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Proof.
Fix ε > 0 \varepsilon>0 and α > 0 \alpha>0 to be chosen later.
It follows from (2) and propositions 2.5, 3.1 that
‖ φ − ψ ‖ L ∞ ( X ) ≤ ε + C 1 [ C a p ω ( | φ − ψ | > ε ) ] α / n . ||\varphi-\psi||_{L^{\infty}(X)}\leq\varepsilon+C_{1}\left[Cap_{\omega}(|\varphi-\psi|>\varepsilon)\right]^{\alpha/n}.
Applying the refined version of lemma 2.2 which involves
the uniform bound on ‖ φ ‖ L ∞ ( X ) , ‖ ψ ‖ L ∞ ( X ) ||\varphi||_{L^{\infty}(X)},||\psi||_{L^{\infty}(X)}
(see inequality (3)), we obtain
C a p ω ( | φ − ψ | > ε ) ≤ C 2 ε n + 2 / q ∫ X | φ − ψ | 2 / q ( f + g ) ω n . Cap_{\omega}(|\varphi-\psi|>\varepsilon)\leq\frac{C_{2}}{\varepsilon^{n+2/q}}\int_{X}|\varphi-\psi|^{2/q}(f+g)\omega^{n}.
It follows thus from Hölder’s inequality that
C a p ω ( | φ − ψ | > ε ) ≤ C 3 ‖ f + g ‖ L p ε n + 2 / q [ ‖ φ − ψ ‖ L 2 ( ω n ) ] 2 / q . Cap_{\omega}(|\varphi-\psi|>\varepsilon)\leq\frac{C_{3}||f+g||_{L^{p}}}{\varepsilon^{n+2/q}}\left[||\varphi-\psi||_{L^{2}(\omega^{n})}\right]^{2/q}.
Choose now ε := ‖ φ − ψ ‖ L 2 ω \varepsilon:=||\varphi-\psi||_{L^{2}}^{\omega} where
0 < γ < 2 / ( 2 + n q ) 0<\gamma<2/(2+nq) . Then
C a p ω ( | φ − ψ | > ε ) ≤ C 4 [ ‖ φ − ψ ‖ L 2 ] 2 / q − γ ( n + 2 / q ) . Cap_{\omega}(|\varphi-\psi|>\varepsilon)\leq C_{4}\left[||\varphi-\psi||_{L^{2}}\right]^{2/q-\gamma(n+2/q)}.
We infer
‖ φ − ψ ‖ L ∞ ( X ) ≤ | | φ − ψ | | L 2 γ + C 5 ‖ φ − ψ ‖ L 2 γ ′ , where γ ′ = α n [ 2 / q − γ ( n + 2 / q ) ] . ||\varphi-\psi||_{L^{\infty}(X)}\leq||\varphi-\psi||_{L^{2}}^{\gamma}+C_{5}||\varphi-\psi||_{L^{2}}^{\gamma^{\prime}},\;\text{ where }\gamma^{\prime}=\frac{\alpha}{n}\left[2/q-\gamma(n+2/q)\right].
We finally choose α > 0 \alpha>0 so large that γ ≤ γ ′ \gamma\leq\gamma^{\prime}
and adjust the value of the constant C C : this yields the desired
estimate.
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