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  • $\begingroup$ Do you mean you define $F^pH^k(X,\mathbb C)=ker(d:F^pA^k\to F^{p+1}A^k)/im(d:F^{p-1}A^k\to F^pA^k)$? $\endgroup$ Commented Jul 24, 2021 at 14:24
  • $\begingroup$ Almost, but there is no shift because $d(F^p)\subset F^p$ $\endgroup$ Commented Jul 24, 2021 at 14:50
  • $\begingroup$ Sorry that I made a mistake, I mean $F^pH^k(X,\mathbb C)=ker(d:F^pA^k\to F^pA^{k+1})/im(d:F^pA^{k-1}\to F^pA^k)$. If we define the filtration like this, I don't think it obvious that we will have of decompostion of this filtration in BC cohomology, since for the Kahler case, it takes p158-159 for Voisin in her book<Hodge theory and...> to prove the corresponding decomposition. $\endgroup$ Commented Jul 24, 2021 at 15:14
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    $\begingroup$ Voisin's argument assumes the Kahler condition, so it won't work here. But see the ref. in the edited version. $\endgroup$ Commented Jul 24, 2021 at 18:42
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    $\begingroup$ Technically, it gives an isomorphism $F^pH^k(X)= H^{p,k-p}\oplus\ldots$ with Dolbeault cohomology. Now use the fact, you stated, that BC and Dolbeault cohomologies are isomorphic. $\endgroup$ Commented Jul 25, 2021 at 14:36