Proof.
Let E be an object of D_\mathit{QCoh}(\mathcal{O}_ X). To prove (1) we have to show that Rf_*E has quasi-coherent cohomology sheaves. This question is local on Y, hence we may assume Y is quasi-compact. Pick N = N(X, Y, f) as in Cohomology of Spaces, Lemma 69.8.1. Thus R^ pf_*\mathcal{F} = 0 for all quasi-coherent \mathcal{O}_ X-modules \mathcal{F} and all p \geq N. Moreover R^ pf_*\mathcal{F} is quasi-coherent for all p by Cohomology of Spaces, Lemma 69.3.1. These statements remain true after base change.
First, assume E is bounded below. We will show (1) and (2) and (3) hold for such E with our choice of N. In this case we can for example use the spectral sequence
R^ pf_*H^ q(E) \Rightarrow R^{p + q}f_*E
(Derived Categories, Lemma 13.21.3), the quasi-coherence of R^ pf_*H^ q(E), and the vanishing of R^ pf_*H^ q(E) for p \geq N to see that (1), (2), and (3) hold in this case.
Next we prove (2) and (3). Say H^ m(E) = 0 for m > 0. Let V be an affine object of Y_{\acute{e}tale}. We have H^ p(V \times _ Y X, \mathcal{F}) = 0 for p \geq N, see Cohomology of Spaces, Lemma 69.3.2. Hence we may apply Lemma 75.5.8 to the functor \Gamma (V \times _ Y X, -) to see that
R\Gamma (V, Rf_*E) = R\Gamma (V \times _ Y X, E)
has vanishing cohomology in degrees \geq N. Since this holds for all V affine in Y_{\acute{e}tale} we conclude that H^ m(Rf_*E) = 0 for m \geq N.
Next, we prove (1) in the general case. Recall that there is a distinguished triangle
\tau _{\leq -n - 1}E \to E \to \tau _{\geq -n}E \to (\tau _{\leq -n - 1}E)[1]
in D(\mathcal{O}_ X), see Derived Categories, Remark 13.12.4. By (2) we see that Rf_*\tau _{\leq -n - 1}E has vanishing cohomology sheaves in degrees \geq -n + N. Thus, given an integer q we see that R^ qf_*E is equal to R^ qf_*\tau _{\geq -n}E for some n and the result above applies.
\square
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