Homogeneous Besov space $\dot B^{-\sigma}(\mathbb R^d)$ is a Banach space

Question

Let $\sigma > 0$ and $\theta \in S(\mathbb R^d)$ (Schwartz function) be fixed, where $\theta$ is such that its Fourier transform $\hat{\theta}$ is compactly supported, $0 \leq \hat{\theta} \leq 1$ and $\hat{\theta} = 1$ in a neighborhood of $0$.

Let $\dot B^{-\sigma}(\mathbb R^d)$ be the subspace of $S'(\mathbb R^d)$ (space of tempered distributions) defined as $$ \dot B^{-\sigma}(\mathbb R^d) = \{u \in S'(\mathbb R^d) \ \mid \ ||u||_{B^{-\sigma}} = \sup_{A > 0} A^{d - \sigma} ||\theta (A \cdot) * u||_{L^{\infty}} < + \infty \}$$

I can prove that $||u||_{B^{-\sigma}}$ is a norm, but I struggle to prove that this space is complete (and why the topology does not depend on the choice of $\theta$ )

If $(u_n)_{n=1}^{\infty} \subset \dot B^{-\sigma}$ is a Cauchy sequence, then $A^{d - \sigma} \theta (A \cdot) * u_n$ (which is in $C^{\infty} \cap L^{\infty}$) converges uniformly (both in $x$ and $A$) to some $f_A \in C^0 \cap L^{\infty}$.

In particular, we have convergence in $S'(\mathbb R^d)$. I think we should use the Fourier transform on $A^{d - \sigma} \theta (A \cdot) * u_n$ and then carry on with something, but I don't know what.

Answer 1

This answer is about the space being Banach. It misses some details, I hope I did not overlook anything important.

Call $A^d \theta(A\cdot)*u=:u_A$. As you said, it's a smooth tempered distribution and we have $u_A\to u$ in $\mathscr S'(\mathbb R^d)$ as $A\to\infty$.

Let $(u_n)$ be a Cauchy sequence. Then, since the norm of $u$ bounds the $L^\infty$ norm of all $u_A$ with some fixed power of $A$, we have that $u_{n,A}$ is a Cauchy sequence in $L^\infty$ for all $A$. In particular, since $L^\infty$ is a Banach space, it converges to some $f_A$ in $L^\infty$, in particular it converges in $\mathscr S'(\mathbb R^d)$. This implies that for fixed $A$, $\widehat \theta(\cdot/A)\widehat u_n$ converges in $\mathscr S'(\mathbb R^d)$ to $\widehat f_A$. But $\widehat \theta(\cdot/A)\widehat u_n$ equals $\widehat u_n$ on a ball of radius proportional to $A$, and $A$ can be taken to be arbitrarily large, so the above implies that $\widehat u_n$ converges to some distribution $z$ in $\mathscr D'(\mathbb R^d)$. Now, by a limiting argument, you know that $f_A=\mathcal F^{-1}[\widehat \theta(\cdot/A)z]$ inherits the $L^\infty$ bounds from $u_{n,A}$, that is, $$ \sup_A A^{-\sigma}\|\mathcal F^{-1}[\widehat \theta(\cdot/A)z]\|_{L^\infty}<\infty.$$ This somehow (I don't see why at the moment) implies that $z$ is indeed a tempered distribution, and from what we said above we should be able to conclude that $u_n\to u:=\mathcal F^{-1} z$ in $\mathscr S'(\mathbb R^d)$. By the above estimate on $z$, we already know that $z\in \dot B^{-\sigma}$ and so what is left is to show that $u_n$ converges to $u$ in $\dot B^{-\sigma}$ (but this simply follows from the fact that $u_A=f_A$.

So, the above is a somewhat direct proof.