$\def\R{\mathbb{R}}$ $\DeclareMathOperator{\supp}{supp}$

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

  • $\begingroup$ I would appreciate even partial solutions or just some useful hints. $\endgroup$
    – Oleg
    Oct 15, 2018 at 14:11
  • $\begingroup$ 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? $\endgroup$ Oct 24, 2018 at 9:34

1 Answer 1


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.

  • $\begingroup$ (i missed a minus sign that gave a silly result...this should be fixed now) $\endgroup$ Oct 15, 2018 at 17:06

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