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YCor
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Regarding basis of Holomorphicholomorphic Hardy space

Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a Holomorphicholomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $$ f=\sum_{j=1}^{\infty}\langle f,\phi_j\rangle\phi_j\qquad \forall f\in H^2(\partial\Omega) $$ My Question: can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$

Regarding basis of Holomorphic Hardy space

Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a Holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $$ f=\sum_{j=1}^{\infty}\langle f,\phi_j\rangle\phi_j\qquad \forall f\in H^2(\partial\Omega) $$ My Question: can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$

Regarding basis of holomorphic Hardy space

Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $$ f=\sum_{j=1}^{\infty}\langle f,\phi_j\rangle\phi_j\qquad \forall f\in H^2(\partial\Omega) $$ My Question: can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$

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Daniele Tampieri
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Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a Holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $\forall f\in H^2(\partial\Omega)$, $f=\sum_{j=1}^{\infty}<f,\phi_j>\phi_j$.

My Question is $$ f=\sum_{j=1}^{\infty}\langle f,\phi_j\rangle\phi_j\qquad \forall f\in H^2(\partial\Omega) $$ My Question: can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$

Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and $H^2(\partial\Omega)$ denotes a Holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $\forall f\in H^2(\partial\Omega)$, $f=\sum_{j=1}^{\infty}<f,\phi_j>\phi_j$.

My Question is can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$

Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a Holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $$ f=\sum_{j=1}^{\infty}\langle f,\phi_j\rangle\phi_j\qquad \forall f\in H^2(\partial\Omega) $$ My Question: can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$

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Regarding basis of Holomorphic Hardy space

Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and $H^2(\partial\Omega)$ denotes a Holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi_j$} where $ j\in\mathbb{N}$ such that $\forall f\in H^2(\partial\Omega)$, $f=\sum_{j=1}^{\infty}<f,\phi_j>\phi_j$.

My Question is can we choose $ \forall j\in\mathbb{N},\;\; \phi_j\in A^{\infty}(\Omega)?$