用户: Solution/ 解答: GTM249 Chapter 2 Part I

2.1 Maximal Functions

2.1.1	(a) Similar to the proof of Theorem 2.1.6, except for the only difference $$D(\mu )\sum _{r=1}^l\mu (B_{j_r})\ge \mu (\bigcup _{i=1}^k{B}_i).$$ (b)$$\underset{r\to 0}{\lim }\frac{1}{\mu (B(x,r))}{\int }_{B(x,r)}f(y)dy=f(x).$$
2.1.2	(a) Select a subset $\mathcal{G}$ of $\mathcal{F}$ with the minimal cardinality such that ${\bigcup }_{J\in \mathcal{G}}J={\bigcup }_{I\in \mathcal{F}}I$. Assume that ${\bigcup }_{J\in \mathcal{G}}J$ is connected, otherwise consider every connected component. Write $\mathcal{G}=\{(a_i,b_i)\mid 1\le i\le n\}$. Let ${\mathcal{G}}_1=\{(a_i,b_i)\mid i\text{ is odd}\}$. We claim that ${\mathcal{G}}_1$ is a family of mutually disjoint intervals. Otherwise there exists $j$ such that $(a_j,b_j)\cap (a_{j+2},b_{j+2})\ne \varnothing$. If $b_{j+1}\le b_{j+2}$, then $(a_{j+1},b_{j+1})\subset (a_j,b_{j+2})=(a_j,b_j)\cup (a_{j+2},b_{j+2})$, so $(a_{j+1},b_{j+1})$ can be removed; if $b_{j+1}>b_{j+2}$, then $(a_{j+2},b_{j+2})\subset (a_{j+1},b_{j+1})$, so $(a_{j+2},b_{j+2})$ can be removed. In both cases the minimality of $\mathcal{G}$ fails. Similarly we can prove that ${\mathcal{G}}_2$ is a family of mutually disjoint intervals. Similar to the proof of Theorem 2.1.6, except for the only difference $$\mu (K)\le \mu (\bigcup _{B_x\in \mathcal{F}}{B}_x)\le \mu (\bigcup _{B_x\in {\mathcal{G}}_1}{B}_x)+\mu (\bigcup _{B_x\in {\mathcal{G}}_2}{B}_x)\le \frac{2}{\alpha }{\int }_{E_{\alpha }}|f|d\mu .$$(b) The first inequality is equivalent to $${\int }_X(|f|-\alpha )({\chi }_A-{\chi }_{|f|>\alpha })d\mu .$$This inequality holds because the integrand is nonpositive whether $|f|>\alpha$ or $|f|\le \alpha$. Set $E_{\alpha }=\{M_{\mu }(f)>\alpha \}$, the following notation follows from the proof of part (a). Let $F_1={\bigcup }_{B_x\in {\mathcal{G}}_1}{B}_x\cap K,F_2={\bigcup }_{B_x\in {\mathcal{G}}_2}{B}_x\cap K$, where $K$ is a compact subset of $E_{\alpha }$. We have$$\begin{array}{rl}\mu (K)+\mu (F_1\cap F_2)& =\mu (F_1)+\mu (F_2)\le \frac{1}{\alpha }({\int }_{F_1}|f|d\mu +{\int }_{F_2}|f|d\mu )\\ & =\frac{1}{\alpha }({\int }_K|f|d\mu +{\int }_{F_1\cap F_2}|f|d\mu ).\end{array}$$Take $A=F_1\cap F_2$ in the first inequality and get$$\frac{1}{\alpha }{\int }_{F_1\cap F_2}|f|d\mu -\mu (F_1\cap F_2)\le \frac{1}{\alpha }{\int }_{\{|f|>\alpha \}\}}|f|d\mu -\mu (\{|f|>\alpha \}).$$Add up the two inequalities, we obtain$$\mu (\{|f|>\alpha \})+\mu (K)\le \frac{1}{\alpha }({\int }_K|f|d\mu +{\int }_{\{|f|>\alpha \}}|f|d\mu ).$$Take the supremum in $K$, and then the required conclusion follows. When $\alpha =1$ and $f(x)=|x{|}^{-\frac{1}{p}}$, $M(f)(x)=a^{-\frac{1}{p}}$, where $a$ is the positive root of$$-\frac{1}{p}{a}^{1-\frac{1}{p}}+|x|\left(\right(1-\frac{1}{p})a^{-\frac{1}{p}}-|x{|}^{-\frac{1}{p}})=0.$$$$\{M(f)(x)>1\}=\{-\frac{1}{p}+|x|((1-\frac{1}{p})-|x{|}^{-\frac{1}{p}})<0\}\triangleq (-A,A).$$So we have$${\int }_{\{M(f)(x)>1\}}(|f|-1)=\frac{p}{p-1}{A}^{1-\frac{1}{p}}-A=-\frac{2}{p-1}=-{\int }_{\{|f|>1\}}(|f|-1).$$(c) Multiplying the last inequality of part (b) by $p{\alpha }^{p-1}$, and integrating with respect to $\alpha$ from $0$ to $\infty$ yields$$\begin{array}{rl}\Vert f{\Vert }_{L^p}^p+\Vert M(f){\Vert }_{L^p}^p& \le \frac{p}{p-1}({\int }_{\mathbb{R}}M(f{)}^{p-1}|f|d\mu +\Vert f{\Vert }_{L^p}^p)\\ & \le \frac{p}{p-1}(\Vert M(f){\Vert }_{L^p}^{p-1}\Vert f{\Vert }_{L^p}+\Vert f{\Vert }_{L^p}^p).\end{array}$$Let $t=\frac{\Vert M(f){\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}$, then$$(p-1)t^p-p{t}^{p-1}-1\le 0.$$So $\Vert f{\Vert }_{L^p\to L^p}\le A_p.$ (d) According to part (b), $\frac{M(f)(x)}{|x{|}^{-\frac{1}{p}}}$ is the positive root of$$(p-1)x^p-p{x}^{p-1}-1,$$which means $M(f)(x)=A_p$. Consider $f_{\epsilon ,N}(x):=|x{|}^{-\frac{1}{p}+\epsilon }{\chi }_N(|x|)$, where ${\chi }_N={\chi }_{[0,N]}$. It’s easy to see that$$\underset{N\to \infty }{\lim }\underset{\epsilon \to 0}{\lim }\frac{\Vert M(f_{\epsilon ,N}){\Vert }_{L^p}}{\Vert f_{\epsilon ,N}{\Vert }_{L^p}}\ge A_p.$$
2.1.3	(a) Obviously, $\mathcal{M}(f)\le M(f).$ For $B(z,r)$ containing $x$,$$\frac{1}{|B(z,r)|}{\int }_{B(z,r)}|f|\le \frac{2^n}{|B(x,2r)|}{\int }_{B(x,2r)}|f|.$$Take the supremum to get $M(f)\le 2^n\mathcal{M}(f).$ For a ball $B$ containing $x$, $C_B$ denotes the minimal cube containing $B$, then$$\frac{1}{|B|}{\int }_B|f|\le \frac{|C_B|}{|B|}\cdot \frac{1}{|C_B|}{\int }_{C_B}|f|=\frac{2^n}{{\nu }_n}\frac{1}{|C_B|}{\int }_{C_B}|f|.$$Take the supremum to get $M(f)\le \frac{2^n}{{\nu }_n}{M}_c(f)$. For a cube $C$ containing $x$, $B_C$ denotes the minimal ball containing $C$, then$$\frac{1}{|C|}{\int }_C|f|\le \frac{|B_C|}{|C|}\cdot \frac{1}{|B_C|}{\int }_{B_C}|f|=\frac{n^{\frac{n}{2}}{\nu }_n}{2^n}\frac{1}{|B_C|}{\int }_{B_C}|f|.$$Take the supremum to get $M_c(f)\le \frac{n^{\frac{n}{2}}{\nu }_n}{2^n}M(f)$. Similar process can be applied to obtain the third inequality. For $f\in L^1({\mathbb{R}}^n)$,$$\alpha d_{M_c(f)}(\alpha )\le \alpha d_{M(f)}\left(\frac{2^n\alpha }{n^{\frac{n}{2}}{\nu }_n}\right)\le \frac{n^{\frac{n}{2}}{\nu }_n}{2^n}\Vert M(f){\Vert }_{L^{1,\infty }}.$$So $M_c(f)\in L^{1,\infty }({\mathbb{R}}^n)$. For $f\in L^p({\mathbb{R}}^n)$,$$\Vert M_c(f){\Vert }_{L^p}\le \frac{n^{\frac{n}{2}}{\nu }_n}{2^n}\Vert M(f){\Vert }_{L^p}.$$So $M_c(f)\in L^p({\mathbb{R}}^n)$. Similarly, ${\mathcal{M}}_c$ has the same property.
2.1.4	(a) Write $f=f{\chi }_{\{|f|>\alpha /2\}}+f{\chi }_{\{|f|\le \alpha /2\}}:=f_1+f_2$. Notice that $M(f)\le M(f_1)+M(f_2)\le M(f_1)+\alpha /2$, so $\{M(f)>\alpha \}\subset \{M(f_1)>\alpha /2\}$. From Theorem 2.1.6,$$\mu (\{M(f)>\alpha \})\le \mu (\{M(f_1)>\alpha /2\})\le \frac{3^n}{\alpha /2}\int |f_1|dy=\frac{3^n}{\alpha /2}{\int }_{\{|f|>\alpha /2\}}|f|dy.$$So $$\begin{array}{r}{\alpha }^p\mu (\{M(f)>\alpha \})=2^p\cdot 3^n(\alpha /2{)}^{p-1}{\int }_{\{|f|>\alpha /2\}}|f|dy\le 2^p\cdot 3^n\Vert f{\Vert }_{L^p}^p,\end{array}$$and hence$$\Vert M(f){\Vert }_{L^{p,\infty }}\le 2\cdot 3^{n/p}\Vert f{\Vert }_{L^p}.$$(b) Write $f=f{\chi }_{\{|f|>1/2\}}+f{\chi }_{\{|f|\le 1/2\}}:=f_1+f_2$. Taking $p=1$ in part (a) yields $\alpha d_{M(f_1)}(\alpha )\le 2\cdot 3^n\Vert f_1{\Vert }_{L^1}$, so$$\begin{array}{rl}\Vert M(f_1){\Vert }_{L^1(B)}& ={\int }_1^{\infty }{d}_{M(f_1)}(\alpha )d\alpha +{\int }_0^1{d}_{M(f_1)}(\alpha )d\alpha \\ & \le {\int }_1^{\infty }\frac{3^n}{\alpha /2}{\int }_{\{|f|>1/2\}}|f|dyd\alpha +{\int }_0^1|B|d\alpha \\ & =2\cdot 3^n{\int }_B\left({\int }_1^{\max (1,2|f|)}\frac{d\alpha }{\alpha }\right)|f|dy+|B|\\ & =2\cdot 3^n\Vert f{\log }^{+}(2|f|){\Vert }_{L^1(B)}+|B|.\end{array}$$Then $M(f)$ is also integrable over $B$, as $M(f)\le M(f_1)+M(f_2)\le M(f_1)+1/2$. (c) From Theorem 2.1.20, there exists a collection of disjoint open cubes $Q_j$ such that $\bigcup Q_j\subset \{M_c(f)>\alpha \}$ and ${\bigcup }_j{Q}_j\supset \{f>\alpha \}$, so$${\int }_{\{|f|>\alpha \}}|f|\le \sum {\int }_{Q_j}|f|\le 2^n\alpha \sum |Q_j|.$$From Exercise 2.1.3,$$|\{M(f)>c_n\alpha \}|\ge |\{M_c(f)>\alpha \}|\ge \sum |Q_j|\ge \frac{1}{2^n\alpha }{\int }_{\{|f|>\alpha \}}|f|.$$(d) Let $x^{\prime }={\rho }^2|x{|}^{-2}x$ for some $\rho <|x|<2\rho$. For $y\in B(0,\rho )$, we have$$|x-y{|}^2-|x^{\prime }-y{|}^2\ge (|x|-|x^{\prime }|)(|x|+|x^{\prime }|-2|y|)>0.$$Consequently, for $R>|x|-\rho$, $B(x,R)\cap B(0,\rho )\subset B(x^{\prime },R)$, and thus$${\int }_{B(x,R)}|f|={\int }_{B(x,R)\cap B(0,\rho )}|f|\le {\int }_{B(x^{\prime },R)}|f|,$$which implies that $$\mathcal{M}(f)(x)\le \mathcal{M}(f)({\rho }^2|x{|}^{-2}x).$$Using polar coordiantes, the determinant of the map $x\mapsto {\rho }^2|x{|}^{-2}x$ is $\pm (\rho /|x|{)}^{2n}$. Thus we have$$\begin{array}{rl}{\int }_{B(0,2\rho )}\mathcal{M}(f)(x)dx& \le {\int }_{B(0,\rho )}\mathcal{M}(f)(x)dx+{\int }_{B(0,2\rho )\backslash B(0,\rho )}\mathcal{M}(f)({\rho }^2|x{|}^{-2}x)dx\\ & ={\int }_{B(0,\rho )}\mathcal{M}(f)(x)dx+{\int }_{B(0,\rho )\backslash B(0,\rho /2)}\mathcal{M}(f)(y)(\frac{|y|}{\rho }{)}^{2n}dy\\ & \le (4^n+1){\int }_{B(0,\rho )}\mathcal{M}(f)(x)dx.\end{array}$$From Exercise 2.1.3, we conclude that $${\int }_{B(0,2\rho )}M(f)(x)dx\le 2^n(4^n+1){\int }_{B(0,\rho )}M(f)(x)dx.$$(e) For $x\notin 2B$,$$M(f)(x)=\underset{x\in \tilde{B},\mathrm{diam}\tilde{B}>\rho }{\sup }\frac{1}{|\tilde{B}|}{\int }_{\tilde{B}}|f|dx<\frac{2^n}{|B|}\Vert f{\Vert }_{L^1}={\lambda }_0.$$Therefore we have$$\begin{array}{rl}{\int }_{2B}M(f)(x)dx& \ge {\int }_{c_n^{-1}{\lambda }_0}^{\infty }|\{x\in 2B\mid M(f)(x)>c_n\alpha \}|d\alpha \\ & ={\int }_{c_n^{-1}{\lambda }_0}^{\infty }|\{M(f)>c_n\alpha \}|d\alpha \\ & \ge {\int }_{c_n^{-1}{\lambda }_0}^{\infty }\frac{2^{-n}}{\alpha }{\int }_{\{|f|>\alpha \}}|f(y)|dyd\alpha \\ & =2^{-n}{\int }_{\{|f|>\alpha \}}\left({\int }_{c_n^{-1}{\lambda }_0}^{\max (|f|,c_n^{-1}{\lambda }_0)}\frac{d\alpha }{\alpha }\right)|f(y)|dy\\ & =2^{-n}\Vert f{\log }^{+}({\lambda }_0^{-1}{c}_n|f|){\Vert }_{L^1},\end{array}$$where the second inequality is from part (c). From part (d),$${\int }_{2B}M(f)(x)dx\le 2^n(4^n+1){\int }_{B}M(f)(x)dx,$$and then the required conclusion follows.
2.1.5	We have$$\begin{array}{rl}& {\int }_A|S(f){|}^{q}dx\\ =& {\int }_0^{\infty }q{\alpha }^{q-1}|A\cap \{S(f)>\alpha \}|d\alpha \\ \le & {\int }_0^{\frac{B}{|A|}\Vert f{\Vert }_{L^1}}q{\alpha }^{q-1}|A|d\alpha +{\int }_{\frac{B}{|A|}\Vert f{\Vert }_{L^1}}^{\infty }q{\alpha }^{q-1}\frac{B}{\alpha }\Vert f{\Vert }_{L^1}d\alpha \\ \le & (\frac{B}{|A|}\Vert f{\Vert }_{L^1}{)}^q|A|+\frac{q}{1-q}(\frac{B}{|A|}\Vert f{\Vert }_{L^1}{)}^{q-1}B\Vert f{\Vert }_{L^1}\\ =& \frac{1}{1-q}{B}^q\Vert f{\Vert }_{L^1}^q|A{|}^{1-q}.\end{array}$$The second result follows from taking $S=M$, then $B=3^n$.
2.1.6	(a) Define $$M_1^{(i)}:=\frac{1}{b_i-a_i}\underset{(a_i,b_i)\ni x_i}{\sup }{\int }_{a_i}^{b_i}|f(y_1,\cdots ,y_n)|d{y}_i.$$It is obvious that $M_s(f)(x)=M_1^{(1)}\circ \cdots \circ M_1^{(n)}(f)(x)$. Thus, $$\begin{array}{rl}{\int }_{{\mathbb{R}}^n}{M}_s(f)(x{)}^{p}dx& \le {\int }_{{\mathbb{R}}^n}{M}_1^{(1)}\circ \cdots \circ M_1^{(n)}(f)(x{)}^{p}d{x}_1\cdots d{x}_n\\ & \le A_p^p{\int }_{{\mathbb{R}}^{n-1}}{M}_1^{(2)}\circ \cdots \circ M_1^{(n)}(f)(x{)}^{p}d{x}_2\cdots d{x}_n\\ & \le \cdots \\ & \le A_p^{pn}\Vert f{\Vert }_{L^p}^p.\end{array}$$(b) Consider $g_{\epsilon ,N}(x):=\prod _{j=1}^n{f}_{\epsilon ,N}(x_j)$ where $f_{\epsilon ,N}(x_j):=|x_j{|}^{-\frac{1}{p}+\epsilon }{\chi }_N(|x_j|)$. From Exercise 2.1.2 (c), $\Vert M_s\Vert \ge A_p^n$, so $\Vert M_s\Vert =A_p^n$. (c) Consider $f_m=m^n{\chi }_{[0,\frac{1}{m}{]}^n}$, then $\Vert f{\Vert }_{L^1}=1$. For $x\in E_m:=\{\prod _{i=1}^n{x}_i\le 1,x_i\ge \frac{1}{m},i=1,\cdots ,n\}$, write $R_x=\prod _{i=1}^n[0,x_i]$, we have$$M_s(f_m)(x)\ge \frac{1}{|R_x|}{\int }_{R_x}{f}_m(x)dx=\frac{1}{|R_x|}\Vert f_m{\Vert }_{L^1}\ge 1.$$Thus$$\frac{\Vert M_s(f_m){\Vert }_{L^{1,\infty }}}{\Vert f_m{\Vert }_{L^1}}\ge m(\{M_s(f_m)\ge 1\})\ge m(E_m).$$Yet let $m\to \infty$, by the Lebesgue monotone convergence theorem and Tonelli’s theorem, $$\underset{m\to \infty }{\lim }m(E_m)={\int }_{\prod _{i=1}^n{x}_i\le 1,x_i>0,i=1,\cdots ,n}\mathrm{d}{x}_1\cdots \mathrm{d}{x}_n={\int }_{{\mathbb{R}}^{n-1}}\frac{1}{x_2\cdots x_n}\mathrm{d}{x}_2\cdots \mathrm{d}{x}_n=({\int }_{\mathbb{R}}\frac{1}{x}\mathrm{d}x{)}^{n-1}=\infty .$$So $M_s$ is not of weak type $(1,1)$.
2.1.7	Write $\psi (x)=(1+\left|x\right|{)}^{-1-\epsilon },x\in \mathbb{R}$, and $T(f):=\underset{t_1,\dots ,t_n>0}{\sup }\left|f\ast {\phi }_{t_1,\dots ,t_n}\right|$. Notice that$$\begin{array}{rl}|f\ast {\phi }_{t_1,\dots ,t_n}|& \le A\left|f\ast ((1+\left|x_1\right|{)}^{-1-\epsilon }\cdots (1+\left|x_n\right|{)}^{-1-\epsilon })\right|\\ & \le A\Vert \psi {\Vert }_{L^1}{M}^{(1)}(f\ast ((1+\left|x_2\right|{)}^{-1-\epsilon }\cdots (1+\left|x_n\right|{)}^{-1-\epsilon }))\\ & \le \cdots \\ & \le A\Vert \psi {\Vert }_{L^1}^n{M}^{(1)}\circ \cdots \circ M^{(n)}(f)=A\Vert \psi {\Vert }_{L^1}^n{M}_s(f).\end{array}$$
2.1.8	Let $f_0$ be the characteristic function of the unit ball $B$ in ${\mathbb{R}}^n$. Note that$$M(f_0)(x)\ge \frac{1}{|B_x|}{\int }_{B_x}{f}_0(y)dy,$$where $B_x=B(\frac{1}{2}(|x|-|x{|}^{-1})\frac{x}{|x|},\frac{1}{2}(|x|+|x{|}^{-1}))$ for $|x|>1$. For $z\in B(0,1)$,$$\begin{array}{rl}z\in B_x& \iff |z-\frac{1}{2}(|x|-|x{|}^{-1})\frac{x}{|x|}{|}^2<\left|\frac{1}{2}\right(|x|+|x{|}^{-1}){|}^2\\ & \iff |z{|}^2-(|x|-|x{|}^{-1})\frac{z\cdot x}{|x|}<\left|\frac{1}{2}\right(|x|+|x{|}^{-1}){|}^2-\left|\frac{1}{2}\right(|x|-|x{|}^{-1}){|}^2\\ & \iff (-|x|+|x{|}^{-1})\frac{z\cdot x}{|x|}<1-|z{|}^2\\ & \Leftarrow z\cdot x>0.\end{array}$$This means half of the unit ball $\{z\in B\mid z\cdot x>0\}$ is contained in $B_x$, implying that$$M(f_0)(x)\ge \frac{|B|/2}{|B_x|}=(\frac{1}{2}(|x|+|x{|}^{-1}){)}^{-n}/2.$$So $$\begin{array}{rl}\frac{\Vert M(f_0){\Vert }_{L^p}^p}{\Vert f_0{\Vert }_{L^p}^p}& \ge \frac{1}{|B|}{\int }_{|x|>1}\left[\right(\frac{1}{2}(|x|+|x{|}^{-1}){)}^{-n}/2{]}^{p}dx\\ & =\frac{n}{2^p}{\int }_1^{\infty }{r}^{n-1}(\frac{1}{2}(r+r^{-1}){)}^{-np}dr\end{array}$$Let $f(r)=r(\frac{1}{2}(r+r^{-1}){)}^{-p}$. Then $f(1)=1$ and $f^{\prime }(1)=1$, so $f(r)>1$ on some interval $(1,r_0)$. Therefore,$$\begin{array}{r}\frac{\Vert M(f_0){\Vert }_{L^p}^p}{\Vert f_0{\Vert }_{L^p}^p}\ge \frac{n}{2^p}{\int }_1^{\infty }\frac{f(r{)}^n}{r}dr>\frac{n}{2^p}{\int }_1^{r_0}\frac{dr}{r}\to \infty \text{as }n\to \infty .\end{array}$$
2.1.9	Consider the Besicovitch Set $E_{ϵ}$ (see Elias M. Stein, Harmonic Analysis, P435), which satisfies (1) $E_{ϵ}$ is a union of $2^N$ rectangles $R_1,\cdots ,R_{2^N}$, each having side lengths $1$ and $2^{-N}$, and $|E_{ϵ}|<ϵ$, and (2) the ${\tilde{R}}_j$’s are mutually disjoint, $j=1,\cdots ,2^N$. Here the ${\tilde{R}}_j$ is defined to be the rectangle obtained by translating $R_j$ by two units along the longer side of $R_j$. So $\Vert {\chi }_{E_{ϵ}}{\Vert }_{L^p}^p<ϵ$, and for $x\in {\tilde{R}}_j$, $M_0({\chi }_{E_{ϵ}})(x)\ge \frac{1}{3}$, which implies that $$\Vert M_0({\chi }_{E_{ϵ}}){\Vert }_{L^p}^p\ge \frac{1}{3}.$$Let $ϵ\to 0$, and then we can see the unboundness. Notice that $M_0$ is weak type $(\infty ,\infty )$, if $M_0$ is weak type $(1,1)$, from the Marcinkiewicz interpolation theorem, $M_0$ is strong type $(2,2)$. Contradiction! (b) Lemma: Let $\{B_1,\cdots ,B_k\}$ be a finite collection of open rectangles in ${\mathbb{R}}^2$, whose eccentricity is $b$. Then there exists a finite subcollection of pairwise disjoint rectangles $\{B_{j_1},\cdots ,B_{j_l}\}$ such that $$16b^2\sum _{r=1}^l|B_{j_r}|\ge |\bigcup _{i=1}^k{B}_i|.$$Proof of lemma: First notice that if two rectangles with eccentricity $b$ intersect, we claim that the smaller one $B$ is contained in the bigger one $A$ scaled $4b$ times. We place the center of $A$ at $0$, the longer edge being parallel to the $x$-axis. Let the length of the edge of $A,B$ be $1,b;c,bc$, where $c<1$. So the abscissa of points in $B$ is no more than $\frac{1}{2}+\mathrm{diam}B\le 4b\cdot \frac{1}{2}$, and the ordinate of points in $B$ is no more than $\frac{b}{2}+\mathrm{diam}B\le 4b\cdot \frac{b}{2}$. This implies $4bA\supset B$. Then follows the proof of Lemma 2.1.5 to prove the lemma. Similar to the proof of Theorem 2.1.6, we see that $$\Vert M_{00}(f){\Vert }_{L^{1,\infty }}\le 16b^2\Vert f{\Vert }_{L^1}.$$(c) Notice that for rectangles $A,B$ satisfying the assumption above, if $A$ and $B$ intersect, $|A|\ge |B|$, then $3A\subset B$. The coefficient $3$ does not depend on the $\{a_2,\cdots ,a_n\}$. Then we can deduce the similar conclusion as in part (b).
2.1.10	(a) Define $O_f(y)=\underset{\epsilon \to 0}{\mathrm{limsup}}|T_{\epsilon }(f)(y)|$, then for $y\in D$,$$O_f(y)\le K((O_{f-g}(y)+O_g(y))=K{O}_{f-g}(y).$$ We claim $\mu \{O_f(y)>\delta \}=0,\forall \delta >0.$ Note that $\forall \eta >0,\exists g\in D,\Vert f-g{\Vert }_{L^p}<\eta .$ So we have $$\begin{array}{rl}\mu \{O_f(y)>\delta \}& \le \mu \{O_{f-g}(y)>\delta /K\}\le \mu \{T_{\ast }(f-g)>\delta /K\}\\ & \le (K/\delta \Vert T_{\ast }(f-g){\Vert }_{L^{q,\infty }}{)}^q\le (K/\delta \Vert T_{\ast }{\Vert }_{L^p\to L^{q,\infty }}\Vert f-g{\Vert }_{L^p}{)}^q.\end{array}$$ Let $\eta \to 0$, we have proved our assertion, which implies that $\underset{\epsilon \to 0}{\mathrm{limsup}}|T_{\epsilon }(f)(y)|=0$. (b) Define$$T_{\epsilon }(f)(x):=\underset{x\in B}{\sup }(\frac{1}{|B|}{\int }_B|f(y)-f(x){|}^{p}dy{)}^{\frac{1}{p}}.$$By Minkowski’s inequality, we have $$({\int }_B|f(y)-f(x)+g(y)-g(x){|}^{p}dy{)}^{\frac{1}{p}}\le ({\int }_B|f(y)-f(x){|}^{p}dy{)}^{\frac{1}{p}}+({\int }_B|g(y)-g(x){|}^{p}dy{)}^{\frac{1}{p}}.$$So we have $T_{\epsilon }(f+g)\le T_{\epsilon }(f)+T_{\epsilon }(g).$ When $p\ge 1$, by Minkowski’s inequality, $$({\int }_B|f(y)-f(x){|}^{p}dy{)}^{\frac{1}{p}}\le ({\int }_B|f(y){|}^{p}dy{)}^{\frac{1}{p}}+|f(x)|\le |f|+M(|f{|}^p{)}^{\frac{1}{p}};$$when $p<1$, by part (c) of Exercise 1.1.5, $$({\int }_B|f(y)-f(x){|}^{p}dy{)}^{\frac{1}{p}}\le ({\int }_B|f(y){|}^p+|f(x){|}^{p}dy{)}^{\frac{1}{p}}\le 2^{\frac{1}{p}-1}(({\int }_B|f(y){|}^{p}dy{)}^{\frac{1}{p}}+|f(x)|)\le 2^{\frac{1}{p}-1}(|f|+M(|f{|}^p){)}^{\frac{1}{p}}.$$Note that $M:L^1\to L^{1,\infty }$ is bounded, we have $${\delta }^p\mu \{M(|f{|}^p{)}^{\frac{1}{p}}\}={\delta }^p\mu \{M(|f{|}^p)>{\delta }^p\}\le \Vert f^p{\Vert }_{L^1}\cdot \Vert M\Vert =\Vert f{\Vert }_{L^p}^p\cdot \Vert M\Vert .$$So $f\mapsto M(|f{|}^p{)}^{\frac{1}{p}}$ is of weak type $(p,p)$, and the norm is less equal to $\Vert M{\Vert }^{\frac{1}{p}}$, which implies $T_{\ast }$ is of weak type $(p,p)$. For $f$ continuous, $T_{\ast }(f)=0$, and $\mathcal{C}({\mathbf{R}}^n)$ is dense in $L^p({\mathbf{R}}^n)$. Then we finish the proof by part (a). (c) Set $Q=[0,1{]}^n$, then $f{\chi }_{Q+m}\in L^1({\mathbf{R}}^n),m\in {\mathbf{Z}}^n$. Apply part (b) to $f{\chi }_{Q+m}$, we have $f(x)=0$, a.e. on every $m+Q$, therefore a.e. on ${\mathbf{R}}^n$.
2.1.11	(a) It’s easy to see $E_{\alpha }:=\{M_R(f)>\alpha \}$ is open. Write $E_{\alpha }=⨄(a_i,b_i)$. For $x\in (a_j,b_j)$, let $$N_x=\{s\mid {\int }_x^s|f|dt>\alpha (s-x)\}\cap (x,b_j].$$The set $N_x$ is nonempty, and $\sup N_x=b_n$. Otherwise, there exists $r>0$ s.t. ${\int }_x^{x+r}|f(t)|dt>\alpha r$. Thus $x+r>b_n$, namely $r>b_n-x$. But as $b_n\notin E_{\alpha }$, ${\int }_{b_n}^{b_n+r^{\prime }}|f(t)|dt\le \alpha r^{\prime }$ for all $r^{\prime }>0$. Take $r^{\prime }=x+r-b_n$, then ${\int }_{b_n}^{x+r}|f(t)|dt\le \alpha (x+r-b_n)$. This leads to ${\int }_x^{b_n}|f(t)|dt>\alpha (b_n-x)$, whence $b_n\in N_x$, contradiction! Let $s_0=\sup N_x$, then $\forall ϵ>0,\exists s_{ϵ}$ such that ${\int }_x^{s_{ϵ}}|f|>\alpha (s-x)$. If $s_0<b_j$, from the definition of $E_{\alpha }$, there exists $s_0^{\prime }>s_0$, such that ${\int }_{s_0}^{s_0^{\prime }}|f|>\alpha (s_0^{\prime }-s_0)$. Let $ϵ\to 0,{\int }_x^{s_0^{\prime }}|f|>\alpha (s_0^{\prime }-x)$, contradictory to the supremum. So $s_0=b_j$. Let $x\to a_j,{\int }_{a_j}^{b_j}|f|\ge \alpha (b_j-a_j)$. If ${\int }_{a_j}^{b_j}|f|>\alpha (b_j-a_j)$, then $a_j\in E_{\alpha }$, a contradiction. Thus,$${\int }_{a_j}^{b_j}|f|=\alpha (b_j-a_j),$$which implies that $$|M_R(f)>\alpha |=\frac{1}{\alpha }{\int }_{\{M(f)>\alpha \}}|f|dt.$$Similar for $M_L(f)$. (b)$$\begin{array}{rl}\Vert M_R(f){\Vert }_{L^p}^p& ={\int }_0^{\infty }p{\alpha }^{p-1}\frac{1}{\alpha }{\int }_{\{M(f)>\alpha \}}|f|dtd\alpha \\ & =\frac{p}{p-1}{\int }_{{\mathbb{R}}^n}M(f{)}^{p-1}|f|dt\\ & \le \frac{p}{p-1}\Vert M_R(f){\Vert }_{L^p}^{p-1}\Vert f{\Vert }_{L^p}.\end{array}$$Here we use Hölder’s inequality for the last inequality. Then we get$$\Vert M_R(f){\Vert }_{L^p}\le \frac{p}{p-1}\Vert f{\Vert }_{L^p}.$$(c) Consider $f_{\epsilon ,N}:=x^{-\frac{1}{p}+\epsilon }{\chi }_{(0,N)}.$ We have $$\Vert f_{\epsilon ,N}{\Vert }_{L^p}^p={\int }_0^N{x}^{-1+\epsilon p}dx=\frac{1}{\epsilon p}{N}^{\epsilon p},M(f)(x)=\frac{1}{x}{\int }_0^x{t}^{-\frac{1}{p}+\epsilon }dt=\frac{x^{-\frac{1}{p}+\epsilon }}{1-\frac{1}{p}+\epsilon }.$$So$$\Vert M(f_{\epsilon ,N}){\Vert }_{L^p}^p\ge {\int }_0^N(\frac{x^{-\frac{1}{p}+\epsilon }}{1-\frac{1}{p}+\epsilon }{)}^{p}dx=(\frac{1}{1-\frac{1}{p}+\epsilon }{)}^p\cdot \frac{1}{\epsilon p}{N}^{\epsilon p}.$$Let $\epsilon \to 0$, then $\Vert M\Vert \ge \frac{p}{p-1}.$ (d)$$\begin{array}{rl}M(f)(x)& \le \underset{x\in (a,b)}{\sup }{\int }_a^b|f|\frac{1}{b-a}\\ & \le \underset{x\in (a,b)}{\sup }\frac{{\int }_a^x|f|}{x-a}\frac{x-a}{b-a}+\frac{{\int }_x^b|f|}{b-x}\frac{b-x}{b-a}\\ & \le \max (M_R,M_L)(f)(x).\end{array}$$Let $a\to x^{-}$, then $\frac{{\int }_x^b|f|}{b-x}\le M(f)(x)$, so $M_R(f)\le M(f)$. Similarly we have $M_L(f)\le M(f)$. We deduce from the three inequalities that $\max (M_R,M_L)=M(f)$. (e)$$\begin{array}{rl}{\int }_{\mathbb{R}}M(f{)}^p+N(f{)}^{p}dx& ={\int }_0^{\infty }p{\alpha }^{p-1}(|\{\max (M_L,M_R)(f)>\alpha \}|+|\{\min (M_L,M_R)(f)>\alpha \}|)d\alpha \\ & ={\int }_0^{\infty }p{\alpha }^{p-1}\frac{1}{\alpha }({\int }_{\{M_L(f)>\alpha \}}|f|dx+{\int }_{\{M_R(f)>\alpha \}}|f|dx)d\alpha \\ & =p{\int }_{\mathbb{R}}|f|({\int }_0^{M_L(f)}{\alpha }^{p-2}d\alpha +{\int }_0^{M_R(f)}{\alpha }^{p-2}d\alpha )dx\\ & =\frac{p}{p-1}{\int }_{\mathbb{R}}|f|(M_L(f{)}^{p-1}+M_R(f{)}^{p-1})dx\\ & =\frac{p}{p-1}{\int }_{\mathbb{R}}|f|(M(f{)}^{p-1}+N(f{)}^{p-1})dx.\end{array}$$(f) Use part (e), we further get$${\int }_{\mathbb{R}}M(f{)}^p+N(f{)}^{p}dx=\frac{p}{p-1}{\int }_{\mathbb{R}}|f|(M(f{)}^{p-1}+N(f{)}^{p-1})dx\le \frac{p}{p-1}({\int }_{\mathbb{R}}|f|M(f{)}^{p-1}+{\int }_{\mathbb{R}}|f{|}^p+{\int }_{\mathbb{R}}N(f{)}^p).$$Here we used Hölder’s inequality. Cancel the $\int N(f{)}^p$, then we obtain the required inequality.
2.1.12	(a) Firstly notice that $\mathcal{F}$, defined to be the set of all dyadic cubes, is countable. So any cover of dyadic cubes has a countable subcover, which is pairwise disjoint. Let $E_{\alpha }=\{M_d(f)>\alpha \}$. $\forall x\in E_{\alpha }$, $\exists Q_x\in \mathcal{F}$ containing $x$, such that ${\int }_{Q_x}|f|>\alpha |Q_x|,Q_x\subset E_{\alpha }$. So $E_{\alpha }=\bigcup _{x\in E_{\alpha }}{Q}_x=\bigcup _{x\in I}{Q}_x$, where $I$ is countable, $\{Q_x{\}}_{x\in I}$ are mutually disjoint. $$|E_{\alpha }|\le |\bigcup _{x\in I}{Q}_x|=\sum _{x\in I}|Q_x|<\frac{1}{\alpha }\sum {\int }_{Q_x}|f|=\frac{1}{\alpha }{\int }_{E_{\alpha }}|f|.$$(b) Apply the result of part (a), we have$$\begin{array}{rl}\Vert M_d(f){\Vert }_{L^p}^p& =p{\int }_0^{\infty }{\alpha }^{p-1}|\{M_d(f)>\alpha \}|d\alpha \\ & \le p{\int }_0^{\infty }{\alpha }^{p-2}{\int }_{\{M_d(f)>\alpha \}}|f|dtd\alpha \\ & =p{\int }_{{\mathbb{R}}^n}{\int }_0^{M_d(f)}{\alpha }^{p-2}d\alpha |f|dt\\ & =\frac{p}{p-1}{\int }_{{\mathbb{R}}^n}|M_d(f){|}^{p-1}|f|dt\\ & \le \frac{p}{p-1}\Vert M_d(f){\Vert }_{L^p}^{p-1}\Vert f{\Vert }_{L^p}.\end{array}$$
2.1.13	Use the result of Exercise 1.1.12 (a), we deduce$$\lambda |\{M(f)>\lambda \}{|}^{\frac{1}{p}}\le 3^n|M(f)>\lambda {|}^{-1+\frac{1}{p}}{\int }_{M(f)>\lambda }|f(y)|dy\le 3^n\left(\frac{p}{p-1}\right)\Vert f{\Vert }_{L^{p,\infty }}.$$Taking the supremum in $\lambda$ yields$$\Vert M(f){\Vert }_{L^{p,\infty }}\le 3^n\left(\frac{p}{p-1}\right)\Vert f{\Vert }_{L^{p,\infty }}.$$
2.1.14	It suffices to prove the inequality for $x=0$, as we may replace $f(t)$ by $f(t+x)$. Define $F,G$ as in Theorem 2.1.10, and let $$A_{{\epsilon }_0}=\underset{\epsilon >{\epsilon }_0}{\sup }\frac{1}{{\epsilon }^n}{\int }_{|y|\le \epsilon }|f(y)|dy=\underset{\epsilon >{\epsilon }_0}{\sup }\frac{1}{{\epsilon }^n}G(\epsilon ).$$Then for any $\epsilon >{\epsilon }_0$,$$\begin{array}{rl}(f\ast K_{\epsilon })(0)& ={\int }_0^{\infty }G(r)d(-K_{\epsilon }(r{e}_1))\\ & \le {\int }_0^{{\epsilon }_0}{\epsilon }_0^n{A}_{{\epsilon }_0}d(-K_{\epsilon }(r{e}_1))+{\int }_{{\epsilon }_0}^{\infty }{r}^n{A}_{{\epsilon }_0}d(-K_{\epsilon }(r{e}_1))\\ & ={\epsilon }_0^n{A}_{{\epsilon }_0}(K_{\epsilon }(0)-K_{\epsilon }({\epsilon }_0{e}_1))+{\epsilon }_0^n{A}_{{\epsilon }_0}{K}_{\epsilon }({\epsilon }_0{e}_1)+{\int }_{{\epsilon }_0}^{\infty }n{r}^{n-1}{A}_{{\epsilon }_0}{K}_{\epsilon }(r{e}_1)dr\\ & =({\epsilon }_0^n{K}_{\epsilon }(0)+\Vert K_{\epsilon }(r{e}_1){\Vert }_{L^1(B(0,{\epsilon }_0{)}^c)}/{\nu }_n)A_{{\epsilon }_0}\\ & \le ({\epsilon }_0^n/{\epsilon }^n+\Vert K_{\epsilon }{\Vert }_{L^1}/{\nu }_n)A_{{\epsilon }_0}\\ & \le (1+\Vert K{\Vert }_{L^1}/{\nu }_n)A_{{\epsilon }_0}:=C_{n,\delta }{A}_{{\epsilon }_0}.\end{array}$$

2.2 The Schwartz Class and the Fourier Transform

2.2.1	(a) $g(x)=\{\begin{array}{cc}{e}^{\frac{1}{|x|-1}}& |x|<1,\\ 0& |x|\ge 1.\end{array}$ (b) $h(x)=\{\begin{array}{cc}0& |x|\le \frac{1}{2},\\ \frac{e^{\frac{1}{1-|x|}}}{e^{\frac{1}{|x|-\frac{1}{2}}}+e^{\frac{1}{1-|x|}}}& \frac{1}{2}<|x|<1,\\ 1& 1\le |x|\le 2,\\ \frac{e^{\frac{1}{|x|-2}}}{e^{\frac{1}{4-|x|}}+e^{\frac{1}{|x|-2}}}& 2<|x|<4,\\ 0& |x|\ge 4.\end{array}$ (c) Take some $\varphi \in \mathcal{S}$ such that ${\varphi }^{\vee }$ is odd, real-valued and compactly supported. For instance, ${\varphi }^{\vee }(\xi )={\xi }_{1}h(\xi )$ where $\xi =({\xi }_1,\cdots ,{\xi }_n)$ and $h$ is defined as in part (b). Then $\varphi =({\varphi }^{\vee }{)}^{\wedge }$ takes purely imaginary values, so $\varphi \ast \tilde{\varphi }$ takes real values. Let$$f=(\varphi \ast \tilde{\varphi }{)}^2=((\varphi \ast \tilde{\varphi }{)}^{\wedge }\ast (\varphi \ast \tilde{\varphi }{)}^{\wedge }{)}^{\vee },$$then$$\hat{f}=(\varphi \ast \tilde{\varphi }{)}^{\wedge }\ast (\varphi \ast \tilde{\varphi }{)}^{\wedge }=(\hat{\varphi }{\varphi }^{\vee })\ast (\hat{\varphi }{\varphi }^{\vee })=(-{\varphi }^{\vee 2})\ast (-{\varphi }^{\vee 2})$$is nonnegative and compactly supported.
2.2.2	For all multi-indices $\alpha ,\beta$,$$\begin{array}{rl}\Vert x^{\alpha }({\partial }^{\beta }(\widehat{f_k}-\hat{f}))(x){\Vert }_{L^{\infty }}& =\frac{(2\pi {)}^{\beta }}{(2\pi {)}^{\alpha }}\Vert ({\partial }^{\alpha }(x^{\beta }(f_k(x)-f(x))){)}^{\wedge }{\Vert }_{L^{\infty }}\\ & \le \frac{(2\pi {)}^{\beta }}{(2\pi {)}^{\alpha }}\Vert {\partial }^{\alpha }(x^{\beta }(f_k(x)-f(x))){\Vert }_{L^1}\to 0.\end{array}$$The last step is because $f_k\to f$ in $\mathcal{S}$ implies $x^{\beta }{f}_k\to x^{\beta }f$ in $\mathcal{S}$, and subsequently ${\partial }^{\alpha }(x^{\beta }{f}_k)\to {\partial }^{\alpha }(x^{\beta }f)$ in $\mathcal{S}$, hence in $L^1$ by Proposition 2.2.6. And $\widehat{\tilde{f_k}}\to \hat{\tilde{f}}$ gives $f_k^{\vee }\to f^{\vee }$.
2.2.3	If there exists nonzero functions $f$ such that $\hat{f}=\lambda f$, then$$f=\widetilde{\stackrel{~}{f}}=\hat{\widehat{\hat{\hat{f}}}}={\lambda }^{4}f,$$so ${\lambda }^4=1$, then $\lambda =1,-1,i,-i$. Indeed $\{1,-1,i,-i\}$ is exactly the spectrum of the Fourier transform.$$(e^{-\pi x^2}{)}^{\wedge }=e^{-\pi x^2}$$shows that $1$ is an eigenvalue.$$(-2\pi x{e}^{-\pi x^2}{)}^{\wedge }=(\partial (e^{-\pi x^2}){)}^{\wedge }=2\pi ix{e}^{-\pi x^2}$$shows that $-i$ is an eigenvalue.$$\begin{array}{rl}(4{\pi }^2{x}^2{e}^{-\pi x^2}-\pi e^{-\pi x^2}{)}^{\wedge }& =(4{\pi }^2{x}^2{e}^{-\pi x^2}-2\pi e^{-\pi x^2}{)}^{\wedge }+\pi (e^{-\pi x^2}{)}^{\wedge }\\ & =({\partial }^2(e^{-\pi x^2}){)}^{\wedge }+\pi e^{-\pi x^2}\\ & =-4{\pi }^2{x}^2{e}^{-\pi x^2}+\pi e^{-\pi x^2}\end{array}$$shows that $-1$ is an eigenvalue.$$\begin{array}{rl}(-8{\pi }^3{x}^3{e}^{-\pi x^2}+6{\pi }^{2}x{e}^{-\pi x^2}{)}^{\wedge }& =(-8{\pi }^3{x}^3{e}^{-\pi x^2}+12{\pi }^{2}x{e}^{-\pi x^2}{)}^{\wedge }+3\pi (-2\pi x{e}^{-\pi x^2}{)}^{\wedge }\\ & =({\partial }^3(e^{-\pi x^2}){)}^{\wedge }+6{\pi }^{2}ix{e}^{-\pi x^2}\\ & =-8{\pi }^{3}i{x}^3{e}^{-\pi x^2}+6{\pi }^{2}ix{e}^{-\pi x^2}\end{array}$$shows that $i$ is an eigenvalue.
2.2.4	Without loss of generality, assume $A=B=1$, $M=N=L$, then the proposition becomes$${\int }_{{\mathbb{R}}^n}(1+|x-y|{)}^{-L}(1+|y|{)}^{-L}dy\le C(1+|x|{)}^{-L}.$$ This is because$$\begin{array}{rl}& {\int }_{{\mathbb{R}}^n}(1+|x|{)}^L(1+|x-y|{)}^{-L}(1+|y|{)}^{-L}dy\\ \le & {\int }_{{\mathbb{R}}^n}(1+|x-y|+1+|y|{)}^L(1+|x-y|{)}^{-L}(1+|y|{)}^{-L}dy\\ \le & {\int }_{{\mathbb{R}}^n}{2}^{L-1}((1+|x-y|{)}^L+(1+|y|{)}^L)(1+|x-y|{)}^{-L}(1+|y|{)}^{-L}dy\\ =& {\int }_{{\mathbb{R}}^n}{2}^{L-1}((1+|y|{)}^{-L}+(1+|x-y|{)}^{-L})dy\\ \le & C.\end{array}$$
2.2.5	Let $L_0^p=\{f\in L^p\mid f=0\text{ a.e. outside some compact set }K\}$. When $1\le p<\infty$, let $k_{\epsilon }(x)={\epsilon }^{-n}k({\epsilon }^{-1}x)$ where $k\in {\mathcal{C}}_0^{\infty }$, then $\{k_{\epsilon }{\}}_{\epsilon >1}$ is a smooth approximate identity all the supports of which are contained in $\mathrm{supp}k$. For $f\in L_0^p$, we have $f\ast k_{\epsilon }\in {\mathcal{C}}_0^{\infty }$ by the Lebesgue dominated convergence theorem, and $f\ast k_{\epsilon }\stackrel{L^p}{\to }f$ by Theorem 1.2.19 (a), which shows that ${\mathcal{C}}_0^{\infty }$ is dense in $L_0^p$, hence dense in $L^p$. When $p<1$, by the $p=1$ case ${\mathcal{C}}_0^{\infty }$ is dense in $L_0^1$ with respect to the $L^1$ norm and therefore also with respect to the $L^p$ norm because for $f\in L_0^1$ we have$$\Vert f\ast k_{\epsilon }-f{\Vert }_{L^p}\le m(\mathrm{support}f+\mathrm{support}k{)}^{\frac{1}{p}-1}\Vert f\ast k_{\epsilon }-f{\Vert }_{L^1}.$$Moreover, $L_0^1$ is dense in $L_0^p$ since for $f\in L_0^p$ we have $f{\chi }_{|f|\le n}\in L_0^1$, hence dense in $L^p$. When $p=\infty$, any sequence of ${\mathcal{C}}_0^{\infty }$ functions converges with respect to the $L^{\infty }$ norm to a continuous function, so ${\mathcal{C}}_0^{\infty }$ is not dense in $L^{\infty }$.
2.2.6	(a) $\Vert \hat{f}-{\tau }^y\hat{f}{\Vert }_{L^{\infty }}$ $=\Vert \hat{f}-(e^{2\pi ixy}f{)}^{\wedge }{\Vert }_{L^{\infty }}$ by Proposition 2.2.11 (6) $\le \Vert f-e^{2\pi ixy}f{\Vert }_{L^1}\to 0$ as $y\to 0$ by the Lebesgue dominated convergence theorem. (b)$$\begin{array}{rl}& {\int }_{{\mathbb{R}}^n}f(x)\hat{g}(x)dx={\int }_{{\mathbb{R}}^n}f(x)({\int }_{{\mathbb{R}}^n}g(y)e^{-2\pi ixy}dy)dx\\ =& {\int }_{{\mathbb{R}}^n}({\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi ixy}dx)g(y)dy={\int }_{{\mathbb{R}}^n}\hat{f}(y)g(y)dy\end{array}$$by Fubini’s theorem. (c) Use (2.2.15). But now the left-hand side of (2.2.15) may not converge to $f(t)$ pointwise; only is known that the left-hand side converges to $f(t)$ with respect to the $L^1$ norm. Yet by the Riesz–Fischer theorem, there is a sequence of ${\epsilon }_n\to 0$ as $n\to \infty$ such that the left-hand side converges to $f(t)$ a.e. if $\epsilon$ is taken to be ${\epsilon }_n$, and the right-hand side still converges to $(\hat{f}{)}^{\vee }(t)$ after the restriction of $\epsilon$, so we obtain $f=(\hat{f}{)}^{\vee }$ a.e.
2.2.7	(a) Let $g(x)=e^{-\pi |\epsilon x{|}^2}$ in Exercise 2.2.6 (b), then $\hat{g}(x)={\epsilon }^{-n}{e}^{-\pi |x/\epsilon {|}^2}$, which is an approximate identity, so $$\underset{\epsilon \to 0}{\lim }{\int }_{{\mathbb{R}}^n}\hat{f}(x)e^{-\pi |\epsilon x{|}^2}dx=\underset{\epsilon \to 0}{\lim }{\int }_{{\mathbb{R}}^n}f(x){\epsilon }^{-n}{e}^{-\pi |x/\epsilon {|}^2}dx=f(0)$$because $f$ is continuous at $0$ and ${\epsilon }^{-n}{e}^{-\pi |x/\epsilon {|}^2}$ is an approximate identity. (b) Suppose that $\hat{f}\notin L^1$, then ${\int }_K\hat{f}(x)dx>f(0)$ for some bounded area $K$, so $$\underset{\epsilon \to 0}{\lim }{\int }_{{\mathbb{R}}^n}\hat{f}(x)e^{-\pi |\epsilon x{|}^2}dx\ge \underset{\epsilon \to 0}{\lim }{\int }_K\hat{f}(x)e^{-\pi |\epsilon x{|}^2}dx={\int }_K\hat{f}(x)dx>f(0),$$contradicting the result of part (a). So $\hat{f}\in L^1$, and it follows that $\Vert \hat{f}{\Vert }_{L^1}=f(0)$ by the Lebesgue dominated convergence theorem.
2.2.8	Let $h=f\ast \overline{\stackrel{~}{f}}$, then $h\in L^1({\mathbb{R}}^n)\cap L^{\infty }({\mathbb{R}}^n)$ because$$\begin{array}{rl}\Vert h{\Vert }_{L^1}& \le \Vert f{\Vert }_{L^1}\Vert \overline{\stackrel{~}{f}}{\Vert }_{L^1}=\Vert f{\Vert }_{L^1}^2<\infty \\ \mathrm{and}\Vert h{\Vert }_{L^{\infty }}& \le \Vert f{\Vert }_{L^2}\Vert \overline{\stackrel{~}{f}}{\Vert }_{L^2}=\Vert f{\Vert }_{L^2}^2<\infty \end{array}$$by Theorem 1.2.12,$$\hat{h}=\hat{f}\widehat{\bar{\stackrel{~}{f}}}\stackrel{(5)}{=}\hat{f}\stackrel{ˉ}{\widetilde{\hat{\stackrel{~}{f}}}}\stackrel{(4)}{=}\hat{f}\stackrel{ˉ}{\widetilde{\stackrel{~}{\hat{f}}}}=\hat{f}\bar{\hat{f}}\ge 0$$where (5) and (4) stands for Proposition 2.2.11 (5) and (4), and $f$ is continuous at $0$ because$$\begin{array}{rl}|h(y)-h(0)|& =|(({\tau }^{-y}f-f)\ast \overline{\stackrel{~}{f}})(0)|\le \Vert ({\tau }^{-y}f-f)\ast \overline{\stackrel{~}{f}}{\Vert }_{L^{\infty }}\\ & \le \Vert {\tau }^{-y}f-f{\Vert }_{L^2}\Vert \overline{\stackrel{~}{f}}{\Vert }_{L^2}\to 0\text{as }y\to 0.\end{array}$$Exercise 2.2.7 (b) yields$$\Vert \hat{f}{\Vert }_{L^2}^2=\Vert \hat{h}{\Vert }_{L^1}=h(0)={\int }_{{\mathbb{R}}^n}f(x)\overline{\tilde{f}(-x)}dx=\Vert f{\Vert }_{L^2}^2.$$
2.2.9	(a) For $t\in [2n\pi ,(2n+2)\pi ]$ where $n\in \mathbb{N}$,$$\begin{array}{rl}{\int }_0^t\frac{\sin (\xi )}{\xi }d\xi & \le {\int }_0^{(2n+1)\pi }\frac{\sin (\xi )}{\xi }d\xi \\ & ={\int }_0^{\pi }\frac{\sin (\xi )}{\xi }d\xi +\sum _{k=1}^n{\int }_0^{\pi }(\frac{\sin (\xi +(2k-1)\pi )}{\xi +(2k-1)\pi }+\frac{\sin (\xi +2k\pi )}{\xi +2k\pi })d\xi \\ & ={\int }_0^{\pi }\frac{\sin (\xi )}{\xi }d\xi +\sum _{k=1}^n{\int }_0^{\pi }(-\frac{\sin (\xi )}{\xi +(2k-1)\pi }+\frac{\sin (\xi )}{\xi +2k\pi })d\xi \\ & \le {\int }_0^{\pi }\frac{\sin (\xi )}{\xi }d\xi \le \pi .\end{array}$$And for similar reasons, $${\int }_0^t\frac{\sin (\xi )}{\xi }d\xi \ge \underset{n\in \mathbb{N}}{\sup }{\int }_0^{2n\pi }\frac{\sin (\xi )}{\xi }d\xi \ge 0.$$So$$\left|{\int }_{\epsilon }^t\frac{\sin (\xi )}{\xi }d\xi \right|\le \pi .$$ (b) $f$ is odd implies that $\hat{f}(\xi )=-i{\int }_{\mathbb{R}}f(x)\sin (2\pi x\xi )dx$, so by part (a)$$\begin{array}{rl}& \left|{\int }_{\epsilon }^t\frac{\hat{f}(\xi )}{\xi }d\xi \right|=|{\int }_{\epsilon }^t{\int }_{\mathbb{R}}f(x)\frac{\sin (2\pi x\xi )}{\xi }dxd\xi |\\ =& |{\int }_{\mathbb{R}}f(x){\int }_{2\pi \epsilon \xi }^{2\pi t\xi }\frac{\sin (2\pi x\xi )}{2\pi x\xi }d(2\pi x\xi )dx|\le 4\Vert f{\Vert }_{L^1}.\end{array}$$ (c) Suppose that such $f$ exists, then by part (b)$$4\Vert f{\Vert }_{L^1}\ge {\int }_2^t\frac{g(\xi )}{\xi }d\xi ={\int }_2^t\frac{d\xi }{\xi \log (\xi )}=\log (\log (\xi )){|}_2^t\to +\infty ,$$a contradiction.
2.2.10	By substituting $x$ with $-\frac{1}{x}$ which increases from 0 to $+\infty$ then from $-\infty$ to $0$ as $x$ ranges from $-\infty$ to $+\infty$, we get$$\begin{array}{rl}{\int }_{-\infty }^{+\infty }f(x-\frac{1}{x})dx& =\frac{1}{2}\left({\int }_{-\infty }^{+\infty }f\right(x-\frac{1}{x})dx+{\int }_{-\infty }^{+\infty }f(-\frac{1}{x}+x\left)d\right(-\frac{1}{x}\left)\right)\\ & =\frac{1}{2}{\int }_{-\infty }^{+\infty }f(x-\frac{1}{x})d(x-\frac{1}{x}).\end{array}$$ Then since $u=x-\frac{1}{x}$ increases from $-\infty$ to $+\infty$ twice as $x$ ranges from $-\infty$ to $+\infty$, $$\frac{1}{2}{\int }_{-\infty }^{+\infty }f(x-\frac{1}{x})d(x-\frac{1}{x})={\int }_{-\infty }^{+\infty }f(u)du.$$
2.2.11	(a) Use Exercise 2.2.10 with $f(x)=e^{-t{x}^2}$ to obtain that$$\begin{array}{rl}{\int }_0^{\infty }{e}^{-y-\frac{t^2}{y}+2t}\frac{dy}{\sqrt{y}}& =2{\int }_0^{\infty }f(\sqrt{\frac{y}{t}}-\sqrt{\frac{t}{y}})d\left(\sqrt{t}\sqrt{\frac{y}{t}}\right)\\ & ={\int }_{-\infty }^{+\infty }f(u)d(\sqrt{t}u)\\ & ={\int }_{-\infty }^{+\infty }{e}^{-t{u}^2}d(\sqrt{t}u)\\ & =\sqrt{\pi }.\end{array}$$ (b) Set $t=\pi |x|$ in part (a) to get that$$\begin{array}{rl}(e^{-2\pi |x|}{)}^{\wedge }(\xi )& ={\int }_{{\mathbb{R}}^n}{e}^{-2\pi |x|}{e}^{-2\pi i\xi x}dx\\ & =\frac{1}{\sqrt{\pi }}{\int }_{{\mathbb{R}}^n}{\int }_0^{\infty }{e}^{-y-\frac{{\pi }^2|x{|}^2}{y}-2\pi i\xi x}\frac{dy}{\sqrt{y}}dx\\ & =\frac{1}{\sqrt{\pi }}{\int }_0^{\infty }{\int }_{{\mathbb{R}}^n}{e}^{-(1+|\xi {|}^2)y-\frac{{\pi }^2|x{|}^2}{y}-2\pi i\xi x+|\xi {|}^{2}y}dx\frac{dy}{\sqrt{y}}.\end{array}$$ For all $1\le j\le n$, use a rectangle contour to lift the $x$-axis and get that$${\int }_{\mathbb{R}}{e}^{-\frac{{\pi }^2{x}_j^2}{y}-2\pi i{\xi }_j{x}_j+{\xi }_j^{2}y}d{x}_j=\frac{\sqrt{y}}{\sqrt{\pi }}{\int }_{\mathbb{R}}{e}^{-\pi (\frac{\sqrt{\pi }{x}_j}{\sqrt{y}}+\frac{i{\xi }_j\sqrt{y}}{\sqrt{\pi }}{)}^2}d\frac{\sqrt{\pi }{x}_j}{\sqrt{y}}=\frac{\sqrt{y}}{\sqrt{\pi }}$$where $x=(x_1,\cdots ,x_n)$ and $\xi =({\xi }_1,\cdots ,{\xi }_n)$, plugging into above$$\begin{array}{rl}& =\frac{1}{{\pi }^{\frac{n+1}{2}}}{\int }_0^{\infty }{e}^{-(1+|\xi {|}^2)y}{y}^{\frac{n-1}{2}}dy\\ & =\frac{\Gamma (\frac{n+1}{2})}{{\pi }^{\frac{n+1}{2}}}\frac{1}{(1+|\xi {|}^2{)}^{\frac{n+1}{2}}}.\end{array}$$
2.2.12	(a) For all $x\in \mathbb{R}$,$$|f(x){|}^2=|{\int }_{-\infty }^x(f^2{)}^{\prime }(t)dt|=2|{\int }_{-\infty }^{x}f(t)f^{\prime }(t)dt|\le 2\Vert f{\Vert }_{L^p}\Vert f^{\prime }{\Vert }_{L^{p^{\prime }}}.$$ (b) For $x=(x_1,\cdots ,x_n)\in {\mathbb{R}}^n$, write $\Omega =\{(t_1,\cdots ,t_n)\in {\mathbb{R}}^n\mid t_1\le x_1,\cdots ,t_n\le x_n\}$. We have$$\begin{array}{rl}|f(x){|}^2& =|{\int }_{-\infty }^{x_n}\cdots {\int }_{-\infty }^{x_1}{\partial }_1\cdots {\partial }_n(f^2)(t)d{t}_1\cdots d{t}_n|=|{\int }_{\Omega }{\partial }^{(1,\cdots ,1)}(f^2)(t)dt|\\ & \le \Vert \sum _{\alpha +\beta =(1,\cdots ,1)}({\partial }^{\alpha }f)({\partial }^{\beta }f){\Vert }_{L^1}\le \sum _{\alpha +\beta =(1,\cdots ,1)}\Vert {\partial }^{\alpha }f{\Vert }_{L^p}\Vert {\partial }^{\beta }f{\Vert }_{L^{p^{\prime }}}.\end{array}$$
2.2.13	Let $y\in {\mathbb{R}}^n$. Integrate by parts and apply the Cauchy–Schwarz inequality and Plancherel’s identity, we get$$\begin{array}{rl}\Vert f{\Vert }_{L^2}^2& =\frac{1}{n}{\int }_{{\mathbb{R}}^n}f(x)\overline{f(x)}\sum _{j=1}^n{\partial }_j(x_j-y_j)dx\\ & =\frac{1}{n}{\int }_{{\mathbb{R}}^n}\sum _{j=1}^n{\partial }_j(f(x)\overline{f(x)})(x_j-y_j)dx\\ & =\frac{1}{n}[{\int }_{{\mathbb{R}}^n}\sum _{j=1}^n({\partial }_{j}f)(x)\overline{f(x)}(x_j-y_j)dx+{\int }_{{\mathbb{R}}^n}\sum _{j=1}^n\overline{({\partial }_{j}f)(x)}f(x)(x_j-y_j)dx]\\ & \le \frac{1}{n}\left[\right({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|{\partial }_{j}f(x){|}^{2}dx{)}^{\frac{1}{2}}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|\overline{f(x)}(x_j-y_j){|}^{2}dx{)}^{\frac{1}{2}}\\ & +({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|\overline{({\partial }_{j}f)(x)}{|}^{2}dx{)}^{\frac{1}{2}}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|f(x)(x_j-y_j){|}^{2}dx{)}^{\frac{1}{2}}]\\ & =\frac{2}{n}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|{\partial }_{j}f(x){|}^{2}dx{)}^{\frac{1}{2}}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|f(x)(x_j-y_j){|}^{2}dx{)}^{\frac{1}{2}}\\ & =\frac{2}{n}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|\widehat{{\partial }_{j}f}(\xi ){|}^{2}d\xi {)}^{\frac{1}{2}}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|f(x)(x_j-y_j){|}^{2}dx{)}^{\frac{1}{2}}\\ & =\frac{4\pi }{n}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|{\xi }_j\hat{f}(\xi ){|}^{2}dx{)}^{\frac{1}{2}}({\int }_{{\mathbb{R}}^n}\sum _{j=1}^n|f(x)(x_j-y_j){|}^{2}dx{)}^{\frac{1}{2}}\\ & =\frac{4\pi }{n}({\int }_{{\mathbb{R}}^n}|\xi {|}^2|\hat{f}(\xi ){|}^{2}d\xi {)}^{\frac{1}{2}}({\int }_{{\mathbb{R}}^n}|f(x){|}^2|x-y{|}^{2}dx{)}^{\frac{1}{2}}.\end{array}$$Then replace $f(x)$ by $f(x)e^{2\pi ixz}$ in the inequality above and the desired inequality follows from$$\sum _{j=1}^n|({\partial }_j(f(\xi )e^{2\pi i\xi z}{)}^{\wedge }{|}^2=(2\pi {)}^2\sum _{j=1}^n|({\xi }_j-z_j)\hat{f}(\xi ){|}^2=(2\pi {)}^2|\xi -z{|}^2|\hat{f}(\xi ){|}^2.$$
2.2.14	Let $B$ be the unit ball $\{x\in {\mathbb{R}}^n\mid |x|<1\}$ and $B^c$ its complement. By the Cauchy–Schwartz inequality,$$\begin{array}{rl}& \Vert g{\Vert }_{L^1}=\Vert g{\Vert }_{L^1(B)}+\Vert g{\Vert }_{L^1(B^c)}\\ \le & \Vert |x{|}^{-\alpha }{\Vert }_{L^2(B)}\Vert |x{|}^{\alpha }g(x){\Vert }_{L^2(B)}+\Vert |x{|}^{-\beta }{\Vert }_{L^2(B^c)}\Vert |x{|}^{\beta }g(x){\Vert }_{L^2(B^c)}\\ \le & C(\Vert |x{|}^{\alpha }g(x){\Vert }_{L^2}+\Vert |x{|}^{\beta }g(x){\Vert }_{L^2}).\end{array}$$Then repalce $g(x)$ by $g(\lambda x)$ for $\lambda >0$ and get$$\begin{array}{rl}& \Vert g{\Vert }_{L^1}={\lambda }^n\Vert g(\lambda x){\Vert }_{L^1}\\ \le & C{\lambda }^n(\Vert |x{|}^{\alpha }g(\lambda x){\Vert }_{L^2}+\Vert |x{|}^{\beta }g(\lambda x){\Vert }_{L^2})\\ =& C({\lambda }^{n/2-\alpha }\Vert |x{|}^{\alpha }g(x){\Vert }_{L^2}+{\lambda }^{n/2-\beta }\Vert |x{|}^{\beta }g(x){\Vert }_{L^2})\\ =& C\Vert |x{|}^{\alpha }g(x){\Vert }_{L^2}^{\frac{\beta -n/2}{\beta -\alpha }}\Vert |x{|}^{\beta }g(x){\Vert }_{L^2}^{\frac{n/2-\alpha }{\beta -\alpha }}\mathrm{by}\mathrm{taking}\mathrm{suitable}\lambda .\end{array}$$

2.3 The Class of Tempered Distributions

2.3.1	Such $\mu$ can be identified with the functional$$f\mapsto {\int }_{{\mathbb{R}}^n}fd\mu$$which is a tempered distribution because the condition of Proposition 2.3.4 (b) is satisfied: in fact,$$\begin{array}{rl}\left|{\int }_{{\mathbb{R}}^n}fd\mu \right|& ={\int }_{{\mathbb{R}}^n}|f(x)|(1+|x|{)}^k\frac{d\mu }{(1+|x|{)}^k}\\ & \le M\underset{x\in {\mathbb{R}}^n}{\sup }|f(x)|(1+|x|{)}^k\\ & \le M{C}_k\sum _{|\alpha |\le k}\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }f(x)|\text{for all }f\in \mathcal{S},\end{array}$$where the last inequality is due to (2.2.3). The Lebesgue measure as a tempered distribution coincides with $1$ since$$⟨f,1⟩={\int }_{{\mathbb{R}}^n}f(x)dx\text{for all }f\text{ in }\mathcal{S}.$$
2.3.2	Pick any nonzero vector $v\in {\mathbb{R}}^n$, and it is easily shown that $(x+v{)}^{{\alpha }_1}{\phi }_{\epsilon }$ is $b{v}^{{\alpha }_1}$ times an approximate identity for any multi-index ${\alpha }_1$ as $\phi \in \mathcal{S}$. Then for all multi-indices $\alpha ,\beta$,$$\begin{array}{rl}& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }({\phi }_{\epsilon }\ast f-bf)|\\ =& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }({\phi }_{\epsilon }\ast {\partial }^{\beta }f-b{\partial }^{\beta }f)|\text{by Proposition 2.2.7}\\ \le & \sum _{{\alpha }_1+{\alpha }_2=\alpha }\left(\genfrac{}{}{0ex}{}{\alpha }{{\alpha }_1}\right)\underset{x\in {\mathbb{R}}^n}{\sup }|((x+v{)}^{{\alpha }_1}{\phi }_{\epsilon })\ast ((x-v{)}^{{\alpha }_2}{\partial }^{\beta }f)-b{v}^{{\alpha }_1}(x-v{)}^{{\alpha }_2}{\partial }^{\beta }f|\\ \to & 0\text{by Proposition 1.2.21 (b)}.\end{array}$$In the last step $(x-v{)}^{{\alpha }_2}{\partial }^{\beta }f\in \mathcal{S}$ and is hence uniformly continuous on ${\mathbb{R}}^n$.
2.3.3	We have$$\begin{array}{rl}& ⟨({\delta }^{a}f)\ast ({\delta }^{a}u),g⟩\\ =& ⟨{\delta }^{a}u,\widetilde{{\delta }^{a}f}\ast g⟩\\ =& ⟨u,a^{-n}{\delta }^{1/a}(\widetilde{{\delta }^{a}f}\ast g)⟩\\ =& ⟨u,a^{-n}{\int }_{{\mathbb{R}}^n}f(-ay)g(a^{-1}x-y)dy⟩\\ =& ⟨u,a^{-n}{\int }_{{\mathbb{R}}^n}f(-y)g(a^{-1}x-a^{-1}y)d(a^{-1}y)⟩\\ =& ⟨a^{-n}u,a^{-n}\tilde{f}\ast {\delta }^{1/a}g⟩\\ =& ⟨a^{-n}f\ast u,a^{-n}{\delta }^{1/a}g⟩\\ =& ⟨a^{-n}{\delta }^a(f\ast u),g⟩\text{for all }g\in \mathcal{S}.\end{array}$$ Or check that the two distributions have identical Fourier transforms:$$\begin{array}{rl}(({\delta }^{a}f)\ast ({\delta }^{a}u){)}^{\wedge }& =({\delta }^{a}f{)}^{\wedge }({\delta }^{a}u{)}^{\wedge }=a^{-n}{\delta }^{1/a}\hat{f}{a}^{-n}{\delta }^{1/a}\hat{u}=a^{-2n}{\delta }^{1/a}(\hat{f}\hat{u}),\\ (a^{-n}{\delta }^a(f\ast u){)}^{\wedge }& =a^{-n}(a^{-n}{\delta }^{1/a}(f\ast u{)}^{\wedge })=a^{-2n}{\delta }^{1/a}(\hat{f}\hat{u}).\end{array}$$
2.3.4	(a)$$⟨\partial {\chi }_{[a,b]},f⟩=-⟨{\chi }_{[a,b]},f^{\prime }⟩=-(f(b)-f(a))=⟨{\delta }_a-{\delta }_b,f⟩.$$(b) To simplify notations, let $j=1$. Sure enough, everything is essentially the same for $j=2$.$$\begin{array}{rl}⟨{\partial }_1{\chi }_{B(0,1)},f⟩& =-⟨{\chi }_{B(0,1)},{\partial }_{1}f⟩\\ & =-{\int }_{B(0,1)}{\partial }_{1}fdx\\ & =-{\int }_{-1}^1{\int }_{-\sqrt{1-x_1^2}}^{\sqrt{1-x_1^2}}{\partial }_{1}fd{x}_{1}d{x}_2\\ & =-{\int }_{-1}^{1}f(\sqrt{1-x^2},x)-f(-\sqrt{1-x^2},x)dx.\end{array}$$(c)$$\begin{array}{rl}⟨(\sin x{)}^{\wedge },f⟩& =⟨\sin x,\hat{f}⟩=⟨(e^{ix}-e^{-ix})/2i,\hat{f}⟩\\ & =((\hat{f}{)}^{\vee }\left(\frac{1}{2\pi }\right)-(\hat{f}{)}^{\vee }(-\frac{1}{2\pi }\left)\right)/2i\\ & =\left(f\right(\frac{1}{2\pi })-f(-\frac{1}{2\pi }\left)\right)/2i\\ & =⟨({\delta }_{\frac{1}{2\pi }}-{\delta }_{-\frac{1}{2\pi }})/2i,f⟩.\end{array}$$(d)$$\begin{array}{rl}⟨\partial \log |x|,\phi ⟩& =-⟨\log |x|,{\phi }^{\prime }⟩=-{\int }_{\mathbb{R}}\log |x|{\phi }^{\prime }(x)dx\\ & =-\underset{\epsilon \to 0}{\lim }{\int }_{\epsilon \le |x|}\log |x|{\phi }^{\prime }(x)dx\\ & =\underset{\epsilon \to 0}{\lim }{\int }_{\epsilon \le |x|}\phi (x)\frac{dx}{x}+(\phi (-\epsilon )-\phi (\epsilon ))\log |\epsilon |\\ & =\underset{\epsilon \to 0}{\lim }{\int }_{\epsilon \le |x|}\phi (x)\frac{dx}{x}\text{since }\phi (-\epsilon )-\phi (\epsilon )=O(\epsilon ).\end{array}$$
2.3.5	(a) For all multi-indices $\alpha ,\beta$ and $|h|<1$,$$\begin{array}{rl}& \underset{x\in {\mathbb{R}}^n}{\sup }\left|x^{\alpha }{\partial }^{\beta }\right(\frac{{\tau }^{-h{e}_j}f-f}{h}-{\partial }_{j}f\left)\right|\\ =& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }\frac{({\partial }^{\beta }f)(x+h{e}_j)-({\partial }^{\beta }f)(x)}{h}-{\partial }_j{\partial }^{\beta }f(x)|\\ =& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }_j{\partial }^{\beta }f(x+\theta h{e}_j)-{\partial }_j{\partial }^{\beta }f(x)|\text{where}\theta =\theta (x)\in (0,1)\\ \le & h\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }_j^2{\partial }^{\beta }f(x+{\theta }^{\prime }h{e}_j)|\text{where}{\theta }^{\prime }={\theta }^{\prime }(x)\in (0,\theta (x))\\ \le & h\underset{x\in {\mathbb{R}}^n}{\sup }|\max \{(x+e_j{)}^{\alpha },(x-e_j{)}^{\alpha }\}{\partial }_j^2{\partial }^{\beta }f(x)|\\ \to & 0\text{as}h\to 0,\end{array}$$where the last step is because $|(x\pm e_j{)}^{\alpha }|$ are both dominated by a finite sum of some $|x^{\gamma }|$’s. In part (b) and (c), assume $f$ is identically equal to $1$ in $B(0,r)$. (b) For all multi-indices $\alpha ,\beta$,$$\begin{array}{rl}& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }({\phi }_{k}f-f)|\\ =& \underset{x\in {\mathbb{R}}^n}{\sup }|({\phi }_k-1)x^{\alpha }{\partial }^{\beta }f+\sum _{\gamma \le \beta ,\gamma \ne 0}{x}^{\alpha }{\partial }^{\gamma }{\phi }_k{\partial }^{\beta -\gamma }f|\\ \le & \Vert {\phi }_k-1{\Vert }_{L^{\infty }}\underset{|x|>kr}{\sup }|x^{\alpha }{\partial }^{\beta }f|+\sum _{\gamma \le \beta ,\gamma \ne 0}\frac{\Vert {\partial }^{\gamma }\phi {\Vert }_{L^{\infty }}}{k^{|\gamma |}}\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta -\gamma }f|.\end{array}$$Noticing that $|x_i|>kr/\sqrt{n}$ for some $1\le i\le n$ when $|x|>kr$, we have$$\underset{|x|>kr}{\sup }|x^{\alpha }{\partial }^{\beta }f|\le \underset{1\le i\le n}{\max }\frac{\underset{|x|>kr}{\sup }|x^{\alpha +e_i}{\partial }^{\beta }f|}{kr/\sqrt{n}}$$where $e_i$ is the multi-index whose $i$-th component is $1$ and the others $0$, so $\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }({\phi }_{k}f-f)|\to 0$ as $k\to \infty$. (c) By part (b) $({\phi }_{k}f{)}^{\wedge }\to \hat{f}$ in $\mathcal{S}$ since the Fourier transform is a homeomorphism (Corollary 2.2.15), which on one hand reduces the proposition to ${\phi }_k({\phi }_{k}f{)}^{\wedge }-({\phi }_{k}f{)}^{\wedge }\to 0$ in $\mathcal{S}$, and on the other hand implies that for all $k$ and multi-indices $\alpha ,\beta$,$$\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }({\phi }_{k}f{)}^{\wedge }|\le C_{\alpha ,\beta }<\infty .$$ Then for all multi-indices $\alpha ,\beta$,$$\begin{array}{rl}& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }(({\phi }_k-1)({\phi }_{k}f{)}^{\wedge })|\\ =& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }({\phi }_k-1){\partial }^{\beta }(({\phi }_{k}f{)}^{\wedge })+\sum _{\gamma \le \beta ,\gamma \ne 0}{\partial }^{\gamma }({\phi }_k-1)x^{\alpha }{\partial }^{\beta -\gamma }(({\phi }_{k}f{)}^{\wedge })|\\ \le & \Vert {\phi }_k-1{\Vert }_{L^{\infty }}\underset{|x|>kr}{\sup }|x^{\alpha }{\partial }^{\beta }(({\phi }_{k}f{)}^{\wedge })|+\sum _{\gamma \le \beta ,\gamma \ne 0}\frac{\Vert {\partial }^{\gamma }\phi {\Vert }_{L^{\infty }}}{k^{|\gamma |}}{C}_{\alpha ,\beta -\gamma }.\end{array}$$ For the same reason as part (b), $$\underset{|x|>kr}{\sup }|x^{\alpha }{\partial }^{\beta }(({\phi }_{k}f{)}^{\wedge })|\le \underset{1\le i\le n}{\max }\frac{C_{\alpha +1_i,\beta }}{kr/\sqrt{n}},$$ so $\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }(({\phi }_k-1)({\phi }_{k}f{)}^{\wedge })|\to 0$ as $k\to \infty$.
2.3.6	The Fourier transform of a nonzero ${\mathcal{C}}_0^{\infty }$ function, as a distribution with compact support, is a real analytic function by Theorem 2.3.21, which cannot be a ${\mathcal{C}}_0^{\infty }$ function for it would otherwise be identically $0$ due to the uniqueness of real analytic functions (the uniqueness is obtained by showing that the interior of the zero set of a real analytic function is both open and closed).
2.3.7	For all $h\in \mathcal{S}$,$$\begin{array}{rl}⟨g_N,h⟩-⟨\hat{f},h⟩& =⟨(f{\chi }_{B(0,N)}{)}^{\wedge },h⟩-⟨\hat{f},h⟩\\ & =⟨f{\chi }_{B(0,N)}-f,\hat{h}⟩\\ & \le \Vert f{\chi }_{B(0,N)}-f{\Vert }_{L^p}\Vert \hat{h}{\Vert }_{L^{p^{\prime }}}\to 0\text{as }N\to \infty .\end{array}$$
2.3.8	Let$$u_N=\sum _{|k|\le N}{c}_k{\delta }_k,u=\sum _{k\in {\mathbb{Z}}^n}{c}_k{\delta }_k.$$Then $u_k\to u$ in ${\mathcal{S}}^{\prime }$ because for all $f\in \mathcal{S}$,$$\begin{array}{rl}|⟨u-u_N,f⟩|& =|\sum _{|k|>N}{c}_{k}f(k)|\\ & \le A\sum _{|k|>N}(1+|k|{)}^M|f(k)|\\ & \le A_N\underset{|k|>N}{\sup }(1+|k|{)}^{M+n+1}|f(k)|\\ & \le A_N\underset{x\in {\mathbb{R}}^n}{\sup }(1+|x|{)}^{M+n+1}|f(x)|\\ & \le A_N\sum _{|\beta |\le M}{C}_{n,M}\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\beta }f(x)|\to 0\text{as}N\to \infty ,\end{array}$$where $A_N=A\sum _{|k|>N}(1+|k|{)}^{-n-1}\to 0$ as $N\to \infty$ and in the last inequality (2.2.3) is used. Since $\widehat{u_N}=h_N$ (see Example 2.3.9), for all $f\in \mathcal{S}$,$$⟨\hat{u}-h_N,f⟩=⟨\hat{u}-\widehat{u_N},f⟩=⟨u-u_N,\hat{f}⟩\to 0\text{as}N\to \infty .$$
2.3.9	(a) The usual definition for functions (still denoted by $u$) homogeneous of degree $\gamma \in \mathbb{C}$ is ${\delta }^{\lambda }u={\lambda }^{\gamma }u$ for all $\lambda >0$, which agrees with the one given in the problem because for functions $u$,$$\begin{array}{rl}& {\delta }^{\lambda }u={\lambda }^{\gamma }u\text{for all }\lambda >0\\ \iff & ⟨{\delta }^{\lambda }u,\phi ⟩=⟨{\lambda }^{\gamma }u,\phi ⟩\text{for all }\lambda >0,\phi \in \mathcal{S}\\ \iff & {\lambda }^{-n}⟨u,{\delta }^{1/\lambda }\phi ⟩=⟨{\lambda }^{\gamma }u,\phi ⟩\text{for all }\lambda >0,\phi \in \mathcal{S}\\ \iff & ⟨u,{\delta }^{1/\lambda }\phi ⟩=(1/\lambda {)}^{-n-\gamma }⟨u,\phi ⟩\text{for all }\lambda >0,\phi \in \mathcal{S}.\end{array}$$ (b) This is because$$⟨{\delta }_0,{\delta }^{\lambda }\phi ⟩=⟨{\delta }_0,\phi ⟩=\phi (0).$$ (c) Since$$\begin{array}{rl}⟨{\partial }^{\alpha }u,{\delta }^{\lambda }\phi ⟩& =(-1{)}^{|\alpha |}⟨u,{\partial }^{\alpha }({\delta }^{\lambda }\phi )⟩=(-1{)}^{|\alpha |}⟨u,{\lambda }^{|\alpha |}{\delta }^{\lambda }({\partial }^{\alpha }\phi )⟩={\lambda }^{|\alpha |}(-1{)}^{|\alpha |}⟨u,{\delta }^{\lambda }({\partial }^{\alpha }\phi )⟩\\ & ={\lambda }^{|\alpha |}(-1{)}^{|\alpha |}{\lambda }^{-n-\gamma }⟨u,{\partial }^{\alpha }\phi ⟩={\lambda }^{-n-(\gamma -|\alpha |)}⟨{\partial }^{\alpha }u,\phi ⟩.\end{array}$$ (d) Since$$\begin{array}{r}⟨\hat{u},{\delta }^{\lambda }\phi ⟩=⟨u,\widehat{{\delta }^{\lambda }\phi }⟩=⟨u,{\lambda }^{-n}{\delta }^{{\lambda }^{-1}}\hat{\phi }⟩={\lambda }^{-n}⟨u,{\delta }^{{\lambda }^{-1}}\hat{\phi }⟩={\lambda }^{-n}(\frac{1}{\lambda }{)}^{-n-\gamma }⟨u,\hat{\phi }⟩={\lambda }^{-n-(-n-\gamma )}⟨\hat{u},\phi ⟩.\end{array}$$ The converse follows similarly.
2.3.10	(a) $e^{inx}$ and $e^{-inx}$ converge to $0$ in ${\mathcal{S}}^{\prime }$ and ${\mathcal{D}}^{\prime }$ as $n\to \infty$ as a consequence of Proposition 2.2.17, from which we know that the multiplication of distributions is not a continuous operation even when it is defined, for $(e^{inx},e^{-inx})\to (0,0)$ in ${\mathcal{S}}^{\prime }\times {\mathcal{S}}^{\prime }$ yet $e^{inx}{e}^{-inx}=1$. (b) $\sqrt{n}(1+n|x{|}^2{)}^{-1}\to \pi {\delta }_0$ in ${\mathcal{D}}^{\prime }$. This is because for all $f\in {\mathcal{C}}_0^{\infty }$, by the Lebesgue dominated convergence theorem,$$\begin{array}{rl}& \underset{n\to \infty }{\lim }{\int }_{\mathbb{R}}\frac{\sqrt{n}}{1+n{x}^2}f(x)dx\\ =& \underset{n\to \infty }{\lim }{\int }_{\mathbb{R}}\frac{1}{1+x^2}f\left(\frac{x}{\sqrt{n}}\right)dx\\ =& {\int }_{\mathbb{R}}\frac{1}{1+x^2}f(0)dx\\ =& \pi f(0).\end{array}$$
2.3.11	Let $h$ be a Schwartz function in ${\mathbb{R}}^n$ whose Fourier transform is equal to $1$ on the ball $B(0,1)$ and vanishes outside the ball $B(0,2)$ (which can be constructed as the inverse Fourier transform of a Schwartz function with the properties that the desired Fourier transform of $h$ is supposed to have). Since$$(f\ast R^{-n}{\delta }^{R}h{)}^{\wedge }=\hat{f}(R^{-n}{\delta }^{R}h{)}^{\wedge }=\hat{f}{\delta }^{\frac{1}{R}}\hat{h}=\hat{f},$$we establish the identity $f=f\ast R^{-n}{\delta }^{R}h$. Then$$\begin{array}{rl}\Vert {\partial }^{\alpha }f{\Vert }_{L^{\infty }}=& \Vert f\ast {\partial }^{\alpha }(R^{-n}{\delta }^{R}h){\Vert }_{L^{\infty }}\\ \le & \Vert f{\Vert }_{L^{\infty }}\Vert {\partial }^{\alpha }(R^{-n}{\delta }^{R}h){\Vert }_{L^1}\\ =& \Vert f{\Vert }_{L^{\infty }}{R}^{|\alpha |}\Vert {\partial }^{\alpha }h{\Vert }_{L^1}.\end{array}$$
2.3.12	(a) $(\hat{\Phi }(\xi /2^N)-1)\hat{f}\to 0$ in $\mathcal{S}$ for $f\in \mathcal{S}$ as $N\to \infty$ because for all multi-indices $\alpha ,\beta$,$$\begin{array}{rl}& \underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta }((\hat{\Phi }(\xi /2^N)-1)\hat{f})|\\ =& \underset{x\in {\mathbb{R}}^n}{\sup }|(\hat{\Phi }(\xi /2^N)-1)x^{\alpha }{\partial }^{\beta }\hat{f}+\sum _{\gamma \le \beta ,\gamma \ne 0}|{\partial }^{\gamma }((\hat{\Phi }(\xi /2^N)-1)x^{\alpha }{\partial }^{\beta -\gamma }\hat{f}|\\ \le & \Vert \hat{\Phi }{\Vert }_{L^{\infty }}\underset{|x|>2^N}{\sup }|x^{\alpha }{\partial }^{\beta }\hat{f}|+\sum _{\gamma \le \beta ,\gamma \ne 0}\frac{\Vert {\partial }^{\gamma }\hat{\Phi }{\Vert }_{L^{\infty }}}{2^{N|\gamma |}}\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha }{\partial }^{\beta -\gamma }\hat{f}|,\end{array}$$and for the same reason as Exercise 2.3.5, $$\underset{|x|>2^N}{\sup }|x^{\alpha }{\partial }^{\beta }\hat{f}|\le \underset{1\le i\le n}{\max }\frac{\underset{x\in {\mathbb{R}}^n}{\sup }|x^{\alpha +1_i}{\partial }^{\beta }\hat{f}|}{2^N/\sqrt{n}}.$$Thus for all $f\in \mathcal{S}$,$$\begin{array}{rl}|⟨(\hat{\Phi }(\psi /2^N)\hat{u}{)}^{\vee }-u,f⟩|=& |⟨\hat{\Phi }(\psi /2^N)\hat{u}-\hat{u},\hat{f}⟩|\\ =& |⟨\hat{u},(\hat{\Phi }(\psi /2^N)-1)\hat{f}⟩|\to 0\text{as }N\to \infty .\end{array}$$ (b) Suppose that $\hat{\Theta }$ is equal to $0$ in $B(0,r)$, then $1-\hat{\Theta }$ is equal to $1$ in $B(0,r)$. Apply part (a), which is also true when $B(0,1)$ is replaced by $B(0,r)$, and get$$\left(\right(1-\hat{\Theta }(\xi /2^N))\hat{u}{)}^{\vee }\to u\text{in }{\mathcal{S}}^{\prime }\text{ as }N\to \infty ,$$namely,$$(\hat{\Theta }(\xi /2^N)\hat{u}{)}^{\vee }\to 0\text{in }{\mathcal{S}}^{\prime }\text{ as }N\to \infty .$$
2.3.13	Assume the converse. Then for all $f\in L^p({\mathbb{R}}^n)$, $\hat{f}$ is locally integrable and hence the map $f\to \hat{f}$ is a well defined operator from $L^p({\mathbb{R}}^n)$ to $L^1(B(0,M))$ for all $M>0$ (i.e. $\Vert \hat{f}{\Vert }_{L^1(B(0,M))}<\infty$ for all $f\in L^p({\mathbb{R}}^n)$, since $B(0,M)$ is covered by a finite number of neighborhoods of $x\in \overline{B(0,M)}$ on which $\hat{f}$ is integrable). Then we shall use the closed graph theorem to deduce that the operator is bounded, or in other words, $\Vert \hat{f}{\Vert }_{L^1(B(0,M))}\le C_M\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}$. Suppose $(f_n,\widehat{f_n})\to (f,g)$ in $L^p({\mathbb{R}}^n)\times L^1(B(0,M))$, then by Proposition 2.3.22, $f_n\to f$ in ${\mathcal{S}}^{\prime }$, so $\widehat{f_n}\to \hat{f}$ in ${\mathcal{S}}^{\prime }$. Note that $\widehat{f_n}\to g$ in $L^1(B(0,M))$ implies convergence in ${\mathcal{S}}^{\prime }$, so we can conclude that $g$ is indeed $\hat{f}$ once we check that: if two $L^1$ functions are identical as tempered distributions, then they are the same as functions. In fact, if for all $\phi \in \mathcal{S}$, ${\int }_{{\mathbb{R}}^n}\hat{f}\phi ={\int }_{{\mathbb{R}}^n}g\phi$, then so will be true for $\phi ={\chi }_{\{\hat{f}\ge g\}}-{\chi }_{\{\hat{f}<g\}}$, which is justified by finding a sequence of Schwartz functions all bounded by $1$ converging pointwise to $\phi$ and making use of the Lebesgue dominated convergence theorem — such Schwartz functions can be taken as a sequence of step functions ${\chi }_{A_n}-{\chi }_{B_n}$ mollified at the endpoints approximating $\phi$, where the $A_n$’s and $B_n$’s are each a finite combination of intervals contained in $\{\hat{f}\ge g\}$ and $\{\hat{f}<g\}$ with $m(\{\hat{f}\ge g\}\backslash A_n)\to 0$ and $m(\{\hat{f}<g\}\backslash B_n)\to 0$. Then we get $\hat{f}=g$ in the usual sense of functions. To violate the previous inequality $\Vert \hat{f}{\Vert }_{L^1(B(0,M))}\le C_M\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}$ whenever $p>2$, take $f_N(x)=t^{-n}{e}^{-\pi t^{-2}|x{|}^2}$ where $t=(1+iN{)}^{1/2}$. Then for $\mathrm{Re}(t^{-2})=\frac{1}{1+N^2}=|t{|}^{-4}$,$$\Vert f_N{\Vert }_{L^p({\mathbb{R}}^n)}=|t{|}^{-n}({\int }_{{\mathbb{R}}^n}{e}^{-\pi (|t{|}^{-2}|x|{)}^{2p}}dx{)}^{\frac{1}{p}}=|t{|}^{(-1+\frac{2}{p})n}\Vert e^{-\pi |x{|}^2}{\Vert }_{L^p({\mathbb{R}}^n)}.$$ And $\widehat{f_N}(\xi )=(t^{-n}{\delta }^{t^{-1}}(e^{-\pi |x{|}^2}){)}^{\wedge }(\xi )=({\delta }^{t}f)(\xi )=e^{-\pi |\xi {|}^2(1+iN)}$ using Proposition 2.2.11 and Example 2.2.9, so $\Vert \widehat{f_N}{\Vert }_{L^1(B(0,M))}=\Vert e^{-\pi |\xi {|}^2}{\Vert }_{L^1(B(0,M))}$ is a constant. Therefore the inequality cannot be true, for $|t{|}^{(-1+\frac{2}{p})n}\to 0$ as $N\to \infty$.