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3 added 291 characters in body

This is still true, although as Francesco says in his comment above, it is trivially so in general : in complex dimension 2 and more, a generic almost complex structure has only constant holomorphic functions, even locally.

Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\,df$, then (obviously) $d(df\circ J)=0$.

But the second order operator $f\mapsto (d(df\circ J))^{1,1}$ from functions to $(1,1)$-forms has the "same" principal symbol at each point as in the integrable case (the "plurisubharmonic Hessian", so to speak)speak, perhaps up to some $-2i$ factor).

In particular you can compose it with contraction by a positive smooth $(1,1)$ form (given by any hermitian metric) to obtain a "Laplace operatoroperator", which satisfies the maximum principle. EDIT (after comment by OP): it is important to observe that the operator vanishes on constants to derive the maximum principle -- locally it writes $\sum g_{jk}(x) \partial_j\partial_k +\sum b_i(x) \partial_i$, with $g_{jk}$ symmetric positive definite.

2 edited body; added 124 characters in body

This is still true, although as Francesco says in his comment above, it is vacuously trivially so in general : in complex dimension 2 and more, a generic almost complex structure has only constant holomorphic functions, even locally.

Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\,df$, then (obviously) $d(df\circ J)=0$.

But the second order operator $f\mapsto (d(df\circ J))^{1,1}$ from functions to $(1,1)$-forms has the "same" principal symbol at each point as in the integrable case (the "plurisubharmonic Hessian", so to speak).

In particular you can compose it with contraction by a positive smooth $(1,1)$ form (given by any hermitian metric) to obtain a Laplace operator, which satisfies the maximum principle.

1

This is still true, although as Francesco says in his comment above, it is vacuously so in general.

Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\,df$, then (obviously) $d(df\circ J)=0$.

But the second order operator $f\mapsto (d(df\circ J))^{1,1}$ from functions to $(1,1)$-forms has the "same" principal symbol at each point as in the integrable case (the "plurisubharmonic Hessian", so to speak).

In particular you can compose it with contraction by a positive smooth $(1,1)$ form (given by any hermitian metric) to obtain a Laplace operator, which satisfies the maximum principle.