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\markboth{}{Foundations of Mathematics \--- Dr. Russell E. Goodman}
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\bf\emph{\Large{Scrambled Proof \#2}}
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\noindent Consider the following theorem and the ``pieces" of the proof you've been given. Please put the pieces in logical order to form an understandable proof.

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\noindent \textbf{Theorem:} Let $\mathcal{H}$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$ and let $T$ be a bounded linear operator on $\mathcal{H}$.  Then there exists a bounded linear operator $T^{*}$ on $\mathcal{H}$ such that, for all $x$ and $y$ in $\mathcal{H}$, $\langle Tx,y\rangle=\langle x,T^{*}y\rangle$.

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\noindent\textbf{Proof:}

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\item Finally, to show that $T^{*}$ is linear, a computation shows that for any complex numbers $\alpha$ and $\beta$ and $y_{1}$ and $y_{2}$ in $\mathcal{H}$, the vector $\alpha T^{*}(y_{1})+\beta T^{*}(y_{2})$ satisfies $\langle Tx,\alpha y_{1}+\beta y_{2}\rangle=\langle x,\alpha T^{*}(y_{1})+\beta T^{*}(y_{2})\rangle$ for all $x$, and using uniqueness, therefore, $T^{*}(\alpha y_{1}+\beta y_{2})=\alpha T^{*}(y_{1})+\beta T^{*}(y_{2})$ as required.
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\item Therefore, $f_{y}$ is a bounded linear functional on $\mathcal{H}$ and since $\mathcal{H}$ is self-dual, there exists a unique vector $y^{*}$ in $\mathcal{H}$ such that $f_{y}(x)=\langle x,y^{*}\rangle$ for all $x$.
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\item Thus, the candidate for $T^{*}$ is satisfactory, and the proof is complete.
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\item Observe that this function $f_{y}$ is linear in $x$ by standard properties of the inner product and that it maps $\mathcal{H}$ to the complex numbers.
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\item Next, observe that by construction, $T^{*}$ satisfies $\langle Tx,y\rangle=\langle x,T^{*}y\rangle$ for all $x$ and $y$ in $\mathcal{H}$.
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\item We will construct the required $T^{*}$ vector by vector.
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\item First, clearly $T^{*}$ so constructed maps $\mathcal{H}$ to $\mathcal{H}$.
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\item Let $y$ be a fixed vector in $\mathcal{H}$ and consider the function $f_{y}$ defined on $\mathcal{H}$ by $f_{y}(x)=\langle Tx,y\rangle$ for all values of $x$.
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\item Further, $f_{y}$ is bounded, since for any $x$, $|f_{y}(x)|=|\langle Tx,y\rangle|\le\|T\|\|x\|\|y\|$.
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\item Further, to show $T^{*}$ is bounded, note that $\|T^{*}y\|=\sup{\{|\langle x,T^{*}y\rangle|:x\in\mathcal{H},\|x\|\le1\}}=\sup{\{|\langle Tx,y\rangle|:x\in\mathcal{H},\|x\|\le1\}}\le\sup{\{\|T\|\|x\|\|y\|:x\in\mathcal{H},\|x\|\le1\}}=\|T\|\|y\|$.
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\item Let $\mathcal{H}$ and $T$ be as in the hypothesis.
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\item Our candidate for $T^{*}$ is to set, for each $y$ in $\mathcal{H}$, $T^{*}(y)=y^{*}$, where $y^{*}$ is as constructed above for $y$.
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\noindent\textbf{Key:} 11, 6, 8, 4, 9, 2, 12, 5, 7, 10, 1, 3\\

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