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Let $\mathbb{T}^3:=(\mathbb{R}/\mathbb{Z})^3$ be the $3$-torus and $C^\infty(\mathbb{T}^3,\mathbb{R}^3)$ be the Frechet space of smooth periodic vector fields on $\mathbb{T}^3$.

By Helmholtz decomposition (or Hodge's Theorem), each $f \in C^\infty(\mathbb{T}^3,\mathbb{R}^3)$ is uniquely written as \begin{equation} f(x)=a+g(x)+h(x) \end{equation} where $a \in \mathbb{R}^3$ is a constant vector while $g, h \in C^\infty(\mathbb{T}^3,\mathbb{R}^3)$ satisfy

  1. $\int_{\mathbb{T}^3}g=\int_{\mathbb{T}^3}h=0$
  2. $\nabla \times g = 0$, that is $g$ is curl-free
  3. $\nabla \cdot h=0$, that is $h$ is divergence-free

Now, the Leray projection $\mathbb{P}$ is a linear map on $C^\infty(\mathbb{T}^3,\mathbb{R}^3)$ defined by \begin{equation} \mathbb{P}f= h \end{equation}

Now, I wonder if this $\mathbb{P}$ is continuous with respect to the Frechet topology of $C^\infty(\mathbb{T}^3,\mathbb{R}^3)$. I am concerned with this issue because the seminorms on $C^\infty(\mathbb{T}^3,\mathbb{R}^3)$ contain supremums, and usual Fourier expansion is not (equi-)continuous with respect to $L^\infty$-norm..

Could anyone pleaes clarify for me?

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    $\begingroup$ Yes it is. It is enough to look at the operator $f \mapsto \mathrm{grad}(\triangle^{-1}(\mathrm{div}(f)))$. This operator maps the Sobolev space $H^k$ to itself boundedly. You have that $C^k$ embeds continuously in $H^k$, and $H^k$ embeds into $C^{k-2}$ by Sobolev embedding. So this shows that if $f_j\to f$ in $C^{k}$ you have $\mathbb{P} f_j \to f$ in $C^{k-2}$. $\endgroup$
    – Willie Wong
    Commented Dec 29, 2023 at 1:49
  • $\begingroup$ @WillieWong Thank you. I guess $k \geq 2$ here? $\endgroup$
    – Isaac
    Commented Dec 29, 2023 at 14:13
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    $\begingroup$ For the boundedness of the Leray projection on $H^k$, you don't need $k \geq 2$. But for the second part, you want $k \geq 2$ to ensure $k - 2 \geq 0$. Of course, you actually want $k$ to be arbitrarily large to prove that $\mathbb{P} f_j \to f$ in all of the seminorms generating the topology of $C^\infty$. $\endgroup$
    – Willie Wong
    Commented Dec 29, 2023 at 14:17
  • $\begingroup$ @WillieWong Thank you. Just one thing : do you mean $\mathbb{P}f_j \to \mathbb{P}f$ for $f_j \to f$? $\endgroup$
    – Isaac
    Commented Dec 29, 2023 at 16:58
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    $\begingroup$ Yes, obviously. $\endgroup$
    – Willie Wong
    Commented Dec 29, 2023 at 23:05

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