Exercise $2.3.2$ of Grafakos, Classical Fourier Analysis.

Question

I am trying to prove the exercise $2.3.2$ of "Grafakos-classical Fourier Analysis".

Exercise $2.3.2.$ Let $\varphi,\ f\in \mathcal{S}(\mathbb{R}^n)$, and for $\epsilon>0$ let $\varphi_{\epsilon}(x)=\epsilon^{-n}\varphi(\epsilon^{-1}x)$. Prove that $\varphi_{\epsilon}\ast f\to bf$ in $S$ (schwartz space), where $b$ is the integral of $\varphi$.

The definition of $f_k\to f$ in $\mathcal{S}$ in Grafakos is:

$\lim_{k\to\infty} \sup_{x\in\mathbb{R}^n}\left|x^{\alpha} \partial^{\beta}(f_k-f)(x)\right|=0$, for all $\alpha,\beta$ multi-indices.

What would be an "efficient" way to start this exercise?

Actualization 1: \begin{align}\left|\varphi_\varepsilon \ast f(x) - b f(x)\right| &= \left|\int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(x-y) dy - \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(y) dy \right|\\ &= \left|\int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) [ f(x-y) - f(y)] dy\right|\\ &\leq \left|\int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) |f(x-y) - f(y)| dy\right| \end{align}

$f\in S$ then $f$ is uniformly continuos therefore, for any $|y|\leq \delta$ some $\delta>0$ implies $|f(x-y)-f(y)|<\epsilon$

therefore

\begin{align} \sup_{|x|\leq \delta} \left| x^{\alpha}\partial^{\beta}[ \varphi_{\epsilon}*f(x)-bf(x)]\right| &=\sup_{|x|\leq \delta}\left| x^{\alpha} [\varphi_{\epsilon}*\partial^{\beta}f(x)-b\partial^{\beta}f(x)]\right|\\ &=\sup_{|x|\leq \delta}\left| x^{\alpha} [\varphi_{\epsilon}*g(x)-bg(x)]\right|\\ &\leq \sup_{|x|\leq \delta}\left| x^{\alpha} \int_{R^n}\epsilon^{-n}\varphi({y/\epsilon})|[g(x-y)-g(y)]|dy\right|\\ &\leq \delta^n\int_{R^n}\epsilon^{-n}\varphi(y/\epsilon)\epsilon dy\\ &=\delta^n\left\|\varphi\right\|_{1} \epsilon\to 0 \end{align} where $g(x)=\partial^{\beta}f(x),\ g\in S$

How proves for $|x|>\delta$?

Actualization 2. I fixed the above.

\begin{align} &\sup_{x\in R^n} \left|x^{\alpha} \partial^{\beta}(\varphi_{\epsilon}*f-bf)(x)\right|\\ &=\sup_{x\in R^n} \left|x^{\alpha} (\varphi_{\epsilon}*\partial_{x}^{\beta}f-b\partial_{x}^{\beta}f)(x)\right|\\ &=\sup_{x\in R^n}\left|x^{\alpha}(\varphi_{\epsilon}*g-bg)(x)\right|,\quad g=\partial_{x}^{\beta}f\in S\\ &\leq \sup_{x\in R^n} \left|x^{\alpha}\int_{R^n}\varphi_{\epsilon}(y)|g(x-y)-g(x)|dy\right|\\ &=\sup_{x\in R^n}\left|x^{\alpha}\left[\int_{|y|>\delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy+\int_{|y|\leq \delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy\right]\right| \end{align} Here I don't know how to continue. From what I've seen, I should use the following facts: $\lim_{\epsilon\to 0}\int_{|x|>\delta}\varphi_{\epsilon}(x)dx=0$ for any $\delta>0$ and $f\in S\rightarrow g:=\partial_{x}^{\beta}f\in S$ then $g$ uniformly continuous. But I don't know how to "kill" the $x^{\alpha}$

Actualization 3. \begin{align} &\sup_{x\in R^n}\left|x^{\alpha}\left[\int_{|y|>\delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy\right]\right|\\ &\leq \sup_{x\in R^n}\left|x^{\alpha}\left[\int_{|y|>\delta} \varphi_{\epsilon}(y)(|g(x-y)|+|g(x)|)dy\right]\right|\\ &\sup_{x\in R^n}\left|\int_{|y|>\delta} x^{\alpha}|g(x-y)|\varphi_{\epsilon}(y)dy+\int_{|y|>\delta} x^{\alpha} |g(x)|\varphi_{\epsilon}dy\right|\\ &\leq \left|\int_{|y|>\delta} C(\alpha)\varphi_{\epsilon}dy+\int_{|y|>\delta} C(\alpha)\varphi_{\epsilon}dy\right| \to 0 \quad (|y|>\delta \text{ and } \epsilon\to 0) \end{align} because $|g|$ and $|\tau_{y}g|$ are Schwartz functions because $g$ is Schwartz.

In $x^{\alpha}\int_{|y|\leq \delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy$ I am lost by the $x^{\alpha}$.

Answer 1

Here's a hint to get you going. The key observation is that $$ b = \int_{\mathbb{R}^n} \varphi(y) dy= \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) dy $$ for all $\varepsilon >0$, and hence $$ \varphi_\varepsilon \ast f(x) - b f(x) = \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(x-y) dy - \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(x) dy \\ = \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) [ f(x-y) - f(x)] dy. $$

Answer 2

Maybe my way can help you.

Denote the semi-norm on Schwartz space by $\rho_{\alpha, \beta}$, then we see $$ \begin{aligned} \lim _{\epsilon \rightarrow 0} \rho_{\alpha, \beta}\left(\varphi_\epsilon * f-b f\right) &=\lim _{\epsilon \rightarrow 0} \sup _{x \in \mathbb{R}^n}\left|x^\alpha \partial^\beta\left(\varphi_\epsilon * f-b f\right)\right| \\ &=\lim _{\epsilon \rightarrow 0} \sup _{x \in \mathbb{R}^n}\left|x^\alpha \partial^\beta \int \varphi(y)(f(x-\epsilon y)-f(x)) d y\right| \\ & \leq \lim _{\epsilon \rightarrow 0} \int|\varphi(y)| \sup _{x \in \mathbb{R}^n}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right| d y \\ &=\int|\varphi(y)| \sup _{x \in \mathbb{R}^n} \lim _{\epsilon \rightarrow 0}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right| d y \\ &=0 . \end{aligned} $$

The reason of change limit and supremum is below.

Since $\sup _{x \in \mathbb{R}^n}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right| \leq \rho_{\alpha, \beta} f(\cdot-\epsilon y)+\rho_{\alpha, \beta} f$, we can change the integral and limit by dominated convergence theorem.

Since every Schwartz function is uniformly continuous, we know that $\left.(x-\epsilon y)^\alpha \partial^\beta f(x-\epsilon y)-x^\alpha \partial^\beta f(x)\right)$ is uniformly convergent to 0 . And then we have $$ \begin{aligned} \left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right| & \leq \mid x^\alpha \partial^\beta\left(f(x-\epsilon y)-(x-\epsilon y)^\alpha \partial^\beta(f(x-\epsilon y) \mid\right.\\ &\left.+\mid(x-\epsilon y)^\alpha \partial^\beta f(x-\epsilon y)-x^\alpha \partial^\beta f(x)\right) \mid \\ & \leq\left|x^\alpha-(x-\epsilon y)^\alpha\right|\left|\partial^\beta f(x-\epsilon y)\right| \\ &\left.+\mid(x-\epsilon y)^\alpha \partial^\beta f(x-\epsilon y)-x^\alpha \partial^\beta f(x)\right) \mid \end{aligned} $$ Since $\partial^\beta f(x-\epsilon y)$ is a Schwartz function then we have $$ \left|\partial^\beta f(x-\epsilon y)\right| \leq \tilde{C}_\alpha(1+|x-\epsilon y|)^{2|\alpha|} \leq C_\alpha(1+|x|)^{2|\alpha|} $$ uniformly in $\epsilon$ for fixed $y$. And hence $$ \left|x^\alpha-(x-\epsilon y)^\alpha\right|\left|\partial^\beta f(x-\epsilon y)\right| \leq C_\alpha \frac{\left|x^\alpha-(x-\epsilon y)^\alpha\right|}{(1+|x|)^{2 \alpha}} . $$ By mean value theorem, we have $$ \frac{\left|x^\alpha-(x-\epsilon y)^\alpha\right|}{(1+|x|)^{2 \alpha}}=\frac{\nabla f(\xi) \cdot(\epsilon y)}{(1+|x|)^{2 \alpha}} \leq C_\alpha \frac{|\nabla f(\xi)| \epsilon|y|}{(1+|x|)^\alpha}, $$ where $f=x_1^{\alpha_1} \cdots x_n^{\alpha_n}$. And then $$ |\nabla f(\xi)|^2=\left(\alpha_1 \xi_1^{\alpha_1-1} \cdots \xi_n^{\alpha_n}\right)^2+\cdots+\left(\alpha_n \xi^{\alpha_1} \cdots \xi^{\alpha_n-1}\right)^2 \leq C_{n, \alpha}|x|^{2|\alpha|-2} . $$ And hence $$ \left|x^\alpha-(x-\epsilon y)^\alpha\right|\left|\partial^\beta f(x-\epsilon y)\right| \leq C_{n, \alpha} \frac{|x|^{2|\alpha|-2}}{(1+|x|)^{2|\alpha|}} \epsilon|y| . $$ This tells us that $x^\alpha-(x-\epsilon y)^\alpha \partial^\beta f(x-\epsilon y)$ is uniformly convergent to 0 . So does $x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))$. And then $$ \lim _{\epsilon \rightarrow 0} \sup _{x \in \mathbb{R}^n}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right|=\sup _{x \in \mathbb{R}^n} \lim _{\epsilon \rightarrow 0}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right|=0 . $$

Answer 3

I was working on the same Grafakos problem and found my way here. I had some of the same issues as you, and perhaps found the fix you needed. It seems your only issue is the case $|x| \geq \delta$ and $|y| < \delta$. My first step is to do $$ x^\alpha \int_{|y| < \delta} \epsilon^{-n}\varphi(y/\epsilon)(\partial^\beta \tau^y f(x) - \partial^\beta f(x))dy = \frac{\epsilon^{-n}}{x^\alpha}\int_{|y|<\delta} x^{2\alpha}\varphi(y/\epsilon)(\partial^\beta \tau^y f(x) - \partial^\beta f(x))dy. $$ Since $\varphi$ is a Schwartz function then $\sup_{x \in \mathbb{R}^n}|x^{2\alpha}f(y/\epsilon)| < C_2 < \infty$ for some fixed $C_2$ and for all $|y| < \delta$. Thus we have $$ \begin{align*} \sup_{|x| \geq \delta} \left| \frac{\epsilon^{-n}}{x^\alpha}\int_{|y|<\delta} x^{2\alpha}\varphi(y/\epsilon)(\partial^\beta \tau^y f(x) - \partial^\beta f(x))dy \right| <{}& \epsilon^{-n}\delta^{1 + |\alpha|}v_n C_2 \|\partial^\beta \tau^y f - \partial^\beta f\|_\infty, \end{align*} $$ where $v_n$ is the volume of the unit ball in $\mathbb{R}^n$. As you have mentioned, by making $\delta$ small enough you can let $\|\partial^\beta \tau^y f - \partial^\beta f\|_\infty \rightarrow 0$.