[proofplan] The assertion is pointwise, so we compute at an arbitrary point using a unitary coframe for the Hermitian cotangent space. A basis $(p,q)$-form is a wedge of $p$ holomorphic coframe elements and $q$ antiholomorphic coframe elements. The Hodge star is characterized by the Hermitian wedge-pairing identity, and that identity forces the conjugate of the starred form to supply exactly the missing holomorphic and antiholomorphic factors. Conjugation swaps type, so the starred form itself has type $(n-q,n-p)$. [/proofplan]

[step:Reduce the type statement to a pointwise calculation] Let $x\in X$ be arbitrary. Denote the complexified cotangent space at $x$ by $T_x^*X_{\mathbb C}:=T_x^*X\otimes_{\mathbb R}\mathbb C$. The complex structure gives a direct sum decomposition $T_x^*X_{\mathbb C}=(T_x^{1,0}X)^*\oplus (T_x^{0,1}X)^*$. For $0\le p,q\le n$, define the pointwise space of forms of type $(p,q)$ by $\Lambda_x^{p,q}:=\Lambda^p (T_x^{1,0}X)^*\otimes \Lambda^q (T_x^{0,1}X)^*$. A smooth form $\alpha\in A^{p,q}(X)$ is precisely a smooth section of the vector bundle whose fibre at $x$ is $\Lambda_x^{p,q}$. Therefore it is enough to prove that the pointwise Hodge star $*_x:\Lambda_x^{p,q}\longrightarrow \Lambda_x^{2n-p-q}T_x^*X_{\mathbb C}$ has image contained in $\Lambda_x^{n-q,n-p}$ for every $x\in X$. [/step]

[step:Choose a unitary coframe and write a basis element] Because $g$ is Hermitian, there is a unitary coframe $\theta_1,\dots,\theta_n\in (T_x^{1,0}X)^*$ for the Hermitian [vector space](/page/Vector%20Space) $(T_x^{1,0}X)^*$. Its complex conjugates $\overline{\theta_1},\dots,\overline{\theta_n}\in (T_x^{0,1}X)^*$ form the corresponding antiholomorphic coframe. Let $I=\{i_1<\cdots<i_p\}\subseteq \{1,\dots,n\}$ and $J=\{j_1<\cdots<j_q\}\subseteq \{1,\dots,n\}$. Define $\theta_I:=\theta_{i_1}\wedge\cdots\wedge\theta_{i_p}$ and $\overline{\theta_J}:=\overline{\theta_{j_1}}\wedge\cdots\wedge\overline{\theta_{j_q}}$. Then $e_{I,J}:=\theta_I\wedge\overline{\theta_J}$ is a basis element of $\Lambda_x^{p,q}$, and the elements $e_{I,J}$ form a basis of $\Lambda_x^{p,q}$ as $I$ and $J$ vary. Since $*_x$ is complex-linear under the standard complex Hodge-star convention, it suffices to determine the type of $*_x e_{I,J}$. [/step]

[step:Use the Hodge star identity to count the complementary factors]Let $\operatorname{vol}_x\in \Lambda_x^{n,n}$ denote the positive Hermitian volume form at $x$. The complex Hodge star is characterized by the identity $\beta\wedge \overline{*_x\gamma}=(\beta,\gamma)_x\,\operatorname{vol}_x$ for all complex forms $\beta,\gamma$ of the same total degree, where $(\cdot,\cdot)_x$ is the pointwise Hermitian [inner product](/page/Inner%20Product) on exterior forms induced by $g$. Apply this identity with $\beta=\gamma=e_{I,J}$. Since $e_{I,J}$ is a unit basis element for the induced Hermitian inner product, we have $e_{I,J}\wedge \overline{*_x e_{I,J}}=\operatorname{vol}_x$. The form $e_{I,J}$ contains exactly the holomorphic factors $\theta_i$ with $i\in I$ and the antiholomorphic factors $\overline{\theta_j}$ with $j\in J$. For the wedge product with $\overline{*_x e_{I,J}}$ to be a nonzero multiple of the volume form, the form $\overline{*_x e_{I,J}}$ must contain the missing holomorphic factors $\theta_a$ for $a\notin I$ and the missing antiholomorphic factors $\overline{\theta_b}$ for $b\notin J$. Therefore $\overline{*_x e_{I,J}}\in \Lambda_x^{n-p,n-q}$. Complex conjugation interchanges holomorphic and antiholomorphic type, so $*_x e_{I,J}\in \Lambda_x^{n-q,n-p}$.[/step]

[guided]The point of the Hodge-star identity is that it determines which factors the starred form must contribute to make a top-degree volume form. We use the standard Hermitian convention $\beta\wedge \overline{*_x\gamma}=(\beta,\gamma)_x\,\operatorname{vol}_x$ for forms $\beta$ and $\gamma$ of the same total degree. Here $\operatorname{vol}_x$ has type $(n,n)$, because it is a nonzero multiple of $\theta_1\wedge\cdots\wedge\theta_n\wedge \overline{\theta_1}\wedge\cdots\wedge\overline{\theta_n}$. Now set $\beta=\gamma=e_{I,J}=\theta_I\wedge\overline{\theta_J}$. Since the coframe is unitary, the wedge monomials $e_{I,J}$ are orthonormal for the induced Hermitian inner product, so $(e_{I,J},e_{I,J})_x=1$. Thus the defining identity gives $e_{I,J}\wedge \overline{*_x e_{I,J}}=\operatorname{vol}_x$. This equation forces a type count. The form $e_{I,J}$ already contributes $p$ holomorphic factors and $q$ antiholomorphic factors. A nonzero top-type form of type $(n,n)$ must contain exactly $n$ holomorphic factors and exactly $n$ antiholomorphic factors. Hence $\overline{*_x e_{I,J}}$ must contribute $n-p$ holomorphic factors and $n-q$ antiholomorphic factors. Equivalently, $\overline{*_x e_{I,J}}\in \Lambda_x^{n-p,n-q}$. Finally, complex conjugation swaps the two type indices: the conjugate of a form of type $(a,b)$ has type $(b,a)$. Applying this to $\overline{*_x e_{I,J}}$ gives $*_x e_{I,J}\in \Lambda_x^{n-q,n-p}$. This is exactly the desired type change.[/guided]

[step:Extend the basis calculation to all smooth forms] The basis calculation shows that every basis vector $e_{I,J}\in \Lambda_x^{p,q}$ is sent by $*_x$ into $\Lambda_x^{n-q,n-p}$. Since $*_x$ is complex-linear and the $e_{I,J}$ form a basis of $\Lambda_x^{p,q}$, it follows that $*_x(\Lambda_x^{p,q})\subseteq \Lambda_x^{n-q,n-p}$. The point $x\in X$ was arbitrary. Therefore, for every smooth section $\alpha\in A^{p,q}(X)$, the smooth form $*\alpha$ has value $(*\alpha)_x\in \Lambda_x^{n-q,n-p}$ at every point $x\in X$. Hence $*\alpha\in A^{n-q,n-p}(X)$. Thus $*:A^{p,q}(X)\longrightarrow A^{n-q,n-p}(X)$, as claimed. [/step]