Arbitrarily dilated Triebel-Lizorkin space have equivalence with original Triebel-Lizorkin space?

35 Views Asked by At

The original Triebel-Lizorkin space is defined as follows:

For $\alpha\in\mathbb{R}$, $0<p<\infty$, $0<q<\infty$ and $\phi\in \mathcal{S}(\mathbb{R})$ whose Fourier transform is supported in $\{ \xi\in\mathbb{R}: \frac{1}{2} \leq |\xi|\leq 2\}$ and $|\hat{\phi}(\xi)| \geq c >0$ if $\frac{3}{5} \leq |\xi| \leq \frac{5}{3}$, the Triebel-Lizorkin space $\dot{F}^{\alpha,q}_p$ is the collection of all $f\in \mathcal{S}'/\mathcal{P}(\mathbb{R})$ such that $$ \left\| f \right\|_{\dot{F}^{\alpha,q}_p} = \left\| \left( \sum\limits_{j\in\mathbb{Z}} (2^{j\alpha} |\phi_j * f|)^q \right)^{1/q}\right\|_{L^p}< \infty$$ where $\phi_j = 2^j \phi(2^jx)$.

Now I newly defined an 'arbitrarily dilated' Triebel-Lizorkin space $\dot{F}^{A,\alpha,q}_p$ (A>1) such that for the same $\alpha, p, q$ and $\phi\in \mathcal{S}(\mathbb{R})$ whose Fourier transform is supported in $\{ \xi\in\mathbb{R}: \frac{1}{A^2} \leq |\xi|\leq 1\}$ and $|\hat{\phi}(\xi)| \geq c >0$ if $\frac{1}{A \sqrt{A}} \leq |\xi| \leq \frac{1}{\sqrt{A}}$, the Triebel-Lizorkin space $\dot{F}^{A,\alpha,q}_{p}$ is the collection of all $f\in \mathcal{S}'/\mathcal{P}(\mathbb{R})$ such that $$ \left\| f \right\|_{\dot{F}^{A,\alpha,q}_p} = \left\| \left( \sum\limits_{j\in\mathbb{Z}} (A^{j\alpha} |\phi_{A,j} * f|)^q \right)^{1/q}\right\|_{L^p}< \infty$$ where $\phi_{A,j} = A^j \phi(A^jx)$.

I am curious about the equivalence of these two norms (if so, so do the spaces). I think this will be related to the Littlewood-Paley theory which is almost always taking dyadic dilation, but actually I'm not really accustomed to this theory. I cannot found any reference about arbitrary dilation about Littlewood-Paley theory or Triebel-Lizorkin spaces. If you have any ideas or reference, please share it.

Thanks a lot.