# Exercise $2.3.2$ of Grafakos, Classical Fourier Analysis.

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 $$x^{\alpha}$$.

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 $$\varphi_\varepsilon \ast f(x) - b f(x) = \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(x-y) dy - \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(x) dy \\ = \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) [ f(x-y) - f(x)] dy.$$

• I thought to use that the fourier transform is a homeomorphism in Schwartz's space but I don't know how to go on. $\varphi_{\epsilon}*f\to f in S$ if and only if $(\varphi_{\epsilon}*f)^{\wedge}\to \hat{f}$ in $S$ $(\varphi_{\epsilon})^{\wedge}\hat{f}\to \hat{f}$ in $S$ Therefore I need prove that: $\lim_{\epsilon\to 0}\sup_{x}\left| x^{\alpha}\partial^{\beta} ({\varphi_{\epsilon}}^{\wedge}\hat{f}-\hat{f})(x)\right|=0$ Commented Jan 12, 2020 at 19:51
• I think there is a mistake. You must say $\int \epsilon^{-n}\varphi(y/\epsilon)f(x)dy$ with $f (x)$ instead of $f (y)$ Commented Feb 28, 2020 at 1:57
• Indeed. It's fixed now. Commented Feb 28, 2020 at 12:33

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}
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 $$\lim _{\epsilon \rightarrow 0} \sup _{x \in \mathbb{R}^n}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right|=\sup _{x \in \mathbb{R}^n} \lim _{\epsilon \rightarrow 0}\left|x^\alpha \partial^\beta(f(x-\epsilon y)-f(x))\right|=0 .$$
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 $$\|\partial^\beta \tau^y f - \partial^\beta f\|_\infty \rightarrow 0$$.