**Proof.**
Choose $R \to P \to A$ as in the definition of $\varphi ^!$. Recall that $R \to A$ is a perfect ring map (More on Algebra, Lemma 15.82.4) and hence $A$ is perfect as a $P$-modue (More on Algebra, Lemma 15.82.2). This shows that $\omega _{A/R}^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ by Lemma 47.24.2. To show $\omega _{A/R}^\bullet $ is $R$-perfect it suffices to show it has finite tor dimension as a complex of $R$-modules. This is true because $\omega _{A/R}^\bullet = \varphi ^!(R) = R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]$ maps to $R\mathop{\mathrm{Hom}}\nolimits _ P(A, P)[n]$ in $D(P)$, which is perfect in $D(P)$ (More on Algebra, Lemma 15.74.15), hence has finite tor dimension in $D(R)$ as $R \to P$ is flat. This proves (1).

Proof of (2). The object $R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _{A/R}^\bullet , \omega _{A/R}^\bullet )$ of $D(A)$ maps in $D(P)$ to

\begin{align*} R\mathop{\mathrm{Hom}}\nolimits _ P(\omega _{A/R}^\bullet , R\mathop{\mathrm{Hom}}\nolimits (A, P)[n]) & = R\mathop{\mathrm{Hom}}\nolimits _ P(R\mathop{\mathrm{Hom}}\nolimits _ P(A, P)[n], P)[n] \\ & = R\mathop{\mathrm{Hom}}\nolimits _ P(R\mathop{\mathrm{Hom}}\nolimits _ P(A, P), P) \end{align*}

This is equal to $A$ by the already used More on Algebra, Lemma 15.74.15.

Proof of (3). By Lemma 47.25.1 there is an isomorphism

\[ \omega _{A/R}^\bullet \otimes _ A^\mathbf {L} (A \otimes _ R k) \cong \omega ^\bullet _{A \otimes _ R k/k} \]

and the right hand side is a dualizing complex by Lemma 47.24.3.
$\square$

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