# 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 $$\|F^{-1}\phi_j e^{i\lambda x}\|_{L_\infty}?$$

• I would appreciate even partial solutions or just some useful hints.
– Oleg
Oct 15, 2018 at 14:11
• You've been around but you have let the bounty expire and you haven't accepted my answer. Can I ask you to tell me how to improve it to your satisfaction? Oct 24, 2018 at 9:34

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.