**Proof.**
(A flat morphism of finite presentation is perfect, see More on Morphisms, Lemma 37.59.5.) We begin with a series of preliminary remarks.

We already know that $f^!$ sends $D_{\textit{Coh}}^+(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^+(\mathcal{O}_ X)$, see Lemma 48.17.6.

If $f$ is an open immersion, then (a) and (b) are true because we can take $\overline{X} = Y$ in the construction of $f^!$ and $\mu _ f$. See also Lemma 48.17.1.

If $f$ is a perfect proper morphism, then (b) is true by Lemma 48.13.3.

If there exists an open covering $X = \bigcup U_ i$ and (a) is true for $U_ i \to Y$, then (a) is true for $X \to Y$. Same for (b). This holds because the construction of $f^!$ and $\mu _ f$ commutes with passing to open subschemes.

If $g : Y \to Z$ is a second perfect morphism in $\textit{FTS}_ S$ and (b) holds for $f$ and $g$, then $f^!g^!\mathcal{O}_ Z = Lf^*g^!\mathcal{O}_ Z \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y$ and (b) holds for $g \circ f$ by the commutative diagram of Lemma 48.16.5.

If (a) and (b) hold for both $f$ and $g$, then (a) and (b) hold for $g \circ f$. Namely, then $f^!g^!\mathcal{O}_ Z$ is bounded above (by the previous point) and $L(g \circ f)^*$ has finite cohomological dimension and (a) follows from (b) which we saw above.

From these points we see it suffices to prove the result in case $X$ is affine. Choose an immersion $X \to \mathbf{A}^ n_ Y$ (Morphisms, Lemma 29.39.2) which we factor as $X \to U \to \mathbf{A}^ n_ Y \to Y$ where $X \to U$ is a closed immersion and $U \subset \mathbf{A}^ n_ Y$ is open. Note that $X \to U$ is a perfect closed immersion by More on Morphisms, Lemma 37.59.8. Thus it suffices to prove the lemma for a perfect closed immersion and for the projection $\mathbf{A}^ n_ Y \to Y$.

Let $f : X \to Y$ be a perfect closed immersion. We already know (b) holds. Let $K \in D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$. Then $f^!K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, K)$ (Lemma 48.17.4) and $f_*f^!K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, K)$. Since $f$ is perfect, the complex $f_*\mathcal{O}_ X$ is perfect and hence $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, K)$ is bounded above. This proves that (a) holds. Some details omitted.

Let $f : \mathbf{A}^ n_ Y \to Y$ be the projection. Then (a) holds by repeated application of Lemma 48.17.3. Finally, (b) is true because it holds for $\mathbf{P}^ n_ Y \to Y$ (flat and proper) and because $\mathbf{A}^ n_ Y \subset \mathbf{P}^ n_ Y$ is an open.
$\square$

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