How to prove that a delta function belongs to the Besov space $B^{-1}$? $\def\R{\mathbb{R}}$
$\DeclareMathOperator{\supp}{supp}$
I am trying to understand the definition of the Besov space and to prove that the delta function in $\R^1$. However I got stuck.
First, let us recall that a sequence of smooth functions $\{\phi_i\}_{i\ge-1}$ is called a partition of unity if $\sum_{i=-1}^\infty \phi_i=1$, $\supp(\phi_{-1})\subset\{x\colon |x|\le 2\}$; $\supp(\phi_{i})\subset\{x\colon 2^{i-1}\le |x|\le 2^{i+1}\}$.
The Besov space $B^s_{p,q}$, where $s\in\R$, $p,q\ge1$ is defined as the collection of all functions $f$ such that
$$
\|f\|_{p,q}^s:=\|(2^{sj}\|F^{-1}\phi_j Ff\|_{L_p})_{j\ge-1}\|_{l_q}<\infty.
$$
Here $F$ is the Fourier transform operator.
I would like to prove that for any fixed $x\in\R$ the delta function $\delta_x$ is in $B^{-1}_{\infty,\infty}$. Clearly,
$$
F\delta_x(\lambda)=e^{i\lambda x}.
$$
But now I am confused. How should we estimate 
$$
\|F^{-1}\phi_j e^{i\lambda x}\|_{L_\infty}?
$$
 A: Firstly, the norm doesn't depend on $x$, due to the shift property $ \mathscr F_y [u(y-x)](\lambda)=\mathscr F[u](\lambda)e^{i\lambda x},$ i.e. $$\mathscr F^{-1}_\lambda[\phi_j(\lambda) e^{i\lambda x}](y)=\mathscr F^{-1}_\lambda [\phi_j(\lambda)](y-x).$$  So
$$\|\delta_x\|_{B^s_{\infty,\infty}} = \sup_j 2^{sj}\| \mathscr F^{-1}\phi_j(y-x) \|_{L^\infty_y(\mathbb R)} = \sup_{j} 2^{sj}\|F^{-1}\phi_j\|_{L^\infty(\mathbb R)}.  $$
Secondly, recall that the $\phi_j$ for a Littlewood-Paley decomposition are defined (for $j> -1$) via rescalings of a smooth function $\phi$ supported on some annulus,
$$ \phi_j(\lambda) = \phi(2^{-j}\lambda)$$
Then the scaling property of the Fourier transform gives $F^{-1}[\phi_j](y) = 2^{j} F^{-1}[\phi](2^{j} y)$, and so
$$ \|\delta_x\|_{B^s_{\infty,\infty}} = \max \left( \|F^{-1}[\phi_{-1}]\|_{L^\infty}2^{-s} ,\ \|F^{-1}[\phi]\|_{L^\infty}\sup_{j> -1} 2^{(s+1)j}\right)$$
this is finite as long as $s\le- 1$; therefore, $\delta_x \in B^s_{\infty,\infty}$ for $s\le -1$.
In fact, since the scaling property of the Fourier transform  in $d$ dimensions is
$$\mathscr F [u(Ky)](\lambda) = \frac1{K^d}\mathscr F [u](\lambda/K) $$
we can also verify with virtually no additional effort that
$$\|\delta_x\|_{B^s_{p,\infty}}=\max \left( \|F^{-1}[\phi_{-1}]\|_{L^p}2^{-s} ,\ \sup_{j> -1}\|F^{-1}[\phi](2^jy)\|_{L^p_y} 2^{(s+d)j}\right)\\=\max \left( \|F^{-1}[\phi_{-1}]\|_{L^p}2^{-s} ,\ \|F^{-1}[\phi]\|_{L^p}\sup_{j> -1} 2^{(s+d-d/p)j}\right)$$
using the scaling property of $L^p$ norms 
$$ \int_{\mathbb R^d} |u(ky)|^p dy = k^{-d}\int_{\mathbb R^d} |u|^p  $$That is, $ \delta_x \in B^{-d+d/p}_{p,\infty}(\mathbb R^d)$ for every $p\in[1,\infty]$, as can be found in e.g. the introduction of this paper by Prof. Hairer.
