Homogeneous Besov space $\dot B^{-\sigma}(\mathbb R^d)$ is a Banach space
This question has been asked several times but no answer.
The question is: How can we find out the limit $u$ of Cauchy sequence $u_{n}$ in $\dot{B}^{-\sigma}(\mathbb{R}^{d})$?
The following is my attemption(I can't justify whether it is right.)
Recently, I am confused by a question from the book "Fourier analysis and nonlinear partial differential equations". In page 30, the authers give us a definition:
Definition: Let $\theta$ be a function in $\mathcal{S}(\mathbb{R}^{d})$ such that $\hat{\theta}$ is compactly supported, has value $1$ near $0$, and satisfies $0\leq\hat{\theta}\leq1$. For $u$ in $\mathcal{S}'(\mathbb{R}^{d})$ and $\sigma>0$, we set $$\|u\|_{\dot{B}^{-\sigma}}:=\sup_{A>0}A^{d-\sigma}\|\theta(A\cdot\ )*u\|_{L^{\infty}}.$$ It is left to the reader that the space$\dot{B}^{-\sigma}$ of tempered distributions $u$ such that$\|u\|_{\dot{B}^{-\sigma}}$ is finite is a Banach space.
My attempting is: Assuming that $\{u_{n}\}$ is a Cauchy sequence, that is to say, for any $\epsilon>0$, there exists $N>0$, such that $\forall m,n>N$, we have $\|u_{n}-u_{m}\|_{\dot{B}^{-\sigma}}<\epsilon$, which says that for any $A>0$, we have $A^{d-\sigma}\|\theta(A\cdot)*(u_{n}-u_{m})\|_{L^{\infty}}<\epsilon$.
I want to know, is $\theta(A\cdot)*u_{n}$ smooth? Yes, it is smooth according to the comment below. Another question is: What should we do to get the limit $u$? How can we use the condition $\hat{\theta}$ has compact support and $\hat{\theta}$ is equal to $1$ near $0$? Any hints are wellcome, thanks a lot ahead!
According to the comment below, I know that in GTM249 page 127 that $\theta(A\cdot)*u_{n}$ is a smooth function. So, we can get that $A^{d}\theta(A\cdot)*u_{n}$ converges to a function, denoted by $u_{A}$, in $L^{\infty}$. Thus in $\mathscr{S}'$, $A^{d}\theta(A\cdot)*u_{n}\to u_{A}$.
The fourier transform of $A^{d}\theta(A\cdot)*u_{n}$ is $\hat{\theta}(\frac{\xi}{A})\widehat{u_{n}}$, because $\hat{\theta}(\frac{\xi}{A})$ has compact support, WLOG, say, $B(0,2A)$ and it equals to $1$ in $B(0,A)$. Then, in $\mathscr{S}'$, we have $\hat{\theta}(\frac{\xi}{A})\widehat{u_{n}}\to \widehat{u_{A}}$
In $B(0,A)$, we have $\widehat{u_{n}}\to \hat{u}$ in $\mathscr{D}'(B(0,A))$
Who make a opposite vote to my question????????????????????????????
