用户: Solution/ 解答: Fourier分析上课习题

1.	推广形式的 Hardy-Littlewood-Sobolev 不等式: 设 $(X,\mu )$, $(Y,\nu )$ 为两个测度空间, $1<q<\infty$, $K(x,y)$ 为 $X\times Y$ 上的可测函数, 且存在 $c>0$ s.t.$$\Vert K(x,\cdot ){\Vert }_{L^{q,\infty }(Y,\nu )}\le c,\text{a.e.}x\in X,$$$$\Vert K(\cdot ,y){\Vert }_{L^{q,\infty }(X,\mu )}\le c,\text{a.e.}y\in Y.$$若 $f\in L^p(Y,\nu )$ ($1\le p<+\infty )$, 则$$Tf(x)={\int }_{Y}K(x,y)f(y)d\nu (y)存在\text{a.e.}x\in X,$$且对于 $1<p<\infty$ 与 $1+\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$, $\exists C(p,q,r,c)>0$ s.t.$$\Vert Tf{\Vert }_{L^{r,\infty }(X,\mu )}\le C(p,q,r,c)\Vert f{\Vert }_{L^p(Y,\nu )},\forall f\in L^p(Y,\nu ).$$
Proof:	分解 $K=K{\chi }_{|K|>\beta }+K{\chi }_{|K|\le \beta }:=K_1+K_2$, 其中 $\beta >0$ 待定. 首先, 估计$$\Vert K_1(x,\cdot ){\Vert }_{L^1(Y,\nu )}={\int }_0^{\beta }{d}_{K_1(x,\cdot )}(\alpha )d\alpha +{\int }_{\beta }^{\infty }{d}_{K_1(x,\cdot )}(\alpha )d\alpha \le \beta c^q{\beta }^{-q}+{\int }_{\beta }^{\infty }{c}^q{\alpha }^{-q}d\alpha =\frac{q}{q-1}{c}^q{\beta }^{1-q},$$同理有$$\Vert K_1(\cdot ,y){\Vert }_{L^1(X,\mu )}\le \frac{q}{q-1}{c}^q{\beta }^{1-q}.$$于是$$\begin{array}{rl}|{\int }_Y{K}_1(x,y)f(y)d\nu |& \le {\int }_Y|K_1(x,y){|}^{\frac{1}{p^{\prime }}}(|K_1(x,y){|}^{\frac{1}{p}}|f(y)|)d\nu 其中\frac{1}{p}+\frac{1}{p^{\prime }}=1\\ & \le ({\int }_Y|K_1(x,y)|d\nu {)}^{\frac{1}{p^{\prime }}}({\int }_Y|K_1(x,y)||f(y){|}^{p}d\nu {)}^{\frac{1}{p}}\\ & \le \left(\frac{q}{q-1}{c}^q{\beta }^{1-q}{)}^{\frac{1}{p^{\prime }}}\right({\int }_Y|K_1(x,y)||f(y){|}^{p}d\nu {)}^{\frac{1}{p}},\end{array}$$由此得到$$\begin{array}{rl}\Vert {\int }_Y{K}_1(x,y)f(y)d\nu {\Vert }_{L^p(X,\mu )}& \le \left(\frac{q}{q-1}{c}^q{\beta }^{1-q}{)}^{\frac{1}{p^{\prime }}}\right({\int }_X{\int }_Y|K_1(x,y)||f(y){|}^{p}d\nu d\mu {)}^{\frac{1}{p}}\\ & =\left(\frac{q}{q-1}{c}^q{\beta }^{1-q}{)}^{\frac{1}{p^{\prime }}}\right({\int }_Y({\int }_X|K_1(x,y)|d\mu )|f(y){|}^{p}d\nu {)}^{\frac{1}{p}}\\ & \le \frac{q}{q-1}{c}^q{\beta }^{1-q}\Vert f{\Vert }_{L^p(Y,\nu )}\text{因 }\frac{1}{p}+\frac{1}{p^{\prime }}=1.\stackrel{◯}{1}\end{array}$$其次, 估计$$\Vert K_2(x,\cdot ){\Vert }_{L^{p^{\prime }}(Y,\nu )}^{p^{\prime }}=p^{\prime }{\int }_0^{\beta }{\alpha }^{p^{\prime }-1}{d}_{K_2(x,\cdot )}(\alpha )d\alpha \le p^{\prime }{\int }_0^{\beta }{\alpha }^{p^{\prime }-1-q}{c}^{q}d\alpha =\frac{p^{\prime }}{p^{\prime }-q}{c}^q{\beta }^{p^{\prime }-q},$$得到$$\Vert {\int }_Y{K}_2(x,y)f(y)d\nu {\Vert }_{L^{\infty }(X,\mu )}\le (\frac{p^{\prime }}{p^{\prime }-q}{c}^q{\beta }^{p^{\prime }-q}{)}^{\frac{1}{p^{\prime }}}\Vert f{\Vert }_{L^p(Y,\nu )}.\stackrel{◯}{2}$$现在用 $\stackrel{◯}{1},\stackrel{◯}{2}$ 来估计 $d_{Tf}(\alpha )$. 由 $Tf(x)={\int }_Y{K}_1(x,y)f(y)d\nu +{\int }_Y{K}_2(x,y)f(y)d\nu :=T_{1}f(x)+T_{2}f(x)$ 得到$$d_{Tf}(\alpha )\le d_{T_{1}f}(\alpha /2)+d_{T_{2}f}(\alpha /2).$$取 $\beta$ s.t. $(\frac{p^{\prime }}{p^{\prime }-q}{c}^q{\beta }^{p^{\prime }-q}{)}^{\frac{1}{p^{\prime }}}\Vert f{\Vert }_{L^p}=\alpha /2$, 则由 $\stackrel{◯}{2}$ 得$$d_{T_{2}f}(\alpha /2)=0.$$于是$$\begin{array}{rl}{d}_{Tf}(\alpha )& \le d_{T_{1}f}(\alpha /2)\\ & \le (\alpha /2{)}^{-p}\Vert {\int }_Y{K}_1(x,y)f(y)d\nu {\Vert }_{L^p(X,\mu )}^p\\ & \le (\alpha /2{)}^{-p}(\frac{q}{q-1}{c}^q{\beta }^{1-q}{)}^p\Vert f{\Vert }_{L^p}^p\text{由 }\stackrel{◯}{1}\\ & =C(p,q,c){\alpha }^{-p}(\alpha \Vert f{\Vert }_{L^p}^{-1}{)}^{\frac{(1-q)p{p}^{\prime }}{p^{\prime }-q}}\Vert f{\Vert }_{L^p}^p\\ & =C(p,q,c)({\alpha }^{-1}\Vert f{\Vert }_{L^p}{)}^r\text{因}-p+\frac{(1-q)p{p}^{\prime }}{p^{\prime }-q}=-\frac{q{p}^{\prime }}{p^{\prime }-q}=-r,\end{array}$$即$$\Vert Tf{\Vert }_{L^{r,\infty }(X,\mu )}\le C(p,q,c)\Vert f{\Vert }_{L^p(Y,\nu )}.$$ 注: 更一般的结论见 Tao 的课程讲义 https://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/ 的 Lemma 43.
2.	Let $1\le p_i\le q_i\le \infty$. Let $T$ be a quasi-linear operator defined on $L^{p_0}(X,\mu )+L^{p_1}(X,\mu )$ and taking values in the set of measurable functions on $(Y,\nu )$, where $K\ge 1$ is the constant that makes $|T(f+g)|\le K(|Tf|+|Tg|)$. Assume that $\Vert Tf{\Vert }_{L^{q_i,\infty }(Y,\nu )}\le A_i\Vert f{\Vert }_{L^{p_i}(X,\mu )}$ for some $A_i$, $i=0,1$. Fix $0<t<1$ and let$$\frac{1}{p}=\frac{1-t}{p_0}+\frac{t}{p_1},\frac{1}{q}=\frac{1-t}{q_0}+\frac{t}{q_1}.$$Then there exists a constant $C=C(p_0,p_1,q_0,q_1,K,A_0,A_1,t)$ such that$$\Vert Tf{\Vert }_{L^q(Y,\nu )}\le C\Vert f{\Vert }_{L^p(X,\mu )}.$$
Proof:	This is the lower triangular form of the Marcinkiewicz interpolation theorem. See Theorem 3.8 of the paper $Interpolation$ $Theorems$ $and$ $Applications$ by Calista Bernard; or see Theorem 27 of Tao’s note from the link given in the previous problem. Note that in both of the references $T$ is assumed to be sublinear instead of quasi-linear, yet for the problem above, we have $d_{T(f+g)}(2K\alpha )\le d_{Tf}(\alpha )+d_{Tg}(\alpha )$ all the same.
3.	A solution to the heat equation on ${\mathbb{R}}^n$$$\{\begin{array}{l}(\frac{\partial }{\partial t}-\Delta )u(t,x)=0,\\ u(0,x)=f(x)\end{array}$$is $e^{t\Delta }f=f\ast P_t$, where $P_t(x)=(4\pi t{)}^{-\frac{n}{2}}{e}^{-\frac{|x{|}^2}{4t}}.$ Prove the following: (a) $\Vert e^{t\Delta }{\Vert }_{L^p\to L^p}\le 1$, $\forall$ $1\le p\le \infty$. (b) $\Vert e^{t\Delta }{\Vert }_{L^1\to L^{\infty }}\le C(n)t^{-\frac{n}{2}}$, $\forall$ $t>0$. (c) $\Vert e^{t\Delta }{\Vert }_{L^p\to L^q}\le C(n)t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{q})}$, $\forall$ $t>0$.
Proof:	(a) By Hölder’s inequality, $$\begin{array}{rl}|e^{t\Delta }f(x)|& =|{\int }_{{\mathbb{R}}^n}f(x-y)P_t(y)\mathrm{d}y|\\ & \le \Vert P_t^{\frac{p-1}{p}}{\Vert }_{L^{\frac{p}{p-1}}}\Vert f(x-\cdot )P_t(\cdot {)}^{\frac{1}{p}}{\Vert }_{L^p}=({\int }_{{\mathbb{R}}^n}|f(x-y){|}^p{P}_t(y)\mathrm{d}y{)}^{\frac{1}{p}}\end{array}$$as $\Vert P_t{\Vert }_{L^1}=1$. Then by Fubini’s theorem,$$\Vert e^{t\Delta }f(x){\Vert }_{L^p}\le ({\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|f(x-y){|}^p{P}_t(y)\mathrm{d}y\mathrm{d}x{)}^{\frac{1}{p}}=\left({\int }_{{\mathbb{R}}^n}\right({\int }_{{\mathbb{R}}^n}|f(x-y){|}^p\mathrm{d}x)P_t(y)\mathrm{d}y{)}^{\frac{1}{p}}=\Vert f{\Vert }_{L^p}.$$(b) For any $x$, we have $|e^{t\Delta }f|\le \Vert P_t{\Vert }_{L^{\infty }}\Vert f{\Vert }_{L^1}=(4\pi t{)}^{-\frac{n}{2}}\Vert f{\Vert }_{L^1}$. (c) The result follows from combining part (a) and (b), using the Riesz-Thorin interpolation theorem. Still, we present a direct proof. By Hölder’s inequality,$$\begin{array}{rl}|e^{t\Delta }f|& \le {\int }_{{\mathbb{R}}^n}|f(x-y){|}^{\alpha }|f(x-y){|}^{1-\alpha }{P}_t(y)\mathrm{d}y\\ & \le \Vert |f{|}^{\alpha }{\Vert }_{L^{\frac{p}{\alpha }}}\Vert |f(x-\cdot ){|}^{1-\alpha }{P}_t(\cdot ){\Vert }_{L^{\frac{p}{p-\alpha }}}=\Vert f{\Vert }_{L^p}^{\alpha }\left({\int }_{{\mathbb{R}}^n}\right(|f(x-y){|}^{1-\alpha }{P}_t(y){)}^{\frac{p}{p-\alpha }}\mathrm{d}y{)}^{\frac{p-\alpha }{p}}.\end{array}$$And by Hölder’s inequality again,$${\int }_{{\mathbb{R}}^n}(|f(x-y){|}^{1-\alpha }{P}_t(y){)}^{\frac{p}{p-\alpha }}\mathrm{d}y\le ({\int }_{{\mathbb{R}}^n}|f(x-y){|}^{(1-\alpha )q}{P}_t(y{)}^{\frac{p}{p-\alpha }}\mathrm{d}y{)}^{\frac{p}{(p-\alpha )q}}({\int }_{{\mathbb{R}}^n}{P}_t(y{)}^{\frac{p}{p-\alpha }}\mathrm{d}y{)}^{1-\frac{p}{(p-\alpha )q}}.$$So we have$$\begin{array}{rl}\Vert e^{t\Delta }f{\Vert }_{L^q}& \le \Vert f{\Vert }_{L^p}^{\alpha }\left({\int }_{{\mathbb{R}}^n}\right({\int }_{{\mathbb{R}}^n}(|f(x-y){|}^{1-\alpha }{P}_t(y){)}^{\frac{p}{p-\alpha }}\mathrm{d}y{)}^{\frac{(p-\alpha )q}{p}}\mathrm{d}x{)}^{\frac{1}{q}}\\ & \le \Vert f{\Vert }_{L^p}^{\alpha }({\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|f(x-y){|}^{(1-\alpha )q}{P}_t(y{)}^{\frac{p}{p-\alpha }}\mathrm{d}y\mathrm{d}x{)}^{\frac{1}{q}}({\int }_{{\mathbb{R}}^n}{P}_t(y{)}^{\frac{p}{p-\alpha }}\mathrm{d}y{)}^{\frac{p-\alpha }{p}-\frac{1}{q}}.\end{array}$$Taking $\alpha =1-\frac{p}{q}$, we see that$$\Vert e^{t\Delta }f{\Vert }_{L^q}\le \Vert f{\Vert }_{L^p}({\int }_{{\mathbb{R}}^n}{P}_t(y{)}^{\frac{1}{1-\frac{1}{p}+\frac{1}{q}}}\mathrm{d}y{)}^{1-\frac{1}{p}+\frac{1}{q}}=C(n)t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{q})}\Vert f{\Vert }_{L^p}.$$
4.	For a semipositive matrix $A\in {\mathbb{R}}_{n\times n}$, define the function $F_A(x)=e^{-⟨Ax,x⟩}$ on ${\mathbb{R}}^n$, which is bounded. Compute $\widehat{F_A}$.
Proof:	For $\epsilon >0$, the matrix $A+\epsilon I$ is positive, hence nondegenerate. Then for all $f\in \mathcal{S}({\mathbb{R}}^n)$, we have $$\begin{array}{rl}& ⟨\widehat{F_A},f⟩\\ =& {\int }_{{\mathbb{R}}^n}{F}_A(x)\hat{f}(x)\mathrm{d}x\\ =& \underset{\epsilon \to 0^{+}}{\lim }{\int }_{{\mathbb{R}}^n}{e}^{-⟨(A+\epsilon I)x,x⟩}\hat{f}(x)\mathrm{d}x\\ =& \underset{\epsilon \to 0^{+}}{\lim }{\int }_{{\mathbb{R}}^n}(e^{-⟨(A+\epsilon I)x,x⟩}{)}^{\wedge }f(x)\mathrm{d}x\\ =& \underset{\epsilon \to 0^{+}}{\lim }{\int }_{{\mathbb{R}}^n}{\pi }^{\frac{n}{2}}\det (A+\epsilon I{)}^{-\frac{1}{2}}{e}^{-{\pi }^2⟨(A+\epsilon I{)}^{-1}x,x⟩}f(x)\mathrm{d}x\\ =& \underset{\epsilon \to 0^{+}}{\lim }{\int }_{{\mathbb{R}}^n}{\pi }^{\frac{n}{2}}{e}^{-{\pi }^2|x{|}^2}f((A+\epsilon I{)}^{\frac{1}{2}}x)\mathrm{d}x\\ =& {\int }_{{\mathbb{R}}^n}{\pi }^{\frac{n}{2}}{e}^{-{\pi }^2|x{|}^2}f(A^{\frac{1}{2}}x)\mathrm{d}x.\end{array}$$The second equality above is from the Lebesgue dominated convergence theorem, and the fifth from substituting $x$ with $(A+\epsilon I{)}^{\frac{1}{2}}x$. To verify the fourth, it suffices to consider the case when $A+\epsilon I$ is a diagonal matrix after a rotation of $x\in {\mathbb{R}}^n$, reducing matters to the multiplication of the ${\mathbb{R}}^1$ case: to be specific, the computation is $$\begin{array}{rl}(e^{-⟨(A+\epsilon I)x,x⟩}{)}^{\wedge }& =(e^{-⟨D(Px),Px⟩}{)}^{\wedge }=(e^{-{\lambda }_1{x}_1^2-\cdots -{\lambda }_n{x}_n^2}{)}^{\wedge }\circ P\\ & ={\pi }^{\frac{n}{2}}({\lambda }_1\cdots {\lambda }_n{)}^{-\frac{1}{2}}{e}^{-{\pi }^2({\lambda }_1^{-1}{x}_1^2+\cdots +{\lambda }_n^{-1}{x}_n^2)}\circ P\\ & ={\pi }^{\frac{n}{2}}\det (A+\epsilon I{)}^{-\frac{1}{2}}{e}^{-{\pi }^2⟨D^{-1}x,x⟩}\circ P\\ & ={\pi }^{\frac{n}{2}}\det (A+\epsilon I{)}^{-\frac{1}{2}}{e}^{-{\pi }^2⟨(A+\epsilon I{)}^{-1}x,x⟩},\end{array}$$where $A+\epsilon I=P^{T}DP$ is the decomposition such that $P$ is orthogonal and $D$ is the diagonal matrix $\mathrm{diag}({\lambda }_1,\cdots ,{\lambda }_n)$. Thus $\widehat{F_A}(x)={\pi }^{\frac{n}{2}}{e}^{-{\pi }^2{x}^2}{\delta }^{A^{\frac{1}{2}}}$, where ${\delta }^{A^{\frac{1}{2}}}f=f\circ A^{\frac{1}{2}}.$