# Exercise 2.3.5 in Grafakos, Classical Fourier Analysis

I got stuck in Exercise $$2.3.5$$ (c) of Grafakos, Classical Fourier Analysis for a few days.

Notation here

$$\mathcal{S}$$ is the Schwartz Class.

$$\mathrm{C}_0^\infty$$ means smooth functions with compact support.

$$\tau^y$$ means translation by $$y$$, that is, $$(\tau^yf)(x):=f(x-y).$$

$$\partial_j$$ means the partial derivative on the $$j$$-th component. For multi-index $$\beta=(\beta_1,\beta_2,\cdots,\beta_n)$$ and $$x=(x_1,\cdots,x_n)\in\mathbb{R}^n$$, $$|\beta|:=\beta_1+\beta_2+\cdots+\beta_n,$$ and $$\partial^\beta f(x):=\frac{\partial^{|\beta|}}{\partial^{\beta_1}_{x_1}\cdots\partial^{\beta_n}_{x_n}}f(x).$$

$$e_j$$ is the elementary vector in $$\mathbb{R}^n$$: $$1$$ on the $$j$$-th component and $$0$$ on the others. And $$e_j$$ can also be viewed as a multi-index.

$$\hat{f}$$ is the Fourier transform of $$f$$.

Problem here

$$2.3.5.$$ Let $$f\in\mathcal{S}(\mathbb{R}^n)$$ and $$\varphi\in\mathrm{C}_0^\infty(\mathbb{R}^n)$$ be identically equal to 1 in a neighbourhood of the origin. Define $$\varphi_k(x)=\varphi(x/k)$$ as in the proof of Proposition $$2.3.23$$.

(a)Prove that $$(\tau^{-he_j}f-f)/h\to\partial_jf$$ in $$\mathcal{S}$$ as $$h\to 0.$$

(b)Prove that $$\varphi_k f\to f$$ in $$\mathcal{S}$$ as $$k\to\infty.$$

(c)Prove that the sequence $$\varphi_k\widehat{\varphi_kf}$$ converges to $$\hat{f}$$ in $$\mathcal{S}$$ as $$k\to\infty.$$

(It doesn't matter if you know nothing about that proposition. Grafakos left some details in that proposition as exercise here. And if you need further explanations for anything above, please comment to let me know.)

The following is what I have done.

To prove (a), for any multi-indices $$\alpha,\beta$$, $$x\in\mathbb{R}^n$$, $$|x^\alpha\partial^\beta(\frac{\tau^{-he_j}f(x)-f(x)}{h}-\partial_jf(x))|=|x^\alpha\cdot\frac{\partial^\beta f(x+he_j)-\partial^\beta f(x)-\partial^{\beta+e_j}f(x)h}{h}|$$
Use the mean value theorem and this semi-norm of $$f$$ in $$\mathcal{S}$$: $$\sup\limits_{x\in\mathbb{R}^n}|x^\alpha\partial^{\beta+2e_j}f(x)|$$ to prove the result.

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To prove (b), suppose $$\varphi$$ equals to $$1$$ on $$B(0,R):=\{x\in\mathbb{R}^n:|x|. Then for multi-indices $$\alpha,\beta$$, we split the semi-norm into two parts after using Leibniz's rule: $$|x^\alpha\partial^\beta(\varphi_kf-f)|=|x^\alpha(\sum\limits_{0<\gamma\leq\beta}\tbinom{\beta}{\gamma} \partial^\gamma\varphi_k ~ \partial^{\beta-\gamma}f)+x^\alpha(\varphi_k-1)\partial^\beta f|.$$ The sum is taken over multi-index $$\gamma$$, where (the partial order on multi-indices) $$\gamma\leq\beta$$ means $$\gamma_j\leq\beta_j$$ for every $$j\in\{1,2,\cdots,n\}$$, and $$\gamma>0$$ means $$\gamma\geq0$$ as multi-index but $$\gamma\not=0.$$

Use the fact that $$\varphi_k-1$$ and $$\partial^\gamma\varphi_k(\gamma>0)$$ vanish on $$B(0,kR)$$ to help prove the result.

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Now prove (c). It is really tempting to prove (c) with (b), because (b) says that the sequence $$\{\varphi_k\}_{k\geq1}$$ approximates the identity map.

So, it seems that $$\widehat{\varphi_kf}$$ is almost $$\hat{f}$$. Then $$\varphi_k\widehat{\varphi_kf}$$ is almost $$\varphi_k\hat{f}$$, and thus almost $$\hat{f}.$$

But actually it doesn't make sense! To be rigorous, consider the sequence with two indices: $$\{\varphi_j(\widehat{\varphi_kf})\}_{k,j\geq1}$$. Fix any one of $$k,j$$ and let the other $$\to\infty$$, for example fix $$k$$ and let $$j\to\infty$$, we get the limit $$\widehat{\varphi_kf}$$, then we let $$k\to\infty$$ to get $$\hat{f}.$$ The same result if we first let $$k\to\infty$$ then let $$j\to\infty$$. With this intepretation, what we care about is the diagonal subsequence where $$k=j$$.

Let's simplify this problem. Consider an infinite diagonal matrix $$(a_{j,k})_{j,k\geq1}$$, where all entries on the diagonal equal $$1$$. Then the limit of every row and column regarded as a sequence is $$0$$ , and the limits of these two sequences of row/column limits are both $$0$$. But the limit of the diagonal sequence is $$1$$. They are not equal...

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Also, I have tried to split $$x^\alpha\partial^\beta(\varphi_k\widehat{\varphi_kf})$$ into two parts as I did in (b), but it points to the same "diagonal sequence" problem.

So I want to know how to prove (c), or how to deal with this "diagonal" problem. Is there anything I've missed or any tool I need? Thanks in advance.

• I'd guess that writing $\|\varphi_k \widehat{\varphi_k f} - \hat{f} \|_{\alpha, \beta} \leq \| \widehat{\varphi_k f} - \hat{f}\|_{\alpha, \beta} + \|\varphi_j \widehat{\varphi_k f} - \widehat{\varphi_k f} \|_{\alpha, \beta} + \|\varphi_j \widehat{\varphi_k f} - \varphi_k \widehat{\varphi_k f} \|_{\alpha, \beta}$ you should be to estimate all three terms on the RHS (though i haven't checked this). By what you wrote, for large $k$ the first one is small and for that fixed $k$ and for large $j$ the second is small. Estimating the third should then just use the form of $\varphi_l$. Commented Nov 29, 2021 at 23:13

Yes, we need to use the conclusion in (b). But we also need a conclusion in Exercise 2.2.2. $$f_j\to f ~~~\text{in}~\mathscr{S}\Longrightarrow \hat{f_j}\to \hat f ~~~\text{in}~\mathscr{S}.$$ And then, For any multi-indices $$\alpha$$ and $$\beta$$, we have \begin{align*} \rho_{\alpha,\beta}(\varphi_k\widehat{\varphi_kf}-\hat{f})&\le \rho_{\alpha,\beta}(\varphi_k(\widehat{\varphi_kf}-\hat{f}))+\rho_{\alpha,\beta}(\varphi_k\hat{f}-\hat{f}) \end{align*} It has been shown in (b) that $$\rho_{\alpha,\beta}(\varphi_k\hat{f}-\hat{f})\to 0$$ as $$k\to\infty$$. So we only need to deal with the first term. We know that \begin{align*} \rho_{\alpha,\beta}(\varphi_k(\widehat{\varphi_kf}-\hat{f}))&=\sup_{x\in\mathbb{R}^n}|x^\alpha\partial^\beta(\varphi_k(\widehat{\varphi_kf}-\hat{f}))|\\ &\le \sum_{\gamma\le \beta}\binom{\beta}{\gamma}\sup_{x\in\mathbb{R}^n}|\partial^\gamma\varphi_k(x^\alpha\partial^{\beta-\gamma}(\widehat{\varphi_kf}-\hat{f}))|\\ &=\sum_{\gamma\le \beta}\binom{\beta}{\gamma}C(\varphi,\beta,\gamma)\rho_{\alpha,\beta-\gamma}(\widehat{\varphi_kf}-\hat{f}) \end{align*} Since we've proved $${\varphi_k f}\to{f}$$ in $$\mathscr{S}$$ in (b), thus $$\widehat{\varphi_kf}\to \hat{f}$$ in $$\mathscr{S}$$. Hence $$\lim_{k\to\infty}\rho_{\alpha,\beta}(\varphi_k\widehat{\varphi_kf}-\hat{f})\to 0,~~~~~~\text{when}~k\to\infty.$$