Solution to GTM249 Chapter 2 Part II

2.4 More About Distributions and the Fourier Transform

2.4.1	Any sequence of functions$$g(x)\frac{e^{-2\pi ix\cdot (\xi +h_n{e}_j)}-e^{-2\pi ix\cdot \xi }}{h_n}$$where $e_j$ is the unit vector with $1$ in the $j$th entry and $h_n\to 0$ as $n\to \infty$ converges pointwise to$$g(x)(-2\pi i{x}_j)e^{-2\pi ix\cdot \xi },$$and is bounded by (use $|e^{ia}-e^{ib}|\le |a-b|$)$$|g(x)(-2\pi i{x}_j)|,$$which is expected to be integrable. Under such assumption, it follows from the Lebesgue dominated convergence theorem that$${\int }_{{\mathbb{R}}^n}g(x)\frac{e^{-2\pi ix\cdot (\xi +h_n{e}_j)}-e^{-2\pi ix\cdot \xi }}{h_n}dx\to {\int }_{{\mathbb{R}}^n}g(x)(-2\pi i{x}_j)e^{-2\pi ix\cdot \xi }dx.$$In other words,$$\frac{\partial \hat{g}}{\partial {\xi }_j}(\xi )={\int }_{{\mathbb{R}}^n}g(x)(-2\pi i{x}_j)e^{-2\pi ix\cdot \xi }dx.$$ Take $g(x)=f(x)(-2\pi ix{)}^{\alpha }$ for all multi–indices $\alpha$ with sizes $|\alpha |=0,\cdots ,M-1$ consecutively, which all satisfy the integrability hypothesis since$$|g(x)(-2\pi i{x}_j)|=|f(x)(-2\pi ix{)}^{\alpha +e_j}|\le \frac{(2\pi {)}^{|\alpha |+1}C}{(1+|x|{)}^{N-(|\alpha |+1)}}$$where $N-(|\alpha |+1)>(M+n)-M=n$. Now we get the expression for ${\hat{f}}^{(M)}$, and therefore $\hat{f}\in {\mathcal{C}}^M$.
2.4.2	A bounded harmonic function, which is surely a tempered distribution, must be a polynomial by Corollary 2.4.3, and is a constant as all polynomials tend to $\infty$ as $x\to \infty$. Suppose that $p(z)$ is a polynomial on $\mathbb{C}$ without any complex root. Then the entire function $\frac{1}{p(z)}$, which is a complex–valued harmonic function on ${\mathbb{R}}^2$, is bounded as it tends to $0$ as $z\to \infty$, and is therefore a constant by Liouville’s theorem for harmonic functions, yielding a contradiction.
2.4.3	$e^x$ is not in ${\mathcal{S}}^{\prime }(\mathbb{R})$, as $e^{-x}\eta \in \mathcal{S}(\mathbb{R})$ where $\eta$ is a smooth function that equals $1$ for $x>0$ and vanishes for $x<-1$, but $⟨e^x,e^{-x}\eta ⟩=\int \eta$ does not converge. For all $f\in \mathcal{S}(\mathbb{R})$,$$|⟨e^x{e}^{i{e}^x},f⟩|=|⟨e^{i{e}^x}/i,f^{\prime }⟩|\le \Vert f^{\prime }{\Vert }_{L^1}\le C\sum _{\alpha \le 3}{\rho }_{\alpha ,1}(f)$$by Proposition 2.2.6, so $e^x{e}^{i{e}^x}$ is in ${\mathcal{S}}^{\prime }(\mathbb{R})$ by Proposition 2.3.4 (b).
2.4.4	Recall that $\mathrm{sech}(x)=\frac{1}{\cosh (x)}=\frac{2}{e^x+e^{-x}}$. We need to show that$$I={\int }_{\mathbb{R}}\frac{2}{e^{\pi x}+e^{-\pi x}}{e}^{-2\pi ix\xi }dx=\frac{2}{e^{\pi \xi }+e^{-\pi \xi }}.$$Let$$f(z)=\frac{2}{e^{\pi z}+e^{-\pi z}}{e}^{-2\pi iz\xi }.$$Integrate $f(z)$ over the rectangle contour ${\gamma }_R$ with corners $(-R,0)$, $(R,0)$, $(-R,i)$ and $(R,i)$, using that $f(z+i)=-e^{2\pi \xi }f(z)$ and that the integral of $f$ over the two vertical sides goes to $0$ as $R\to \infty$ since $|f(z)|\le \frac{2}{e^{\pi R}}{e}^{2\pi |\xi |}$ when $\mathrm{Re}z=\pm R$, we get$$(1-(-e^{2\pi \xi }))I=\underset{R\to \infty }{\lim }{\int }_{{\gamma }_R}f(z)dz=2\pi i\mathrm{Res}(f,\frac{i}{2}),$$and proceed our calculation as follows:$$\begin{array}{rl}I& =\frac{1}{1+e^{2\pi \xi }}2\pi i\underset{z\to \frac{i}{2}}{\lim }\frac{2(z-\frac{i}{2})}{e^{\pi z}+e^{-\pi z}}{e}^{-2\pi i\frac{i}{2}\xi }\\ & =\frac{2}{e^{\pi \xi }+e^{-\pi \xi }}2\pi i\frac{1}{(e^{\pi z}+e^{-\pi z}{)}^{\prime }{|}_{z=\frac{i}{2}}}\\ & =\frac{2}{e^{\pi \xi }+e^{-\pi \xi }}.\end{array}$$
2.4.5	Let $w$ be a smooth function the support of whose Fourier transform is $[-1/2,1/2]$ and let $\varphi$ be a real-valued smooth nonconstant periodic function with period $1$, which we can make further assumptions upon later. Then take a family of Schwartz functions$$f_a(x)=w(x)e^{i\varphi (x-a)}\text{for }a\in (0,1/3).$$Clearly it is an uncountable family of Schwartz functions such that $|f_a|=|f_b|$ for all $a$ and $b$, and we will show that the $f_a$’s are linearly independent, which will be addressed later, and that $|\widehat{f_a}|=|\widehat{f_b}|$. To this end, we expand the function $e^{i\varphi (x)}$ in the Fourier series$$e^{i\varphi (x)}=\sum _{k=-\infty }^{\infty }{c}_k{e}^{2\pi ikx}.$$Then$$f_a(x)=\sum _{k=-\infty }^{\infty }{c}_k{e}^{-2\pi ika}{e}^{2\pi ikx}w(x),$$so$$\widehat{f_a}(\xi )=\sum _{k=-\infty }^{\infty }{c}_k{e}^{-2\pi ika}\hat{w}(\xi -k).$$ Since the functions $\hat{w}(\xi -k)$ are supported in mutually disjoint intervals $[k-1/2,k+1/2)$, we have$$|\widehat{f_a}(\xi )|=|c_k\hat{w}(\xi -k(\xi ))|,$$where $k(\xi )$ is the integer such that $\xi -k(\xi )\in [-1/2,1/2)$, which shows that $|\widehat{f_a}(\xi )|$ is independent of $a$. Now we return to the linear independence of the $f_a$’s, under certain additional assumption on $\varphi$. Suppose that$$\sum _{k=1}^n{d}_k{e}^{i\varphi (x-a_k)}=0$$for some distinct $a_k\in (0,1/3)$. Let $\varphi (x)$ coincide with $\frac{1}{x}$ on $(1/3,1)$, then$$\sum _{k=1}^n{d}_k{e}^{\frac{i}{z-a_k}}=0$$by the uniqueness of meromorphic functions as the left side equals $0$ for $z\in (2/3,1)$ (hence $z-a_k\in (1/3,1)$). Taking $z=i\epsilon +a_k$ where $\epsilon \to 0^{+}$ gives $d_k=0$.
2.4.6	We are going to verify that $(P_y\ast f)(x)$ satisfies the mean value property and is continuous on ${\mathbb{R}}_{+}^{n+1}$. The two properties together imply that $f$ is harmonic. (Sketch: On any ball consider the difference of such a function and the harmonic function solving the Dirichlet problem where the boundary–value is given by its confinement, which supposedly attains its nonzero maximum or minimum on a closed set, at the farthest point away from the center of the ball of which the mean value property fails.) First establish the mean value property: for a ball $B$ in ${\mathbb{R}}_{+}^{n+1}$ centering at $(x_0,y_0)$,$$\begin{array}{rl}{-\int }_B(P_y\ast f)(x)d\bar{x}& ={-\int }_B{\int }_{{\mathbb{R}}^n}{P}_y(x-z)f(z)dzd\bar{x}\\ & ={\int }_{{\mathbb{R}}^n}({-\int }_B{P}_y(x-z)d\bar{x})f(z)dz\\ & ={\int }_{{\mathbb{R}}^n}{P}_{y_0}(x_0-z)f(z)dz\\ & =(P_{y_0}\ast f)(x_0),\end{array}$$where $\bar{x}=(x,y)$ and a slash through an integral denotes an average. Then check its continuity:$$\begin{array}{r}|(P_{y^{\prime }}\ast f)(x^{\prime })-(P_y\ast f)(x)|\le \Vert P_{y^{\prime }}(x^{\prime }-x+\cdot )-P_y(\cdot ){\Vert }_{L^{p^{\prime }}}\Vert f{\Vert }_{L^p}\to 0\end{array}$$as $(x^{\prime },y^{\prime })\to (x,y)$, where $\frac{1}{p}+\frac{1}{p^{\prime }}=1$ and $\Vert P_{y^{\prime }}(x^{\prime }-x+\cdot )-P_y(\cdot ){\Vert }_{L^{p^{\prime }}}\to 0$ is easily confirmed by, say, the Lebesgue dominated convergence theorem. To show that $P_{y_1}\ast P_{y_2}=P_{y_1+y_2}$, it suffices to show that their Fourier transforms agree. Indeed, applying Proposition 2.2.11 (8) and Exercise 2.2.11, $$\widehat{P_y}=(y^{-n}{\delta }^{y}P{)}^{\wedge }={\delta }^y\hat{P}=e^{-2\pi y|\xi |},$$so$$(P_{y_1}\ast P_{y_2}{)}^{\wedge }=\widehat{P_{y_1}}\widehat{P_{y_2}}=e^{-2\pi y_1|\xi |}{e}^{-2\pi y_2|\xi |}=e^{-2\pi (y_1+y_2)|\xi |}=\widehat{P_{y_1+y_2}}.$$
2.4.7	(a) Observe that $v(x;x_0)$ is harmonic is equivalent to $v(x+x_0;x_0)$ is harmonic, where$$\begin{array}{rl}v(x+x_0;x_0)& =(|x_0{|}^2-|x+x_0{|}^2)|x{|}^{-n}\\ & =-2x_0\cdot x|x{|}^{-n}-|x{|}^{2-n}\end{array}$$is a linear combination of the harmonic function $|x{|}^{2-n}$ and its partial derivatives $\frac{\partial }{\partial x_j}|x{|}^{2-n}=(2-n)x_j|x{|}^{-n}$ for $n\ge 3$, and $\frac{\partial }{\partial x_j}\log |x|=x_j|x{|}^{-2}$ for $n=2$. Alternatively, one may calculate directly as follows:$$\begin{array}{rl}\Delta v(x+x_0;x_0)& =\Delta ((1-|x+x_0{|}^2)\cdot |x{|}^{-n})\\ & =\Delta (1-|x+x_0{|}^2)|x{|}^{-n}+2\nabla (1-|x+x_0{|}^2)\cdot \nabla (|x{|}^{-n})+(1-|x+x_0{|}^2)\Delta |x{|}^{-n}\\ & =(-2n)|x{|}^{-n}+2(-2(x+x_0))\cdot (-n|x{|}^{-n-2}x)+(1-|x+x_0{|}^2)2n|x{|}^{-n-2}\\ & =2n|x{|}^{-n-2}(-|x{|}^2+2(x+x_0)\cdot x+|x_0{|}^2-|x+x_0{|}^2)\\ & =0,\end{array}$$in which the calculation of $\Delta |x{|}^{-n}$ deserves elaboration:$$\begin{array}{rl}\Delta |x{|}^{-n}& =\mathrm{div}(-nx|x{|}^{-n-2})=-n(\mathrm{div}(x)|x{|}^{-n-2}+x\cdot \nabla |x{|}^{-n-2})\\ & =-n(n|x{|}^{-n-2}+x\cdot (\frac{-n-2}{2}|x{|}^{-n-4}2x\left)\right)=2n|x{|}^{-n-2}.\end{array}$$ (b) For $\theta \in {\mathbf{S}}^{n-1}$ such that $|\theta -x_0|>\delta$, we have$$\begin{array}{r}v(r{x}_0;\theta )=\frac{1-r^2}{|r{x}_0-\theta {|}^n}=\frac{1-r^2}{|(r-1)x_0+(x_0-\theta ){|}^n}\le \frac{1-r^2}{(\delta -(1-r){)}^n}.\end{array}$$Then$$\begin{array}{r}{\int }_{\begin{array}{c}\theta \in {\mathbf{S}}^{n-1}\\ |\theta -x_0|>\delta \end{array}}v(r{x}_0;\theta )d\theta \le {\omega }_{n-1}\frac{1-r^2}{(\delta -(1-r){)}^n}\end{array}$$tends to $0$ uniformly in $x_0$ as $r\uparrow 1$. (c)$$\frac{1}{{\omega }_{n-1}}{\int }_{{\mathbf{S}}^{n-1}}\frac{1-|x{|}^2}{|x-\theta {|}^n}d\theta ={-\int }_{{\mathbf{S}}^{n-1}}\frac{1-||x|\theta {|}^2}{||x|\theta -\frac{x}{|x|}{|}^n}d\theta =1,$$where the first equality is justified by the illustration below, and the second by the mean value property of the integrand which is the harmonic function $\theta \mapsto v(|x|\theta ;\frac{x}{|x|})$ and $v(0;\frac{x}{|x|})=1$. (d) The harmonicity of $u$ is easily shown by taking the Laplacian inside the integral; the continuity of $u$ at the boundary is a consequence of the observation in part (b) that the Poisson kernel $v(r{x}_0;\theta )$ is massed at $\theta$ as $r\uparrow 1$ — to put it more clearly, by part (c) we have$$\begin{array}{rl}|u(r{x}_0)-f(x_0)|& =\left|{-\int }_{{\mathbf{S}}^{n-1}}\right(f(\theta )-f(x_0))v(r{x}_0;\theta )d\theta |\\ & \le \underset{|\theta -x_0|\le \delta }{\sup }|f(\theta )-f(x_0)|\cdot {-\int }_{{\mathbf{S}}^{n-1}}v(r{x}_0;\theta )d\theta \\ & +\underset{\theta \in {\mathbf{S}}^{n-1}}{\sup }|f(\theta )-f(x_0)|\cdot \frac{1}{{\omega }_{n-1}}{\int }_{\begin{array}{c}\theta \in {\mathbf{S}}^{n-1}\\ |\theta -x_0|>\delta \end{array}}v(r{x}_0;\theta )d\theta ,\end{array}$$first choose $\delta$ small enough so that the first term is vanishingly small by the continuity of $f$, and then for the fixed $\delta$, let $r\uparrow 1$ so that the second term becomes vanishingly small as well by part (b).
2.4.8	(a) In Appendix D.4 the following identity is obtained:$${\int }_{{\mathbf{S}}^{n-1}}\frac{d\theta }{|\theta -e_1{|}^{\alpha }}=\frac{{\omega }_{n-2}}{2^{\frac{\alpha }{2}}}{\int }_{-1}^{+1}(1-s{)}^{\frac{n-3-\alpha }{2}}(1+s{)}^{\frac{n-3}{2}}ds.$$ Since ${\mathbf{S}}^n$ is invariant under the rotation that carries $\eta$ to $e_1$, we get$$\begin{array}{rl}{\int }_{{\mathbf{S}}^n}|\xi -\eta {|}^{-\lambda }d\xi & ={\int }_{{\mathbf{S}}^n}|\xi -e_1{|}^{-\lambda }d\xi \\ & =\frac{{\omega }_{n-1}}{2^{\frac{\lambda }{2}}}{\int }_{-1}^{+1}(1-s{)}^{\frac{n-\lambda }{2}-1}(1+s{)}^{\frac{n}{2}-1}ds\\ & =\frac{{\omega }_{n-1}}{2^{\frac{\lambda }{2}}}{\int }_{-1}^{+1}(2-2t{)}^{\frac{n-\lambda }{2}-1}(2t{)}^{\frac{n}{2}-1}d(2t-1)\\ & ={\omega }_{n-1}{2}^{n-\lambda -1}{\int }_0^1(1-t{)}^{\frac{n-\lambda }{2}-1}{t}^{\frac{n}{2}-1}dt\\ & ={\omega }_{n-1}{2}^{n-\lambda -1}B(\frac{n-\lambda }{2},\frac{n}{2})\\ & =\frac{2{\pi }^{\frac{n}{2}}}{\Gamma (\frac{n}{2})}{2}^{n-\lambda -1}\frac{\Gamma (\frac{n-\lambda }{2})\Gamma (\frac{n}{2})}{\Gamma (n-\frac{\lambda }{2})}\\ & ={\pi }^{\frac{n}{2}}{2}^{n-\lambda }\frac{\Gamma (\frac{n-\lambda }{2})}{\Gamma (n-\frac{\lambda }{2})}.\end{array}$$ (b) Notice that$$\begin{array}{rl}& {\int }_{{\mathbb{R}}^n}|x-y{|}^{-\lambda }(1+|x{|}^2{)}^{\frac{\lambda }{2}}(1+|y{|}^2{)}^{\frac{\lambda }{2}}(1+|x{|}^2{)}^{-n}dx\\ =& {\int }_{{\mathbb{R}}^n}(|\Pi (x)-\Pi (y)|/2{)}^{-\lambda }{J}_{\Pi }(x)/2^{n}dx\\ =& 2^{\lambda -n}{\int }_{{\mathbf{S}}^{n-1}}|\xi -\eta {|}^{-\lambda }d\xi \\ =& \frac{{\pi }^{\frac{n}{2}}\Gamma (\frac{n-\lambda }{2})}{\Gamma (n-\frac{\lambda }{2})},\end{array}$$where $\Pi$ is the stereographic projection in Appendix D.6, in the first equality the so–called interesting formula $|\Pi (x)-\Pi (y)|=2|x-y|(1+|x{|}^2{)}^{-1/2}(1+|y{|}^2{)}^{-1/2}$ is used, and the last equality is by (a), from which we deduce that the Jacobian of $\Pi$ is $J_{\Pi }(x)=(\frac{2}{1+|x{|}^2}{)}^n$ (instead of $J_{\Pi }(x)=(\frac{2}{1+|x{|}^2}{)}^{-n}$ as Appendix D.6 claims). Now we give a proof of the formula for $|\Pi (x)-\Pi (y)|$. Observe that $△exy\backsim △e\Pi (y)\Pi (x)$ with a scale denoted by $1:k$ as$$|ex||e\Pi (x)|=|ey||e\Pi (y)|=|eO||e{e}^{\prime }|=2,$$where $e=(0,\cdots ,0,1)$ and $e^{\prime }=(0,\cdots ,0,-1)$ are the north pole and the south pole. Since $$2=|ex||e\Pi (x)|=k|ex||ey|=k(1+|x{|}^2{)}^{1/2}(1+|y{|}^2{)}^{1/2},$$we get$$|\Pi (x)\Pi (y)|=k|xy|=2|xy|(1+|x{|}^2{)}^{-1/2}(1+|y{|}^2{)}^{-1/2}.$$
2.4.9	Theorem 2.4.6 establishes the formula$${\pi }^{\frac{z+n}{2}}\Gamma (\frac{z+n}{2}{)}^{-1}(|x{|}^z{)}^{\wedge }={\pi }^{-\frac{z}{2}}\Gamma (-\frac{z}{2}{)}^{-1}|x{|}^{-n-z},$$ that is,$$(|x{|}^z{)}^{\wedge }={\pi }^{-z-\frac{n}{2}}\frac{\Gamma (\frac{z+n}{2})}{\Gamma (-\frac{z}{2})}|x{|}^{-n-z}\triangleq c_z|x{|}^{-n-z}.$$ Then$$\begin{array}{rl}& {\int }_{{\mathbb{R}}^n}\frac{dt}{|x-t{|}^{{\alpha }_1}|y-t{|}^{{\alpha }_2}}\\ =& (|t{|}^{-{\alpha }_1}\ast |t{|}^{-{\alpha }_2})(x-y)\\ =& ((|t{|}^{-{\alpha }_1}{)}^{\wedge }(|t{|}^{-{\alpha }_2}{)}^{\wedge }{)}^{\vee }(x-y)\\ =& c_{-{\alpha }_1}{c}_{-{\alpha }_2}(|t{|}^{-n+{\alpha }_1-n+{\alpha }_2}{)}^{\vee }(x-y)\\ =& c_{-{\alpha }_1}{c}_{-{\alpha }_2}{c}_{-2n+{\alpha }_1+{\alpha }_2}|x-y{|}^{n-{\alpha }_1-{\alpha }_2},\end{array}$$where$$\begin{array}{rl}& c_{-{\alpha }_1}{c}_{-{\alpha }_2}{c}_{-2n+{\alpha }_1+{\alpha }_2}\\ =& {\pi }^{{\alpha }_1-\frac{n}{2}}\frac{\Gamma (\frac{n-{\alpha }_1}{2})}{\Gamma (\frac{{\alpha }_1}{2})}{\pi }^{{\alpha }_2-\frac{n}{2}}\frac{\Gamma (\frac{n-{\alpha }_2}{2})}{\Gamma (\frac{{\alpha }_2}{2})}{\pi }^{2n-{\alpha }_1-{\alpha }_2-\frac{n}{2}}\frac{\Gamma (\frac{n-2n+{\alpha }_1+{\alpha }_2}{2})}{\Gamma (-\frac{-2n+{\alpha }_1+{\alpha }_2}{2})}\\ =& {\pi }^{\frac{n}{2}}\frac{\Gamma (\frac{n-{\alpha }_1}{2})\Gamma (\frac{n-{\alpha }_2}{2})\Gamma (\frac{{\alpha }_1+{\alpha }_2-n}{2})}{\Gamma (\frac{{\alpha }_1}{2})\Gamma (\frac{{\alpha }_2}{2})\Gamma (n-\frac{{\alpha }_1+{\alpha }_2}{2})}.\end{array}$$
2.4.10	(a) This is because for all rotations $A$ and $\xi \in {\mathbb{R}}^n$, by Proposition 2.2.11 (13),$$\hat{f}(A\xi )=\widehat{f\circ A}(\xi )=\hat{f}(\xi ).$$ (b) Now the identity above holds for all rotations $A$ which keep $e_1$ fixed.
2.4.11	The translation $(x,y)\mapsto (x+z,y+z)$ and a dilation with a rotation that carries $y$ to $e_1$ reduce matters to the case that $z=0$ and $y=e_1$. We prove the identity by showing that the Fourier transforms in $\mathbf{x}$ of both sides coincide. First take the Fourier transform on the right side, setting aside its coefficient, and get$$\begin{array}{rl}& (|x-e_1{|}^{d_1-n}|x{|}^{d_3-n}{)}^{\wedge }(\xi )\\ =& c_{d_1-n}{c}_{d_3-n}(|\xi {|}^{-d_1}{e}^{-2\pi i\xi e_1})\ast |\xi {|}^{-d_3}\\ =& c_{d_1-n}{c}_{d_3-n}{\int }_{{\mathbb{R}}^n}|t{|}^{-d_1}{e}^{-2\pi it\cdot e_1}|\xi -t{|}^{-d_3}dt\\ =& c_{d_1-n}{c}_{d_3-n}|\xi {|}^{-d_1-d_3+n}{\int }_{{\mathbb{R}}^n}|t{|}^{-d_1}{e}^{-2\pi i|\xi |t\cdot e_1}|\xi /|\xi |-t{|}^{-d_3}dt.\end{array}$$Let $A_{\xi }$ be a rotation that transforms $e_1$ to $\xi /|\xi |$, then the expression above is equal to$$\begin{array}{rl}=& c_{d_1-n}{c}_{d_3-n}|\xi {|}^{-d_1-d_3+n}{\int }_{{\mathbb{R}}^n}|A_{\xi }^{-1}t{|}^{-d_1}{e}^{-2\pi i|\xi |A_{\xi }^{-1}t\cdot A_{\xi }^{-1}{e}_1}|e_1-A_{\xi }^{-1}t{|}^{-d_3}d(A_{\xi }^{-1}t)\\ =& c_{d_1-n}{c}_{d_3-n}|\xi {|}^{-d_1-d_3+n}\hat{h}(A_{\xi }^{-1}{e}_1),\end{array}$$where $h(t)=|t{|}^{-d_1}|t-e_1{|}^{-d_3}$. Next take the Fourier transform on the left side, and get$$(h\ast |t{|}^{-d_2}{)}^{\wedge }(\xi )=c_{-d_2}|\xi {|}^{n-d_2}\hat{h}(\xi ).$$ Since $-d_1-d_3+n=n-d_2$ and $$\begin{array}{rl}& c_{-d_2}/(c_{d_1-n}{c}_{d_3-n})\\ =& {\pi }^{d_2-\frac{n}{2}}\frac{\Gamma (\frac{n-d_2}{2})}{\Gamma (\frac{d_2}{2})}/\left({\pi }^{\frac{n}{2}-d_1}\frac{\Gamma (\frac{d_1}{2})}{\Gamma (\frac{n-d_1}{2})}{\pi }^{\frac{n}{2}-d_3}\frac{\Gamma (\frac{d_3}{2})}{\Gamma (\frac{n-d_3}{2})}\right)\\ =& {\pi }^{\frac{n}{2}}\prod _{j=1}^3\frac{\Gamma (\frac{n-d_j}{2})}{\Gamma (\frac{d_j}{2})},\end{array}$$it remains to show that $\hat{h}(\xi )=\hat{h}(A_{\xi }^{-1}{e}_1)$, which explains what part (b) of the preceding exercise is for.
2.4.12	(a) On $P_2$ we have $\left|e^{i(R{e}^{i\theta }{)}^2}\right|=e^{-R^2\sin (2\theta )}\le e^{-\frac{4}{\pi }{R}^2\theta }$, and hence$$\left|{\int }_{P_2}{e}^{i{z}^2}dz\right|\le {\int }_0^{\frac{\pi }{4}}{e}^{-\frac{4}{\pi }{R}^2\theta }|R{e}^{i\theta }d\theta |\le \frac{1}{R}{\int }_0^{\infty }{e}^{-\frac{4}{\pi }{R}^2\theta }d(R^2\theta )\to 0\text{as }R\to \infty .$$Then we get from ${\int }_{P_1\cup P_2\cup P_3}{e}^{i{z}^2}dz=0$ that$$\begin{array}{rl}\underset{R\to \infty }{\lim }{\int }_0^R{e}^{i{x}^2}dx& =\underset{R\to \infty }{\lim }{\int }_{P_1}{e}^{i{z}^2}dz=-\underset{R\to \infty }{\lim }{\int }_{P_3}{e}^{i{z}^2}dz\\ & =-\underset{R\to \infty }{\lim }{\int }_0^R{e}^{-r^2}{e}^{i\frac{\pi }{4}}dr=e^{i\frac{\pi }{4}}\frac{\sqrt{\pi }}{2}=\frac{\sqrt{2\pi }}{4}(1+i).\end{array}$$ (b) Enough to deal with the case of $n=1$, and the general case is its mere multiplication. When $n=1$,$$\begin{array}{rl}{\int }_{\mathbb{R}}{e}^{i\pi x^2}{e}^{-2\pi ix\xi }dx& =e^{-i\pi {\xi }^2}{\int }_{\mathbb{R}}{e}^{i\pi (x-\xi {)}^2}dx\\ & =e^{-i\pi {\xi }^2}2{\int }_0^{\infty }{e}^{i\pi x^2}dx\\ & =e^{-i\pi {\xi }^2}{e}^{i\frac{\pi }{4}}.\end{array}$$

2.5 Convolution Operators on $L^p$ Spaces and Multipliers

2.5.1	By the Lebesgue dominated convergence theorem, $\forall \epsilon >0,\exists N$ s.t.$$\Vert f-f{\chi }_{B(0,N)}{\Vert }_{L^q}<\epsilon .$$Let $g=f{\chi }_{B(0,N)}$. Then for $|h|>2N$, the supports of ${\tau }^h(g)$ and $g$ are disjoint, so$$\begin{array}{rl}|\Vert {\tau }^h(f)+f{\Vert }_{L^q}-2^{\frac{1}{q}}\Vert f{\Vert }_{L^q}|& =|\Vert {\tau }^h(f)+f{\Vert }_{L^q}-\Vert {\tau }^h(g)+g{\Vert }_{L^q}-2^{\frac{1}{q}}(\Vert f{\Vert }_{L^q}-\Vert g{\Vert }_{L^q})|\\ & \le \Vert {\tau }^h(f-g)+(f-g){\Vert }_{L^q}+2^{\frac{1}{q}}|\Vert f{\Vert }_{L^q}-\Vert g{\Vert }_{L^q}|\\ & \le \Vert {\tau }^h(f-g){\Vert }_{L^q}+\Vert (f-g){\Vert }_{L^q}+2^{\frac{1}{q}}\Vert f-g{\Vert }_{L^q}\\ & \le (2+2^{\frac{1}{q}})\epsilon .\end{array}$$Let $\epsilon$ tend to $0$ to come to the required conclusion.
2.5.2	To verify Proposition 2.5.14, we have$$\begin{array}{rl}& \Vert {\tau }^x(m){\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert ({\tau }^x({\tau }^{-x}(\hat{f})m){)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\sup \frac{\Vert (\widehat{e^{-2\pi i(\cdot )\cdot x}f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert e^{-2\pi i(\cdot )\cdot x}f{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\Vert m{\Vert }_{{\mathcal{M}}_p}.\\ & \Vert {\delta }^h(m){\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert ({\delta }^h({\delta }^{h^{-1}}(\hat{f})m){)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\sup \frac{\Vert (\widehat{{\delta }^{h}f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert {\delta }^{h}f{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\Vert m{\Vert }_{{\mathcal{M}}_p}.\\ & \Vert \tilde{m}{\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert (\hat{f}\tilde{m}{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{\tilde{f}}m{)}^{\vee }{\Vert }_{L^p}}{\Vert \tilde{f}{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\Vert m{\Vert }_{{\mathcal{M}}_p}.\\ & \Vert e^{2\pi i(\cdot )\cdot x}m{\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert (\hat{f}{e}^{2\pi i(\cdot )\cdot x}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\sup \frac{\Vert (\widehat{{\tau }^{-x}f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert {\tau }^{-x}f{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\Vert m{\Vert }_{{\mathcal{M}}_p}.\\ & \Vert m\circ A{\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert (\widehat{f\circ A^{-1}}\circ A\cdot m\circ A{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\sup \frac{\Vert (\widehat{f\circ A^{-1}}m{)}^{\vee }\circ A{\Vert }_{L^p}}{\Vert f\circ A^{-1}{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\Vert m{\Vert }_{{\mathcal{M}}_p}.\end{array}$$Note that ${\delta }_j^n\hat{f}(\xi )=\frac{1}{h}\widehat{{\delta }_j^{h^{-1}}}$, then the proof is similar with that of ${\delta }^h$.
2.5.3	(a)$$\Vert \psi m{\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert (\hat{f}\psi m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\sup \frac{\Vert (\hat{f}m{)}^{\vee }\ast {\psi }^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}.$$By Young’s inequality,$$\Vert \psi m{\Vert }_{{\mathcal{M}}_p}\le \sup \frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}\Vert {\psi }^{\vee }{\Vert }_{L^1}=\Vert m{\Vert }_{{\mathcal{M}}_p}\Vert {\psi }^{\vee }{\Vert }_{L^1}.$$(b) Since $m$ is essentially bounded, by Fubini’s theorem, $$\begin{array}{rl}{T}_{\psi \ast m}(f)(x)& ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\psi (y)m(\xi -y)\mathrm{d}y\hat{f}(\xi )e^{2\pi ix\cdot \xi }\mathrm{d}\xi \\ & ={\int }_{{\mathbb{R}}^n}\psi (y){\int }_{{\mathbb{R}}^n}m(\xi -y)\hat{f}(\xi )e^{2\pi ix\cdot \xi }\mathrm{d}\xi \mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^n}\psi (y)(m(\cdot -y)\hat{f}(\cdot ){)}^{\vee }(x)\mathrm{d}y.\end{array}$$Therefore $$\begin{array}{rl}\Vert T_{\psi \ast m}(f){\Vert }_{L^p}& \le {\int }_{{\mathbb{R}}^n}|\psi (y)|\Vert ({\tau }^{y}m\hat{f}{)}^{\vee }{\Vert }_{L^p}\mathrm{d}y\\ & \le {\int }_{{\mathbb{R}}^n}\left|\psi (y)\right|\Vert m{\Vert }_{{\mathcal{M}}_p}\Vert f{\Vert }_{L^p}\mathrm{d}y\\ & =\Vert \psi {\Vert }_{L^1}\Vert m{\Vert }_{{\mathcal{M}}_p}\Vert f{\Vert }_{L^p},\end{array}$$where in the last inequality the first property of Proposition 2.5.14 is used.
2.5.4	(a) Notice that $\widehat{T(f)}=\widehat{f\ast {\partial }^{\gamma }{\delta }_0}=\hat{f}\widehat{{\partial }^{\gamma }{\delta }_0}=\hat{f}(2\pi ix{)}^{\gamma }$, which shows that the operator $\mathcal{F}\circ T$ is continuous as it is bounded. But $\mathcal{F}$ is a homeomorphism on $\mathcal{S}$, which implies that $T$ is continuous. (b) When $p>q$, ${\mathcal{M}}_{p,q}=0$, implying that there is no nontrivial element. When $q\ge p$, for $f\in \mathcal{S}$,$$\frac{\Vert T({\delta }^{\lambda }f){\Vert }_{L^q}}{\Vert {\delta }^{\lambda }f{\Vert }_{L^p}}=\frac{\Vert {\delta }^{\lambda }f\ast {\partial }^{\gamma }{\delta }_0{\Vert }_{L^q}}{\Vert {\delta }^{\lambda }f{\Vert }_{L^q}}=\frac{\Vert {\partial }^{\gamma }{\delta }^{\lambda }f\ast {\delta }_0{\Vert }_{L^q}}{\Vert {\delta }^{\lambda }f{\Vert }_{L^q}}=\frac{|\lambda {|}^{|\gamma |}|\lambda {|}^{\frac{n}{q}}}{|\lambda {|}^{\frac{n}{p}}}\frac{\Vert T(f){\Vert }_{L^q}}{\Vert f{\Vert }_{L^p}}=|\lambda {|}^{|\gamma |+\frac{n}{q}-\frac{n}{p}}\frac{\Vert T(f){\Vert }_{L^q}}{\Vert f{\Vert }_{L^p}}.$$By the condition, the coefficient $|\lambda {|}^{|\gamma |+\frac{n}{q}-\frac{n}{p}}$ is unbounded when $\lambda$ varies.
2.5.5	Let $r=\frac{n}{n-\gamma }$. From Theorem 1.4.25,$$\frac{\Vert f\ast K_{\gamma }{\Vert }_{L^q}}{\Vert f{\Vert }_{L^p}}\le B\Vert K_{\gamma }{\Vert }_{L^{r,\infty }}=B{\nu }_n.$$
2.5.6	(a) We only need to consider the case of $1\le p\le 2$. Note that $\Vert m_j{\Vert }_{L^{\infty }}\le \Vert m_j{\Vert }_{{\mathcal{M}}_p}\le C.$ By the Lebsegue dominated convergence theorem,$$T_{m_j}(f)(x)={\int }_{{\mathbb{R}}^n}\hat{f}(\xi )m_j(\xi )e^{2\pi ix\cdot \xi }d\xi \to {\int }_{{\mathbb{R}}^n}\hat{f}(\xi )m(\xi )e^{2\pi ix\cdot \xi }d\xi =T_m(f)(x).$$By Fatou’s lemma,$${\int }_{{\mathbb{R}}^n}|T_m(f){|}^{p}dx\le \underset{j\to \infty }{\mathrm{liminf}}{\int }_{{\mathbb{R}}^n}|T_{m_j}(f){|}^{p}dx\le \underset{j\to \infty }{\mathrm{liminf}}\Vert m_j{\Vert }_{{\mathcal{M}}_p}^p\Vert f{\Vert }_{L^p}^p.$$Therefore, $\Vert m{\Vert }_{{\mathcal{M}}_p}\le \underset{j\to \infty }{\mathrm{liminf}}\Vert m_j{\Vert }_{{\mathcal{M}}_p}\le C.$ (b) Let $m_j={\delta }^{\frac{1}{j}}m,j\ge 1$, then $m_j\to m_0$. Note that $\Vert m_j{\Vert }_{{\mathcal{M}}_p}=\Vert m{\Vert }_{{\mathcal{M}}_p}$, so by part (a), $\Vert m_0{\Vert }_{{\mathcal{M}}_p}\le \Vert m{\Vert }_{{\mathcal{M}}_p}.$ (c) Obviously, $m_0(x)=\{\begin{array}{l}m(0+),x>0,\\ m(0-),x<0.\end{array}$ Therefore,$$\Vert m_0{\Vert }_{{\mathcal{M}}_p}\ge \Vert m_0{\Vert }_{{\mathcal{M}}_2}=\Vert m_0{\Vert }_{L^{\infty }}=\max (|m(0+)|,|m(0-)|).$$ (d) By Fubini’s theorem and Minkowski’s inequality, we see that$$\Vert m{\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert {\int }_0^{\infty }(m_t\hat{f}{)}^{\vee }\frac{dt}{t}{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}\le {\int }_0^{\infty }\Vert m_t{\Vert }_{{\mathcal{M}}_p}\frac{dt}{t}.$$
2.5.7	$$\Vert f{\Vert }_{L^p}=\Vert (\hat{f}{m}^{-1}m{)}^{\vee }{\Vert }_{L^p}\le \Vert m{\Vert }_{{\mathcal{M}}_p}\Vert (\hat{f}{m}^{-1}{)}^{\vee }{\Vert }_{L^p}=\Vert m{\Vert }_{{\mathcal{M}}_p}\Vert T(f){\Vert }_{L^p}.$$The lower bound for $|m|$ is to guarantee that $\hat{f}{m}^{-1}\in L^1({\mathbb{R}}^n)$ so that $(\hat{f}{m}^{-1}{)}^{\vee }$ is indeed a function.
2.5.8	(a) Using Exercise 1.2.9, we have$$\Vert m{\Vert }_{{\mathcal{M}}_p}=\sup \frac{\Vert f\ast m^{\vee }{\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}=\Vert m^{\vee }{\Vert }_{L^1}.$$(b)$$(m_2{m}_1^{-1}{)}^{\vee }(\xi )={\int }_{-1}^1(1-|x|)e^{-2\pi ix\xi }dx=\frac{1}{2(\pi \xi {)}^2}(\cos (2\pi \xi )-1)\le 0.$$By part (a),$$\Vert m_2{m}_1^{-1}{\Vert }_{{\mathcal{M}}_p}=\Vert (m_2{m}_1^{-1}{)}^{\vee }{\Vert }_{L^1}={\int }_{\mathbb{R}}\frac{1-\cos x}{\pi x^2}dx.$$Consider the integral of $f(z)=\frac{e^{iz}-1}{z^2}$ along the path below, which is the boundary of half of an annulus. Since $e^{iz}=1+iz+O(|z{|}^2)$ near $0$, $${\int }_{{\gamma }_2}\frac{e^{iz}-1}{\pi z^2}dz={\int }_{{\gamma }_2}\frac{i}{\pi z}+O(1)dz=\frac{i}{\pi }(-i\pi )+O(r)=1+O(r).$$Since $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ on the upper half–plane,$$\left|{\int }_{{\gamma }_4}\frac{e^{iz}-1}{\pi z^2}dz\right|\le {\int }_{{\gamma }_4}\frac{|e^{iz}|+1}{\pi |z{|}^2}|dz|\le {\int }_{{\gamma }_4}\frac{2}{\pi R^2}|dz|=\frac{2}{R}.$$As $f$ has no singularity in the domain enclosed, we have$${\int }_{\mathbb{R}}\frac{1-\cos x}{\pi x^2}dx=-\mathrm{Re}\underset{\begin{array}{c}r\to 0,\\ R\to \infty \end{array}}{\lim }{\int }_{{\gamma }_1\cup {\gamma }_3}\frac{e^{iz}-1}{\pi z^2}dz=\mathrm{Re}\underset{\begin{array}{c}r\to 0,\\ R\to \infty \end{array}}{\lim }{\int }_{{\gamma }_2\cup {\gamma }_4}\frac{e^{iz}-1}{\pi z^2}dz=1.$$Hence $\Vert m_2{m}_1^{-1}{\Vert }_{{\mathcal{M}}_p}=1$. Then$$\Vert m_2{\Vert }_{{\mathcal{M}}_p}\le \Vert m_2{m}_1^{-1}{\Vert }_{{\mathcal{M}}_p}\Vert m_1{\Vert }_{{\mathcal{M}}_p}=\Vert m_1{\Vert }_{{\mathcal{M}}_p}.$$Note that $m_1(\xi )=\underset{R\to \infty }{\lim }{m}_2\left(\frac{\xi }{R}\right)$, then by Exercise 2.5.6 (b), $$\Vert m_1{\Vert }_{{\mathcal{M}}_p}\le \Vert m_2{\Vert }_{{\mathcal{M}}_p}.$$
2.5.9	$$\Vert (\sum _m{a}_m{e}^{-2\pi imx}\hat{f}(x){)}^{\vee }{\Vert }_{L^p}=\Vert (\sum _m{a}_m\widehat{{\tau }^{m}f}{)}^{\vee }{\Vert }_{L^p}=\Vert \sum _m{a}_m{\tau }^{m}f{\Vert }_{L^p}=\Vert \sum _m{a}_{m}f(x-m){\Vert }_{L^p},$$from which we know (a) is equivalent to (b). For a sequence $\{a_m\}$ that satisfies (c), it is easy to see $\{a_m\}\in l^p({\mathbb{Z}}^n)$. Let $K$ be the operator$$L^p({\mathbb{R}}^n)\to L^p({\mathbb{R}}^n),f\mapsto \sum _m{a}_{m}f(x-m).$$Write the cube $Q=[-\frac{1}{2},\frac{1}{2}{]}^n$, $|Q|=1$. Note that$$\Vert f{\Vert }_{L^p}^p={\int }_Q\sum _n|f(x+n){|}^{p}dx,$$so $\{f(x-m){\}}_{m\in {\mathbb{Z}}^n}\in l^p({\mathbb{Z}}^n),a.e.x$, so $Kf$ is defined almost everywhere.$${\int }_{{\mathbb{R}}^n}|Kf{|}^{p}dx={\int }_Q\sum _n|\sum _m{a}_{m}f(a+n-m){|}^{p}dx\le {\int }_Q{A}^p\sum _n|f(x+n){|}^{p}dx=A^p\Vert f{\Vert }_{L^p}^p.$$This implies $\Vert K\Vert \le A.$ For a sequence $\{c_m\}$, take $f(x)=\sum _m{c}_m{\chi }_Q(x-m)$. Then $$\begin{array}{rl}& \frac{\left(\sum _n\right|\sum _m{a}_m{c}_{n-m}{|}^p{)}^{\frac{1}{p}}}{(\sum _n|c_n{|}^p{)}^{\frac{1}{p}}}=\frac{\left(\sum _n\right|{\int }_{n+Q}\sum _m{a}_{m}f(x-m)dx{|}^p{)}^{\frac{1}{p}}}{(\sum _n|c_n{|}^p{)}^{\frac{1}{p}}}\\ \le & \frac{\left(\sum _n{\int }_{n+Q}\right|\sum _m{a}_{m}f(x-m){|}^{p}dx{)}^{\frac{1}{p}}}{\Vert f{\Vert }_{L^p}}=\frac{\Vert \sum _m{a}_{m}f(x-m){\Vert }_{L^p}}{\Vert f{\Vert }_{L^p}}\le \Vert K\Vert .\end{array}$$Here we use Hölder’s inequality. This implies $A\le \Vert K\Vert$. So (a) is equivalent to (c).
2.5.10	Let $Q=[-\frac{1}{2},\frac{1}{2}{]}^n$, by the invariance of norm under translation, we can assume that $m$ is supported in $Q$. Since $m$ is in $L^{\infty }$ and is compactly supported, it lies in $L^2$ and hence it has a Fourier expansion. Write$$m(\xi )=\sum _{j\in {\mathbb{Z}}^n}{a}_j{e}^{2\pi ij\cdot \xi },\xi \in [0,1{]}^n.$$Then$$M(\xi )=\sum _{j\in {\mathbb{Z}}^n}{a}_j{e}^{2\pi ij\cdot \xi },\xi \in {\mathbb{R}}^n.$$Let $h\in C^{\infty }$ to be determined later. Define the operators$$\begin{array}{rl}E(\{c_j{\}}_j)& =\sum _{k\in {\mathbb{Z}}^n}{c}_k{\chi }_Q(x+k),\\ R(f)(j)& ={\int }_{-j+Q}f(x)\mathrm{d}x.\end{array}$$Consider $$\begin{array}{rl}R{T}_{mh}E(\{c_k{\}}_k)(j)& =R{T}_{mh}(\sum _{k\in {\mathbb{Z}}^n}{c}_k{\chi }_Q(\cdot +k))(j)\\ & =R\left(\right(mh\sum _{k\in {\mathbb{Z}}^n}{c}_k\widehat{{\chi }_Q}{e}^{2\pi i(\cdot )\cdot k}{)}^{\vee })(j)\\ & =\sum _{k\in {\mathbb{Z}}^n}{c}_k{q}_{j-k},\end{array}$$where $$\begin{array}{rl}{q}_j& ={\int }_Q(mh\widehat{{\chi }_Q}{)}^{\vee }(x-j)\mathrm{d}x\\ & ={\int }_Q{\int }_{Q}m(\xi )h(\xi )\widehat{{\chi }_Q}(\xi )e^{2\pi i(x-j)\cdot \xi }\mathrm{d}\xi \mathrm{d}x\\ & =\sum _{k\in {\mathbb{Z}}^n}{a}_k{\int }_Q(\widehat{{\chi }_Q}{)}^2(\xi )h(\xi )e^{2\pi i(k-j)\cdot \xi }\mathrm{d}\xi .\end{array}$$(The exchange of sum and integral is from the convergence of the series in $L^2$.) Thus $q_j=a_j$ if we pick a $C_c^{\infty }({\mathbb{R}}^n)$ function $h$ equal to $$\begin{array}{rl}(\widehat{{\chi }_Q}{)}^{-2}(\xi )& =\prod _{j=1}^n({\int }_{-1/2}^{1/2}{e}^{-2\pi i{x}_j{\xi }_j}d{x}_j{)}^{-2}\\ & =\prod _{j=1}^n(\frac{e^{-\pi i{\xi }_j}-e^{\pi i{\xi }_j}}{-2\pi i{\xi }_j}{)}^{-2}\\ & =\prod _{j=1}^n{\pi }^2{\xi }_j^2{\sin }^{-2}(\pi {\xi }_j)\end{array}$$on $Q$. Therefore the operator $R{T}_{mh}E$ is given by convolution with the sequence $\{a_j{\}}_j$. It is obvious that $\Vert E{\Vert }_{l^p\to L^p}\le 1$, and by Hölder’s inequality, $\Vert R{\Vert }_{L^p\to l^p}\le 1$. So we have$$\Vert R{T}_{mh}E{\Vert }_{l^p\to l^p}\le \Vert T_{mh}{\Vert }_{L^p\to L^p}\le \Vert h^{\vee }{\Vert }_{L^1}\Vert m{\Vert }_{{\mathcal{M}}_p}.$$The last inequality comes from Exercise 2.5.3, and the required conclusion comes from Exercise 2.5.9.
2.5.11	By Proposition 2.4.8, $\hat{u}\in C^{\infty }({\mathbb{R}}^n\backslash \{0\})$, and $\hat{u}$ is homogenous of degree $-i\tau$. Let $m=\hat{u}$, then$$\Vert m{\Vert }_{L^{\infty }({\mathbb{R}}^n\backslash \{0\})}=\Vert m{\Vert }_{L^{\infty }(\partial B(0,1))}<\infty .$$So$$\frac{\Vert f\ast u{\Vert }_{L^2}}{\Vert f{\Vert }_{L^2}}=\frac{\Vert (\hat{f}m{)}^{\vee }{\Vert }_{L^2}}{\Vert f{\Vert }_{L^2}}\le \Vert m{\Vert }_{{\mathcal{M}}_2}=\Vert m{\Vert }_{L^{\infty }}<\infty .$$
2.5.12	Consider $T:(m_1,m_2,f)\mapsto ((m_1\ast m_2)\hat{f}{)}^{\vee }$. Note that$$\Vert T(m_1,m_2,f){\Vert }_{L^2}=\Vert (m_1\ast m_2)f^{\vee }{\Vert }_{L^2}\le \Vert m_1\ast m_2{\Vert }_{L^{\infty }}\Vert f{\Vert }_{L^2}\le \Vert m_1{\Vert }_{L^{\infty }}\Vert m_2{\Vert }_{L^1}\Vert f{\Vert }_{L^2}.$$So $T$ maps $L^{\infty }\times L^1\times L^2$ to $L^2$ with norm no more than $1$.$$\begin{array}{rl}\Vert T(m_1,m_2,f){\Vert }_{L^1}& \le \Vert (m_1\ast m_2{)}^{\vee }{\Vert }_{L^1}\Vert f{\Vert }_{L^1}=\Vert m_1^{\vee }{m}_2^{\vee }{\Vert }_{L^1}\Vert f{\Vert }_{L^1}\\ & \le \Vert f{\Vert }_{L^1}\Vert m_1^{\vee }{\Vert }_{L^2}\Vert m_2^{\vee }{\Vert }_{L^2}=\Vert f{\Vert }_{L^1}\Vert m_1{\Vert }_{L^2}\Vert m_2{\Vert }_{L^2}.\end{array}$$So $T$ maps $L^2\times L^2\times L^1$ with norm no more than $1$. Observe that $\{\begin{array}{l}\frac{1}{r}=\frac{1}{2}\frac{2}{r}\\ \frac{1}{r^{\prime }}=\frac{1}{2}\frac{2}{r}+1(1-\frac{2}{r})\\ \frac{1}{p}=\frac{2}{r}+\frac{1}{2}(1-\frac{2}{r})\end{array}$. By the multilinear interpolation theorem (see GTM250 pp.516-7), we have$$\Vert T(m_1,m_2,f){\Vert }_{L^p}\le \Vert m_1{\Vert }_{L^r}\Vert m_2{\Vert }_{L^{r^{\prime }}}\Vert f{\Vert }_{L^p},$$which implies that $m_1\ast m_2\in {\mathcal{M}}_p({\mathbb{R}}^n).$
2.5.13	By Exercise 2.4.12, $(e^{i|\xi {|}^2}{)}^{\vee }=e^{i\frac{\pi n}{4}}{e}^{-i\pi |x{|}^2}\in L^{\infty }$. Let $m(\xi )=e^{i|\xi {|}^2}$, then$$\Vert T_m(f){\Vert }_{L^{\infty }}=\Vert f\ast m^{\vee }{\Vert }_{L^{\infty }}\le \Vert f{\Vert }_{L^1}\Vert m^{\vee }{\Vert }_{L^{\infty }},$$which implies $T_m$ is of strong $(1,\infty )$ type. Plus,$$\Vert T_m(f){\Vert }_{L^2}=\Vert \hat{f}(\xi )e^{i|\xi {|}^2}{\Vert }_{L^2}=\Vert \hat{f}(\xi ){\Vert }_{L^2}=\Vert f{\Vert }_{L^2},$$which implies $T_m$ is of strong $(2,2)$ type. Consequently by the Riesz–Thorin interpolation theorem, $T_m$ is of strong $(p,p^{\prime })$ type. If $T_m$ were of strong $(p,p)$ type for some $p\ne 2$, for $\Vert m{\Vert }_{{\mathcal{M}}_p}=\Vert m{\Vert }_{{\mathcal{M}}_{p^{\prime }}}$, we could assume that $1\le p<2$. Then by Riesz–Thorin again, $T_m$ is of strong $(p,2)$ type with$$\Vert f{\Vert }_{L^2}=\Vert T_m(f){\Vert }_{L^2}\le \Vert f{\Vert }_{L^p}.$$This is surely not true.

2.6 Oscillatory Integrals

2.6.1	Since $u^{(k)}$ is continuous, it can assumed that $u^{(k)}$ is positive. Replacing $\frac{u}{c_0},c_0\lambda$ with $u,\lambda$ in Proposition 2.6.7 leads to the required conclusion.
2.6.2	Let $u(t)=\{\begin{array}{ll}t+m(k-1),& t\in [2m,2m+1)\\ kt+(m+1)(1-k),& t\in [2m+1,2m+2)\end{array},m\in \mathbb{N}$. It is easy to see$$\left|{\int }_0^{2n}{e}^{i\lambda u(t)}\mathrm{d}t\right|=\left|\right(\sum _{m=0}^{n-1}{e}^{i\lambda (k+1)m})\cdot \frac{1}{\lambda }(e^{i\lambda }-1+\frac{e^{i(k+1)\lambda }-e^{i\lambda }}{k}\left)\right|=\left|\frac{1-e^{i\lambda (k+1)n}}{1-e^{i\lambda (k+1)}}\frac{1}{\lambda }\right(e^{i\lambda }-1+\frac{e^{i(k+1)\lambda }-e^{i\lambda }}{k}\left)\right|.$$Let $\lambda =\pi$, for any $\epsilon >0$, by Kroncker’s lemma (4.2.2), there exists $k>2,k\in \mathbb{Z}$ such that $|1-e^{i\lambda (k+1)}|<\epsilon$, and choose $n\in \mathbb{Z}$ such that $|1-e^{i\lambda (k+1)n}|>1$. Modify in the $\delta$-neighborhood of $\mathbb{Z}$ to make $u$ smooth, choose $\epsilon$ sufficiently small, we get the counterexample.
2.6.3
2.6.4	(a) Apply Exercise 2.6.1 to the intervals $[a,c-\delta ]$ and $[c+\delta ,b]$, and notice that $\left|{\int }_{c-\delta }^{c+\delta }{e}^{i\lambda u(t)}\right|\le 2\delta$, we reach the conclusion. (b) Similar to part (a). (c) Choose $\delta ={\lambda }^{-\frac{1}{k+1}}$.
2.6.5	(a) Notice that$${\int }_{\epsilon \le |t|\le 1}{e}^{i\lambda t}\frac{\mathrm{d}t}{t}={\int }_{\epsilon \le |t|\le 1}2i\sin (\lambda t)\frac{\mathrm{d}t}{t}={\int }_{\lambda \epsilon \le |t|\le \lambda }2i\sin t\frac{\mathrm{d}t}{t}.$$By Exercise 2.2.9, we come to the required conclusion. (b)$$\left|{\int }_{\epsilon \le |t|\le 1}{e}^{i(\lambda t\pm t^k)}\frac{\mathrm{d}t}{t}\right|=|{\int }_{\epsilon \le |t|\le 1}{e}^{i\lambda t}\frac{e^{\pm i{t}^k}-1}{t}+\frac{e^{i\lambda t}}{t}\mathrm{d}t|\le \left|{\int }_{\epsilon \le |t|\le 1}\frac{e^{i\lambda t}}{t}\mathrm{d}t\right|+{\int }_{\epsilon \le |t|\le 1}|t{|}^{k-1}\mathrm{d}t.$$By part (a), we come to the required conclusion. (c) When ${\xi }_1={\xi }_2=0$, it is trivial. When ${\xi }_2=0$, it follows from the same argument in part (a). When ${\xi }_2\ne 0$, let $t=|{\xi }_2{|}^{\frac{1}{k}}s$, and split the domain of integration into the sets $|t|\ge 1$ and $|t|\le 1$. When $|t|\le 1$ use part (b) and in the case $|t|\ge 1$ use Corollary 2.6.8, noting that $$\frac{{\mathrm{d}}^k}{\mathrm{d}t}({\xi }_1|{\xi }_2{|}^{-\frac{1}{k}}t+t^k)=k!\ge 1.$$
2.6.6	(a)$$\left|{\int }_0^{a\lambda }{e}^{i\lambda \log t}\mathrm{d}t\right|=\left|\frac{1}{i\lambda +1}a\lambda e^{i\lambda \log (a\lambda )}\right|\le a.$$(b) Split $(0,b)$ into $(0,1)$ and $(1,b)$. Apply Proposition 2.6.7 with $k=1$ to the former one, and $k=2$ to the other.
2.6.7	We can assume that $\lambda =1$, for otherwise let $s={\lambda }^{\frac{1}{\gamma }}t$. On the interval $(0,\delta )$ apply Proposition 2.6.7 with $k=[\gamma ]+1$, and on the interval $(\delta ,b)$ with $k=[\gamma ]$.