Consider a cube $\Omega = [-1,1]^d$ (we can also call it a torus), and complex bounded periodic functions $g \in L^\infty_{\text{per}}(\Omega)$ such that $\int_\Omega g = 0$. 'Periodic' means that $g(...,-1,...) = g(...,1,...)$. For each $g$, define $f_g$ to be the unique function such that $\int_\Omega f_g = 0$ and \begin{align*} \Delta f_g = g. \end{align*} Can we prove that $f_g \in L^\infty_{\text{per}}(\Omega)$ and \begin{align*} \|f_g\|_{L^\infty_{\rm per}} \le c \|g\|_{L^\infty_{\rm per}} \end{align*} where $c$ does not depend on $g$ ?

For $\Delta f_g = g$, can we prove that $f_g \in L^\infty_{\text{per}}(\Omega)$ and \begin{align*} \|f_g\|_{L^\infty_{\rm per}} \le c \|g\|_{L^\infty_{\rm per}} \end{align*} where $c$ does not depend on $g$ ?

Consider a cube $\Omega = [-1,1]^d$, and complex bounded periodic functions $g \in L^\infty_{\text{per}}(\Omega)$ such that $\int_\Omega g = 0$. For each $g$, define $f_g$ to be the unique function such that $\int_\Omega f_g = 0$ and \begin{align*} \Delta f_g = g. \end{align*} Can we prove that $f_g \in L^\infty_{\text{per}}(\Omega)$ and \begin{align*} \|f_g\|_{L^\infty_{\rm per}} \le c \|g\|_{L^\infty_{\rm per}} \end{align*} where $c$ does not depend on $g$ ?

Consider a cube $\Omega = [-1,1]^d$ (we can also call it a torus), and complex bounded periodic functions $g \in L^\infty_{\text{per}}(\Omega)$ such that $\int_\Omega g = 0$. 'Periodic' means that $g(...,-1,...) = g(...,1,...)$. For each $g$, define $f_g$ to be the unique function such that $\int_\Omega f_g = 0$ and \begin{align*} \Delta f_g = g. \end{align*} Can we prove that $f_g \in L^\infty_{\text{per}}(\Omega)$ and \begin{align*} \|f_g\|_{L^\infty_{\rm per}} \le c \|g\|_{L^\infty_{\rm per}} \end{align*} where $c$ does not depend on $g$ ?

Consider a cube $\Omega = [-1,1]^d$, and complex bounded periodic functions $g \in L^\infty_{\text{per}}(\Omega)$ such that $\int_\Omega g = 0$. For each $g$, define $f_g$ to be the unique function such that $\int_\Omega f_g = 0$ and \begin{align*} \Delta f_g = g. \end{align*} Can we prove that $f_g \in L^\infty_{\text{per}}(\Omega)$ and \begin{align*} \|f_g\|_{L^\infty_{\rm per}} \le c \|g\|_{L^\infty_{\rm per}} \end{align*} where $c$ does not depend on $g$ ? If true, does someone know a reference where it's proved ?

Consider a cube $\Omega = [-1,1]^d$, and complex bounded periodic functions $g \in L^\infty_{\text{per}}(\Omega)$ such that $\int_\Omega g = 0$. For each $g$, define $f_g$ to be the unique function such that $\int_\Omega f_g = 0$ and \begin{align*} \Delta f_g = g. \end{align*} Can we prove that $f_g \in L^\infty_{\text{per}}(\Omega)$ and \begin{align*} \|f_g\|_{L^\infty_{\rm per}} \le c \|g\|_{L^\infty_{\rm per}} \end{align*} where $c$ does not depend on $g$ ?