar-approximate-identity-existence-actual}
Let $A$ be a $C^*$-algebra, then $A$ admits an increasing approximate identity.
\end{corollary}
\begin{proof}
Identify $A$ as a self-adjoint two-sided ideal of its unitisation $\td A$, which is a $C^*$-algebra by \autoref{proposition:c-star-unitisation}. Applying \autoref{theorem:cstar-approximate-identity-existence} to $A$ and $\bracs{x^*|x \in A}$ as left and right ideals, respectively, yields that the net constructed by the theorem is an increasing approximate identity for $A$.
\end{proof}

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The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$.
\end{remark} \end{remark}
\begin{lemma}
\label{lemma:cstar-inversion-order-reversing}
Let $A$ be a unital $C^*$-algebra, $x, y \in G(A)$ be positive elements with $x \le y$, then $x^{-1} \ge y^{-1}$.
\end{lemma}
\begin{proof}
Since $y - x \ge 0$ and $x$ is invertible, $y^{-1/2}(y - x)y^{-1/2} \ge 0$ as well. As such,
\[
y^{-1/2}xy^{-1/2} \le y^{-1/2}yy^{-1/2} = 1
\]
Thus $\sigma_A(y^{-1/2}xy^{-1/2}) \subset \ol{B_\complex(0, 1)}$, and $\sigma_A(y^{1/2}x^{-1}y^{1/2}) \subset \complex \setminus B_\complex(0, 1)$. Hence $y^{1/2}x^{-1}y^{1/2} \ge 1$, and $x^{-1} \ge y^{-1}$.
\end{proof}