# Switching from an approximation of the identity to another

Sometimes one needs to switch from an approximation of the identity to another with some other property which is helpful to the problem in exam. In Stein’s Harmonic Analysis book there’s this nice little result which allows one to do this switch.

Notice that by ${f_\delta(x)}$ we denote the dilated function ${\delta^{-n}f\left(\frac{x}{\delta}\right)}$.

Proposition 1 Let ${\Phi \in \mathcal{S}(\mathbb{R}^n)}$ be such that ${\int{\Phi}\,dx=1}$, and let ${\Psi}$ be another Schwartz function. Then there exist a sequence ${\{\eta_k\}_{k\in\mathbb{N}} \subset \mathcal{S}(\mathbb{R}^n)}$ such that we have the following nice decomposition:

• ${\Psi=\sum_{k\geq 0}{\eta_k \ast \Phi_{2^{-k}}}}$
• the ${\eta_k}$‘s size decreases rapidly in ${k}$, in particular $\displaystyle \|\eta_k\|_{\alpha,\beta}\in O\left(2^{-k N}\right)$

for every ${N>0}$, where ${\|\cdot\|_{\alpha,\beta}}$ are the seminorms of ${\mathcal{S}(\mathbb{R}^n)}$ defining its topology.

Proof: As usual we’ll localize in the frequency domain. Take then a bump function ${\widehat{\varphi}}$ which is ${C^\infty}$, identically ${1}$ on ${B(0,1)}$ and identically zero on ${B(0,2)^c}$ (it is therefore in ${\mathcal{S}(\mathbb{R}^n)}$). Now from ${\varphi}$ define bump functions localized at (big) frequency ${|\xi|\sim 2^k}$ by $\displaystyle \widehat{\psi_k}(\xi):=\widehat{\varphi}(2^{-k}\xi)-\widehat{\varphi}(2^{-(k-1)}\xi),$

where ${k\in\mathbb{N}}$. They have the further properties that $\displaystyle |\partial^{\alpha}_\xi \widehat{\psi_k}(\xi)|\lesssim_\alpha 2^{-k|\alpha|}$

because ${\partial^{\alpha}_\xi \widehat{\psi_k}(\xi)=2^{-k|\alpha|}\left(\partial^\alpha \widehat{\varphi}\right)(2^{-k}\xi)-2^{-(k-1)|\alpha|}\left(\partial^\alpha \widehat{\varphi}\right)(2^{-(k-1)}\xi)}$ and these two terms have disjoint supports and comparable upperbounds; the other property is trivially that $\displaystyle \sum_{k\geq 0}{\widehat{\psi_k}}=1.$

From this second property then $\displaystyle \widehat{\Psi}=\sum_{k\geq 0}{\widehat{\psi_k}\widehat{\Psi}},$

and the ${\eta_k}$ are then defined by $\displaystyle \widehat{\Psi}(\xi)=\sum_{k\geq 0}{\left(\frac{\widehat{\psi_k}(\xi)\widehat{\Psi}(\xi)}{\widehat{\Phi}(2^{-k}\xi)}\right)\widehat{\Phi}(2^{-k}\xi)},$

i.e. ${\widehat{\eta_k(\xi)}=\frac{\widehat{\psi_k}(\xi)\widehat{\Psi}(\xi)}{\widehat{\Phi}(2^{-k}\xi)}}$. Of course we’re cheating a little: we need some lowerbound on ${\widehat{\Phi}}$ for the ${\eta_k}$‘s to be well defined. But this is implicit in the definition since ${1=\int{\Phi}\,dx=\widehat{\Phi}(0)}$, and therefore we can assume ${|\widehat{\Phi}|\geq \frac{1}{2}}$ in the whole ball ${B(0,2)}$ – if the ball were smaller we should just shift the indices ${k}$, as will be apparent in the following. In our case, since ${\widehat{\psi_k}}$ is localized at frequency ${|\xi|\sim 2^k}$ one has therefore ${|\widehat{\Phi}(2^{-k}\xi)|\gtrsim 1}$ in the support of ${\widehat{\eta_k}}$. It is thus evident that, since the denominator is bounded away from zero and all the functions are in ${\mathcal{S}}$, the ${\eta_k}$‘s are in ${\mathcal{S}}$.

What’s left to prove is that the seminorms decay rapidly in ${k}$. This property actually follows from the frequency localization. The proof is a bit long and boring, but here it is anyway: using Hausdorff-Young inequality $\displaystyle \|\eta_k\|_{\alpha,\beta}=\|x^\alpha \partial^\beta \eta_k(x)\|_{L^\infty}\leq \|\left(x^\alpha \partial^\beta \eta_k(x)\right)^\wedge\|_{L^1},$

and by the properties of the Fourier transform ${ \left(x^\alpha \partial^\beta \eta_k(x)\right)^\wedge=C \partial^{\alpha}_\xi \left(\xi^{\beta} \widehat{\eta_k}\right)}$, so that we can estimate the ${L^1}$ norm of the latter. Using Leibniz’s rule $\displaystyle \|\partial^{\alpha}_\xi \left(\xi^{\beta} \widehat{\eta_k}\right)\|_{L^1} \leq \sum_{\gamma\leq \alpha}\binom{\alpha}{\gamma}\|\partial^{\gamma} \xi^\beta \cdot \partial^{\alpha-\gamma}\widehat{\eta_k}\|_{L^1},$

and since the support of ${\widehat{\eta_k}}$ is ${|\xi|\sim 2^k}$ (localization!) the terms ${\partial^\gamma \xi^\beta}$ are essentially bounded in size. We only care about a uniform bound, so we write $\displaystyle \|\partial^{\alpha}_\xi \left(\xi^{\beta} \widehat{\eta_k}\right)\|_{L^1} \lesssim_\alpha \sum_{\gamma \leq \min(\alpha,\beta)}{2^{k|\beta-\gamma|}\|\partial^{\alpha-\gamma}\widehat{\eta_k}\|_{L^1}},$

and again by boundedness of the support one has $\displaystyle \sum_{\gamma \leq \min(\alpha,\beta)}{2^{k|\beta-\gamma|}\|\partial^{\alpha-\gamma}\widehat{\eta_k}\|_{L^1}} \leq 2^{k|\alpha|}\sum_{\gamma \leq \min(\alpha,\beta)}{\left|\{|\xi|\sim 2^k\}\right|\|\partial^{\alpha-\gamma}\widehat{\eta_k}\|_{L^\infty}}$ $\displaystyle \lesssim_n 2^{k(|\alpha|+n)}\sum_{\gamma \leq \min(\alpha,\beta)}{\|\partial^{\alpha-\gamma}\widehat{\eta_k}\|_{L^\infty}}.$

We have therefore to estimate ${\|\partial^{\alpha-\gamma}\widehat{\eta_k}\|_{L^\infty}}$. Again, by Leibniz’s rule (discarding the constants) and writing ${\varepsilon := \alpha-\gamma}$, $\displaystyle \|\partial^{\varepsilon}\widehat{\eta_k}\|_{L^\infty}\lesssim_{\alpha,\beta,\gamma} \sum_{\delta\leq \varepsilon}{\left\|\partial^{\varepsilon-\delta}\left(\frac{\widehat{\psi_k}(\xi)}{\widehat{\Phi}(2^{-k}\xi)}\right)\partial^\delta \widehat{\Psi}(\xi)\right\|_{L^\infty}}.$ ${\Psi\in\mathcal{S}}$, therefore there’s a constant ${C=C_{\delta,\Psi,M}}$ such that ${|\partial^\delta \widehat{\Psi}(\xi)|\leq C |\xi|^{-M}}$ for every ${\xi\in\mathbb{R}^n}$, and once again the localization of frequency to size ${2^k}$ allows us to write $\displaystyle \|\partial^{\varepsilon}\widehat{\eta_k}\|_{L^\infty}\lesssim 2^{-kM}\sum_{\delta\leq \varepsilon}{\left\|\partial^{\varepsilon-\delta}\left(\frac{\widehat{\psi_k}(\xi)}{\widehat{\Phi}(2^{-k}\xi)}\right)\right\|_{L^\infty}}.$

Notice the constants here depend on ${\delta}$ and thus on ${\varepsilon}$, and thus on ${\alpha}$ in the end (other than ${\Psi}$ and ${M}$). Now, by the definition of ${\psi_k}$ we have $\displaystyle \partial^{\varepsilon-\delta}\left(\frac{\widehat{\psi_k}(\xi)}{\widehat{\Phi}(2^{-k}\xi)}\right) = \partial^{\varepsilon-\delta}\left(\frac{\widehat{\varphi}(2^{-k}\xi)-\widehat{\varphi}(2\cdot 2^{-k}\xi)}{\widehat{\Phi}(2^{-k}\xi)}\right)$ $\displaystyle = 2^{-k|\varepsilon-\delta|}\partial^{\varepsilon-\delta}\left(\frac{\widehat{\varphi}(\cdot)-\widehat{\varphi}(2\cdot )}{\widehat{\Phi}(\cdot)}\right)(2^{-k}\xi),$

therefore we can see how these terms are all uniformly bounded by a constant depending only on ${\alpha, \varphi}$ and ${\Phi}$. Then we’ve shown that $\displaystyle \|\eta_k\|_{\alpha,\beta} \lesssim_{\alpha,\beta,n,\varphi,\Phi,N} 2^{-kN}$

for whatever ${N>0}$ we choose.

In the end we go back to the time domain applying the inverse Fourier transform to ${\widehat{\Psi}(\xi)=\sum_{k\geq 0}{\widehat{\psi_k}(\xi)\widehat{\Phi}(2^{-k}\xi)}}$, which becomes $\displaystyle \Psi(x) = \sum_{k\geq 0}{\psi_k(x) \ast \Phi_{2^{-k}}(x)}.$ $\Box$

Notice, in case the ball where ${|\widehat{\Phi}|\gtrsim 1}$ were smaller, say ${B(0,2^{-k_0})}$ for some ${k_0>0}$, it would suffice to define instead $\displaystyle \widehat{\eta_k}(\xi):= \frac{\widehat{\psi_{k-k_0}}(\xi)\widehat{\Psi}(\xi)}{\widehat{\Phi}(2^{-k}\xi)}\;\;\;\mbox{ for }k\geq k_0,$

so that the sum starts from ${k=k_0}$ instead of ${k=0}$.

Notice we assumed very little about ${\Phi}$ and ${\Psi}$, this last one in particular can be any Schwartz function.