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Version 5 |
| \PMlinkescapephrase{decomposition} |
\PMlinkescapephrase{decomposition} |
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{\bf [Not finished]} |
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| Every real valued function $f$ admits a well-known decomposition into its \PMlinkescapetext{positive} and \PMlinkescapetext{negative} parts: $f = f_+ - f_-$. (These functions are defined within the entry Lebesgue integral.) There is an analogous result for self-adjoint elements in a \PMlinkname{$C^*$-algebra}{CAlgebra} that we will now describe. |
Every real valued function $f$ admits a well-known decomposition into its \PMlinkescapetext{positive} and \PMlinkescapetext{negative} parts: $f = f_+ - f_-$. (These functions are defined within the entry Lebesgue integral.) There is an analogous result for self-adjoint elements in a \PMlinkname{$C^*$-algebra}{CAlgebra} that we will now describe. |
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| {\bf Theorem - } Let $\mathcal{A}$ be a $C^*$-algebra and $a \in \mathcal{A}$ a self-adjoint element. Then there are unique positive elements $a_+$ and $a_-$ in $\mathcal{A}$ such that: |
{\bf Theorem - } Let $\mathcal{A}$ be a $C^*$-algebra and $a \in \mathcal{A}$ a self-adjoint element. Then there are unique positive elements $a_+$ and $a_-$ in $\mathcal{A}$ such that: |
| \begin{itemize} |
\begin{itemize} |
| \item $a= a_+ - a_-$ |
\item $a= a_+ - a_-$ |
| \item $a_+a_- = a_-a_+ = 0$ |
\item $a_+a_- = a_-a_+ = 0$ |
| \item Both $a_+$ and $a_-$ belong to $C^*$-subalgebra generated by $a$. |
\item Both $a_+$ and $a_-$ belong to $C^*$-subalgebra generated by $a$. |
| \item $\|a\| = \max\{\|a_+\|, \|a_-\|\}$ |
\item $\|a\| = \max\{\|a_+\|, \|a_-\|\}$ |
| \end{itemize} |
\end{itemize} |
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$\,$ |
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| {\bf Remark - } As a particular case, the result provides a decomposition of each self-adjoint operator $T$ on a Hilbert space as a difference of two positive operators $T=T_+ - T_-$ such that $\mathrm{Ran}\; T_- \subseteq \mathrm{Ker}\; T_+$ and $\mathrm{Ran}\; T_+ \subseteq \mathrm{Ker}\; T_-$, where $\mathrm{Ran}\;$ and $\mathrm{Ker}\;$ denote, respectively, the range and kernel of an operator. |
{\bf Remark - } As a particular case, the result provides a decomposition of each self-adjoint operator $T$ on a Hilbert space as a difference of two positive operators $T=T_+ - T_-$ such that $\mathrm{Ran}\; T_- \subseteq \mathrm{Ker}\; T_+$ and $\mathrm{Ran}\; T_+ \subseteq \mathrm{Ker}\; T_-$, where $\mathrm{Ran}\;$ and $\mathrm{Ker}\;$ denote, respectively, the range and kernel of an operator. |
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$\,$ |
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| {\bf \emph{Proof:}} |
{\bf \emph{Proof:}} |
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| Let us \PMlinkescapetext{fix} some notation first: |
Let us \PMlinkescapetext{fix} some notation first: |
| \begin{itemize} |
\begin{itemize} |
| \item $\sigma(a)$ denotes the spectrum of $a \in \mathcal{A}$. |
\item $\sigma(a)$ denotes the spectrum of $a \in \mathcal{A}$. |
| \item $C^*[a]$ denotes the $C^*$-subalgebra generated by $a$. |
\item $C^*[a]$ denotes the $C^*$-subalgebra generated by $a$. |
| \item $C_0 \big(\sigma(a)\setminus \{0\}\big)$ denotes the algebra of continuous functions in $\sigma(a)\setminus \{0\}$ that vanish at infinity. |
\item $C_0 \big(\sigma(a)\setminus \{0\}\big)$ denotes the algebra of continuous functions in $\sigma(a)\setminus \{0\}$ that vanish at infinity. |
| \end{itemize} |
\end{itemize} |
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| Let $f, f_+, f_- \in C_0\big(\sigma(a)\setminus \{0\}\big)$ be the functions defined by |
Let $f, f_+, f_- \in C_0\big(\sigma(a)\setminus \{0\}\big)$ be the functions defined by |
| \begin{align*} |
\begin{align*} |
| f(t):=t \qquad\qquad |
f(t):=t \qquad\qquad |
| f_+(t) := |
f_+(t) := |
| \begin{cases} |
\begin{cases} |
| t, & $if$\;\; t \geq 0\\ |
t, & $if$\;\; t \geq 0\\ |
| 0, & $if$\;\; t \leq 0 |
0, & $if$\;\; t \leq 0 |
| \end{cases} |
\end{cases} |
| \qquad\qquad f_-(t):= |
\qquad\qquad f_-(t):= |
| \begin{cases} |
\begin{cases} |
| 0, & $if$\;\; t \geq 0\\ |
0, & $if$\;\; t \geq 0\\ |
| -t, & $if$\;\; t \leq 0 |
-t, & $if$\;\; t \leq 0 |
| \end{cases} |
\end{cases} |
| \end{align*} |
\end{align*} |
| Since $a$ is \PMlinkescapetext{self-adjoint}, $\sigma(a) \subseteq \mathbb{R}$, so the above functions are well defined. It is clear that |
Since $a$ is \PMlinkescapetext{self-adjoint}, $\sigma(a) \subseteq \mathbb{R}$, so the above functions are well defined. It is clear that |
| \begin{align} |
\begin{align} |
| f = f_+ - f_- \;\;\;\text{and}\;\;\; f_+f_-=f_-f_+=0 \;\;\;\text{and}\;\;\; f_+, f_- \;\text{are both positive} |
f = f_+ - f_- \;\;\;\text{and}\;\;\; f_+f_-=f_-f_+=0 \;\;\;\text{and}\;\;\; f_+, f_- \;\text{are both positive} |
| \end{align} |
\end{align} |
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| The continuous functional calculus gives an isomorphism $C^*[a] \cong C_0\big(\sigma(a)\setminus \{0\}\big)$ such that the element $a$ corresponds to the function $f$. Let $a_+$ and $a_-$ be the elements corresponding to $f_+$ and $f_-$ respectively. From the \PMlinkescapetext{observations} made in (1) it is now clear that |
The continuous functional calculus gives an isomorphism $C^*[a] \cong C_0\big(\sigma(a)\setminus \{0\}\big)$ such that the element $a$ corresponds to the function $f$. Let $a_+$ and $a_-$ be the elements corresponding to $f_+$ and $f_-$ respectively. From the \PMlinkescapetext{observations} made in (1) it is now clear that |
| \begin{itemize} |
\begin{itemize} |
| \item $a_+$ and $a_-$ are both positive elements. |
\item $a_+$ and $a_-$ are both positive elements. |
| \item $a = a_+ - a_-$ |
\item $a = a_+ - a_-$ |
| \item $a_+a_- = a_-a_+ = 0$ |
\item $a_+a_- = a_-a_+ = 0$ |
| \item Both $a_+$ and $a_-$ belong to $C^*[a]$. |
\item Both $a_+$ and $a_-$ belong to $C^*[a]$. |
| \end{itemize} |
\end{itemize} |
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| From the fact the every $C^*$-isomorphism is isometric (see this \PMlinkname{entry}{HomomorphismsOfCAlgebrasAreContinuous}) and $\|f\| = \max\{\|f_+\|, \|f_-\|\}$ it follows that $\|a\| = \max\{\|a_+\|, \|a_-\|\}$. |
From the fact the every $C^*$-isomorphism is isometric (see this \PMlinkname{entry}{HomomorphismsOfCAlgebrasAreContinuous}) and $\|f\| = \max\{\|f_+\|, \|f_-\|\}$ it follows that $\|a\| = \max\{\|a_+\|, \|a_-\|\}$. |
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The uniqueness of the decomposition follows from the uniqueness of the decomposition of real valued functions in its positive and negative parts $f = f_+-f_-$ (with $f_+f_- = 0$). $\square$
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It remains to prove the uniqueness of the decomposition $(...)$
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