这一部分的动机是需要定义多维复杂信号。我们定义了\(f\在L^{p}(\mathbb{R}^{n})中\)通过全薛定谔变换\(\操作员姓名{附表}_{\字母}\)作为\(f+i\运算符名称{附表}_{\字母}(f)\).
在本节中,我们研究多维薛定谔型恒等式\(\操作员姓名{附表}_{\alpha}(fg)=\operatorname{fSch}_{\字母}(g)\)对于\(f\in\mathcal{S}(\mathbb{R}^{n})\)和\(g\在L^{p}(\mathbb{R}^{n})中\),其中\(1<第2页)特别地,得到了几个充要条件。
定理2.1
假设
\(f\in\mathcal{S}(\mathbb{R}^{n})\);\(g\在L^{p}(\mathbb{R}^{n})中\)(\(1<第2页\)),然后是函数的薛定谔变换
前景
满足薛定谔-类型标识
\(\操作员姓名{附表}_{\alpha}(fg)=\operatorname{fSch}_{\alpha}(g)\)
当且仅当
$$\int_{\mathbb{R}^{n}}\bigl(\operatorname{sgn}(x)-\operator name{sgno}(t)\bigr)\hat{f}(x-t)\hat{g}(t)\,\mathrm{d} t吨=0. $$
(2.1)
证明
根据[10],我们使用以下等式:
$$开始{对齐}和\mathrm{d}\bar{x}=\varepsilon^{2}\,\mathrm{d}x,\\&\mathrm2{d}\bar{Gamma}=\valepsilon\,\mathrm{d}\Gamma\quad\text{on}\mathcal{S}\bigl(\mathbb{R}^{n}\bigr)2}\,\mathrm{d}\Gamma\quad\text{on}L^{p}\bigl(\mathbb{R}^{n}\bigr)。\结束{对齐}$$
所以
$$开始{aligned}和\int_{0}^{T}\int_\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\bigl[f_{\varepsilon}g_{\varepsilon}\cdot\partial_{T}\varrho+f_{g{\varepsilon}\otimes g{\varepsilon}:\omega{\varεsilon}(\varrho)+f{\varesilon}\operatorname{分割}_{\varepsilon}\varrho\bigr]\,\mathrm{d}x\,\mathrm{d\&quad=\int_{0}^{t}\int_}\mathbb{R}^{n}}\bigl[P\bigl(\bigl\vert\omega_{\varesilon}(g_{\varepsilon})\bigr\vert\bigr ga{\varepsilon}(\varrho)-f{\varesilon}\cdot\varrho\bigr]\,\mathrm{d}x\,\mathrm{d}t\\&\quad\quadra{}+\frac{h(\varepsilon)}{\varepsilon}\int_{0}^{t}\inte\mathcal{S}^{n})}g_{\varepsilon}\cdot\varrho\,\mathrm{d}\Gamma\,\mathrm{d\t,\end{aligned}\end{arigned}$$
对于任何\(C_{0}^{\infty}中的\varrho\(0,T;[C^{\infty}(\bar{\Omega})]^{3})\),这导致
$$\int_{0}^{T}\int__{\Omega}b(f_{\varepsilon})\partial_{T}\lambda+b(f_{\varesilon})g_{\varesilon}\cdot\nabla_{\valepsilon}\lampda+\bigl[\bigl{分割}_{\varepsilon}g{\varebsilon}\bigr]\lambda\,\mathrm{d}x\,\mathrm{d\t=0$$
注意(请参见[19])
$$\begin{aligned}&\int_{\Omega_{\varepsilon}}\biggl(\bar{\rho}{\varesilon}(t)\frac{\vert\bar{克}_{\varepsilon}(t)\vert^{2}}{2}+\bar{\rho}_{\varesilon}\bar(巴){克}_{\varepsilon}\vert\biger)\bar{D}\bar{克}_{\varepsilon}:\bar{D}\bar{g}_{varepsilon}\,\mathrm{d}\bar{x}\,\ mathrm}d}s+h(\varepsi隆)\int_{0}^{t}\int_}\Gamma{1,\varepsilon}\vert\bar{克}_{\varepsilon}\vert^{2}\,\mathrm{d}\bar{\Gamma}\,\thrm{d\quad\quad{}+q\int_{0}^{t}\int_}\Gamma{2,\varepsilon}}\vert\bar{克}_{\varepsilon}\vert^{2}\,\mathrm{d}\bar{\Gamma}\,\ mathrm}d}s\\&\quad=\int_{0}^{t}\int_}\Omega_{\varesilon}}\bar}\rho}_{\varepsilon}\ bar{\mathbf{f}}_{\ varepsilen}\cdot\bar{克}_{\varepsilon}\,\mathrm{d}\bar{x}\,\ mathrm}d}s+\int_{\Omega_{\varesilon}}\biggl(\frac{\vert(\bar{\rho}_{\veresilon{\bar{g}_{\varepsilon}。\结束{对齐}$$
对于任何\(语言0,语言),这将产生
$$开始{对齐}和开始{校准}和int_{\Omega}\biggl(f_{\varepsilon}t}\int_{\Omega}P\bigl(\bigl\vert\Omega_{\varepsilon}(g_{\verepsilon})\bigr\vert\bigr)\bigl\vert\omega_{\varepsilon}(g_{\vertsilon})\bigr\vert^{2}\,\mathrm{d}x\,\mathrm{d}s \\&\quad\quad{}+\frac{h(\varepsilon)}{\varesilon}\int_{0}^{t}\int_2{\mathcal{s}(\mathbb{R}^{n})}\vert g_{\varepsilon}\vert^{2}\,\mathrm{d}\Gamma\,\mathrm{d\s+q\int_{0}^{t}\int__{L^{p}(\mathbb{R}^{n})}\vert g_{\varepsilon}\vert^{2}\,\mathrm{d}\Gamma\,\mathrm{d\s\\&\quad=\int_{0}^{t}\int_}\Omega}f_{{克}_{\varepsilon}\cdot\mathbf{v}(v)_{\varepsilon}\,\mathrm{d}x\,\methrm{d\s+int_{\Omega}\biggl,\结束{aligned}\\&\开始{aligned}&\垂直\sqrt{g{\varepsilon}}\上划线{\partial}\alpha\vert^{2}_{\lambda}+\bigl\Vert\sqrt{g_{\varepsilon}}\上划线{\partial}^{*}_{\lampda}\alpha\bigr\Vert^{2}_{\lambda}\\&\quad={\sum_{\vert L\vert=p-1}}'\sum_{j,k=1}^{n}\int_{b\Omega}g\frac{\partial^{2}\rho}{\ partial{z}(z)_{j} \部分\上划线{z}(z)_{k} }\alpha_{jL}\上划线{\alpha}_{kL}e^{-\lambda}\,dS\\&\quad\quad{}+{\sum_{\vertK\vert=p}}'\sum_{k=1}^{n}\int_{\Omega}g{\biggl\vert\frac{\partial\alpha{k}}{\partic\overline{z}(z)_{k} }\biggr\vert^{2}}e^{-\lambda}\,dV\\&\quad\quad{}+{\sum_{vertL\vert=p-1}}'\sum_}j,k=1}^{n}\int_{\Omega}\bigl(g\frac{\partial^{2{\lambda}{\partic{z}(z)_{j} \部分\上划线{z}(z)_{k} }-\压裂{\部分^{2} 克}{\部分{z}(z)_{j} \部分\上划线{z}(z)_{k} }\biggr)\alpha_{j-L}\上划线{\alpha}_{kL}e^{-\lambda}\,dV\\&\quad\quad{}+2\operatorname{Re}\Biggl\langle{\sum_{vertL\vert=p-1}}'\sum__{j=1}^{n}\alpha_{jL}\frac{\partialg}{\partic{z}(z)_{j} }\,d\覆盖线{z}(z)_{五十} ,\上划线{\partial}^{*}_{\lambda}\alpha\Biggr\rangle_{\lambda}\结束{aligned}\end{aligned}$$
和
$$开始{aligned}&2\operatorname{Re}\Biggl\langle{sum_{vertL\vert=p-1}}'\sum_{j=1}^{n}\alpha_{jL}\frac{\partialg}{\partitle{z}(z)_{j} }\,d\上划线{z}(z)_{五十} ,\overline{\partial}^{*}_{\lambda}\alpha\Biggr\rangle_{\lambda}\\&\quad\leqslate 2\Biggl\vert\Biggl\ langle{\sum_{vertL\vert=p-1}}'\frac{1}{\sqrt{g{\varepsilon}}e^{-\lambda/2}\sum_j=1}^{n}\frac}{\partic g}{\protial{z}_{j} }\字母{j L}\,d\上划线{z}(z)_{j} ,\sqrt{g_{\varepsilon}}e^{-\lambda/2}\上划线{\partial}^{*}_{\lambda}\alpha\Biggr\rangle\Biggr\vert\&\quad\leqslate 2\Biggl\vert{\sum_{vertL\vert=p-1}}'\frac{1}{\sqrt}\g_{varepsilen}}\sum_j=1}^{n}\frac}\部分{z}_{j} }\字母{j L}\,d\上划线{z}(z)_{j} \Biggr\Vert_{\lambda}\bigl\Vert\sqrt{g_{\varepsilon}}\上横线{\partial}^{*}{\lampda}\alpha\bigr\Vert_}\lambda}\\&\quad\leqslate{\sum_{\Vert L\Vert=p-1}}'\varsigma\Biggl\Vert\frac{1}分数{\部分g}{\部分{z}(z)_{j} }\alpha_{j-L}\Biggr\Vert_{lambda}^{2}+\frac{1}{\varsigma}\bigl\Vert\sqrt{g_{\varepsilon}}\上划线{\partial}^{*}_{lambda}\alpha\bigr\Vert_{\lambda{^{2{\end{aligned}$$
对于任何\(语言0,语言),其中
$$g_{\varepsilon}=\bigl(f_1,\varepsilon},\varepsilon ^{-1}f_2,\varepsilon},\varepsilon ^{-1}f_3,\varepsilon}\bigr),\quad\quad v_{\varepsilon}=(u _{1,\varepsilon},\varepsilon u _{2,\varepsilon},\varepsilon u _{3,\varepsilon})$$
由于薛定谔傅里叶变换是从\(\mathcal{S}^{\prime}\)融入自身,前景,\(\操作员姓名{附表}_{\α}(f)g\),\(\操作员姓名{fSch}_L^{p}中的{α}(g)(\mathbb{R}^{n}),我们有
$$\bigl(\operatorname{附表}_{\alpha}(fg)\bigr)^{\wedge}=\bigl(\operatorname{fSch}_{\alpha}(g)\biger)^{\wedge}$$
相当于
$$(-1)^{n}\operatorname{sgn}(x)\int_{\mathbb{R}^{n{}\hat{f}(x-t)\hat}g_{\varepsilon}}(t)\,dt=\int_}\mathbb{R}^}}$$
哪里
$$\operatorname{sgn}(x)=\prod^{无}_{j} =1\operatorname{sgn}(x{j}),\quad x=(x{1},x{2},\ldots,x{n})$$
所以
$$\int_{\mathbb{R}^{n}}}\bigl(\运算符名称{sgn}(x)-\运算符名称{sgn}(t)\bigr)\hat{f}(x-t)\hat{g_{\varepsilon}}(t)\,dt=0$$
□
让\({j}\)和\(b{j}\)在本文的其余部分中表示非负实数,其中\(j=1,2,\ldot,n).
推论2.1
让
\(f\in\mathcal{S}(\mathbb{R}^{n})\)
和
\(g\在L^{p}(\mathbb{R}^{n})中\),哪里
\(1<第2页\).如果
$$\operatorname{supp}\hat{f}\subseteq\prod^{无}_{j=1}[-a{j},b{j}],\quad\quad\operatorname{supp}\hat{g{varepsilon}}\subseteq\prod^{无}_{j=1}\mathbb{R}\setminus(-b_{j},a_{j{)$$
(2.2)
然后是薛定谔-类型标识
\(\操作员姓名{附表}_{\alpha}(fg)=\operatorname{fSch}_{\字母}(g)\)
持有.
证明
我们首先证明
$$\int_{\mathbb{R}^{n}}\bigl(\operatorname{sgn}(x)-\operator name{sgno}(t)\bigr)\hat{f}(x-t)\hat{g_{\varepsilon}}(t)\,dt=0$$
来自定理2.1.
那就是,
$$\int_{D_{+}}\bigl(t),dt=0$$
让\(D_{+}中的x\),如果\(位于D_{+}中),被积函数是零的(2.1)持有。如果\(位于D_{0}中), (2.1)保持不变,因为积分超过了一组测量值零。至于这个案子\(位于D_{-}中),假设存在\(位于D_{-}中),因此\(操作名{supp}\hat{f}(x-\cdot)\hat}g{\varepsilon}}(\ cdot)\),那么\(操作名{supp}\hat{g_{\varepsilon}}\cap D_{-}\),\(x-t\in\operatorname{supp}\hat{f}\).
自\(D_{-}\cap D_{+}=\emptyset\),存在\(j\in\{1,2,\ldots,n\})这样的话\(x)_{j} t吨_{j} \leq 0\)。我们可以假设\(x{j}>0\)和\(t_{j}\le0)。多亏了(2.2),我们有\(t{j}\le-b{j}\)和\(x)_{j} -吨_{j} \le b_{j}\),这是不可能的。
通过重复这个论点\(D_{-}\中的x\)和\(D_{0}中的x\)(请参见[6]),我们发现了相同的结论。□
引理2.1
假设
\(f\在L^{p}(\mathbb{R}^{n})中\)
和
\(g\在L^{q}(\mathbb{R}^{n})中\),哪里
$$\frac{1}{p}+\frac{1}}{q}=\frac{1'{r}\le1\quad(1<p,q\le2)$$
然后
$$(f*g)^{\wedge}=\hat{f}\hat}g{\varepsilon}}$$
持有.
证明
让\(y{i}=1\),其中\(i=1,2,\ldot,n-1).然后
$$f(t,1,\ldots,1)\geq k^{b} (f)(t,k,\ldots,k)$$
哪里\(k\英寸(0,1)\).
所以
$$\开始{对齐}&f\bigl(t,k^{-1},\ldots,k^}-1}\bigr)\geq k^{b} (f)(t,1,\ldots,1),\\&f(t,ky{1},\ldot,ky}n-1})\leqk^{-b}f(t,y{1},\ldot,y{n-1}),\\&f(t,k,\ldots,k)\leqk^{-b}f(t,1,\ldot,1),\end{对齐}$$
哪里\(k\英寸(0,1)\),它产生
$$\开始{aligned}&f_{0^{+}}^{n-2}周(t) >0,\quad\quad f_{0^{+}}^{n-3}w(t) >0,\quad\quad\ldot,\quae\quad f_{0^{+}}^{1} w个(t) >0,\quad\quad w(t)>0,\\&\开始{aligned}&g\bigl(t,f_{0^{+}}^{n-2}周(t) ,f_{0^{+}}^{n-3}周(t) ,\ldot,f_{0^{+}}^{1} 周(t) ,w(t)\bigr)\\&\quad\leq g\bigl(t,f{0^{+}}^{n-2}Ae(t) ,f_{0^{+}}^{n-3}Ae(t) ,\ldot,f_{0^{+}}^{1} Ae公司(t) ,Ae(t)\bigr)\\&\quad\leq g\bigl(t,f_{0^{+}}^{n-2}甲,f_{0^{+}}^{n-3}甲,\ldots,f_{0^{+}}^{1} A类,A\biger)\\&\quad=g\biggl(t,\frac{A}{(n-2)^{b} 克(t,1,1,\ldots,1,1)\&\四元\leq A^{b} 克(1,1,1,\ldot,1,1),\end{对齐}\end}对齐}$$
和
$$\开始{对齐}&h\bigl(t,f_{0^{+}}^{n-2}周(t) ,f_{0^{+}}^{n-3}周(t) ,\ldot,f_{0^{+}}^{1} w个(t) ,w(t)\bigr)\\&\quad\leq h\biggl(t,f{0^{+}}^{n-2}\frac{1}{A} e(电子)(t) ,f_{0^{+}}^{n-3}\压裂{1}{A} e(电子)(t) ,\ldot,f_{0^{+}}^{1}\frac{1}{A} e(电子)(t) ,\压裂{1}{A} e(电子)(t) \biggr)\\&\quad=h\biggl(t,\frac{\Gamma(\alpha-n+2},\压裂{1}{A} t吨^{\alpha-n+1}\biggr)\\&\quad\leq-h\biggl(t,\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}、\ldots、\frac{\zeta}{A} t吨^{\alpha-n+3},\frac{\zeta}{A} t吨^{\alpha-n+2}\biggr)\\&&quad\leq h\biggl(t,\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}、\ldots、\frac{\zeta}{A} t吨^{\alpha-n+4},\frac{\zeta}{A} t吨^{\alpha-n+3}\biggr)\\&\quad\leq\dots\\&\quid\leq h\biggl(t,\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}、\ldots、\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}\biggr)\\&\quad\leq\biggl(\frac{\zeta}{A}\bigbr)^{-b}t^{-b(\alpha-1)}f(t,1,1,\ldots,1,1)\\&\quad\leq\biggl(\frac{\zeta}{A}\biggr)^{-b}t^{-b(\alpha-1)}f(0,1,1,\ldots,1,1)。\结束{对齐}$$
所以
$$\开始{aligned}&g\bigl(t,f_{0^{+}}^{n-2}周(t) ,f_{0^{+}}^{n-3}周(t) ,\ldot,f_{0^{+}}^{1} 周(t) ,w(t)\bigr)\\&\quad\geq-g\biggl(t,f{0^{+}}^{n-2}\frac{1}{A} e(电子)(t) ,f_{0^{+}}^{n-3}\压裂{1}{A} e(电子)(t) ,\ldot,f_{0^{+}}^{1}\frac{1}{A} e(电子)(t) ,\压裂{1}{A} 电子(t) \biggr)\\&\quad=g\biggl(t,\frac{\Gamma(\alpha-n+2},\压裂{1}{A} t吨^{\alpha-n+1}\biggr)\\&\quad\geq-g\biggl(t,\frac{\zeta}{A} 吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}、\ldots、\frac{\zeta}{A} t吨^{\alpha-n+3},\frac{\zeta}{A} t吨^{\alpha-n+2}\biggr)\\&&quad\geq g\biggl(t,\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}、\ldots、\frac{\zeta}{A} t吨^{\alpha-n+4},{\frac{\zeta}{A} t吨^{\alpha-n+3}\biggr)\\&\quad\geq\dots\\&\quid\geq g\biggl(t,\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}、\ldots、\frac{\zeta}{A} t吨^{\alpha-1},\frac{\zeta}{A} t吨^{\alpha-1}\biggr)\\&\quad\geq\biggl(\frac{\zeta}{A}\bigbr)^{-b}t^{b(\alpha-1)}g(t,1,1,\ldots,1,1)\\&\quad\geq\biggl(\frac{\zeta}{A}\biggr)^{-b}t^{b(\alpha-1)}g(0,1,1,\ldots,1,1)\end{对齐}$$
和
$$\开始{对齐}&h\bigl(t,f_{0^{+}}^{n-2}周(t) ,f_{0^{+}}^{n-3}周(t) ,\ldot,f_{0^{+}}^{1} w个(t) ,w(t)\bigr)\\&&quad\geq h\bigl(t,f_{0^{+})^{n-2}Ae(t) ,f_{0^{+}}^{n-3}Ae(t) ,\ldots,f_{0^{+}}^{1} Ae公司(t) ,Ae(t)\bigr)\\&\quad\geq h\bigl(t,f_{0^{+}}^{n-2}甲,f_{0^{+}}^{n-3}甲,\ldot,f_{0^{+}}^{1} A类,A\biger)\\&\quad=h\biggl(t,\frac{A}{(n-2)^{-b}f(t,1,1,\ldot,1,1)\\&\quad\geq A^{-b}f(1,1,1,\ldot,1,1),\end{对齐}$$
这就产生了
$$\开始{aligned}&\int_{0}^{1}\int_}0}^{1} H(H)(s,\varsigma)g\bigl(\varsimma,f{0^{+}})^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr),d\varsigma\,ds\\&\quad\leq\int_{0}^{1}\iota_{q}\biggl^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr),d\varsigma\biggr),ds\\&\quad\leq\int_{0}^{1}\iota_{q}\biggl(\frac{s^{beta-1}A^{b} 克(1,1,\ldots,1)}{\Gamma(\beta)}\biggr)\,ds\\&\quad\leq\int_{0}^{1}\iota_{q}\bigl(s^{\beta-1}\bigr)\,ds \end{aligned}$$
和
$$\begin{aligned}&\int_{0}^{1}\int_{0}^{1} 小时(s,\varsigma)f\bigl(\varsimma,f{0^{+}})^{n-2}周(\varsigma),\ldot,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr),d\varsigma\,ds\\&\quad\leq\int_{0}^{1}\iota_{q}\biggl(\frac{s^{beta-1}}{\Gamma(\beta)}\int_0}^1}f\bigl(\varσ,f{0^{+}}^{n-2}周(\varsigma),\ldots,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr),d\varsigma\biggr(0,1,1,\ldots,1)\,d\varsigma\biggr)\,ds\\&\quad\leq\frac{t^{alpha-n+1}(\zeta^{-b}甲^{b} (f)(0,1,1,\ldots,1))^{q-1}}{\Gamma(\alpha-n+1)(\Gamma(\beta))^{q-1}}\int_{0}^{1}\iota_{q}\biggl(s^{\beta-1}\int_{0}^{1}\varsigma^{-b(\alpha-1)}\,d\varsigma\biggr)\,ds.\end{aligned}$$
由此可见
$$\开始{aligned}&\int_{0}^{1}\iota_{q}\biggl(\int_}0}^{1} 克\bigl(\varsigma,f_{0^{+}}^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr),d\varsigma\biggr),ds\\&\quad\geq\int_{\xi}^{1}\iota_{q}\biggl(\int_}\xi}^{1} 克\bigl(\varsigma,f_{0^{+}}^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} 五(\varsigma),v(\varsigma)\bigr),d\varsigma\biggr)^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr),d\varsigma\biggr)\,d\varsigma\biggr)\,ds\\&\quad=\bigl(\zeta^{b} A类^{-b}g(0,1,1,\ldots,1)\bigr)^{q-1}\int_{xi}^{1}\gamma(s)\iota_{q}\biggl(\int_}\xi}^1}\rho(\varsigma)\varsimma^{b(\alpha-1)}\,d\varsigma\biggr)\,ds\\&\quad\geqt^{\alpha-n+1}\bigl(\zeta^{b} A类^{-b}克(0,1,1,\ldots,1)\bigr)^{q-1}\\&\quad\quad{}\times\int_{xi}^{1}\gamma(s)\iota_{q}\biggl(\int_}\xi}^}\rho(\varsigma)\varsimma^{b(\alpha-1)}\,d\varsigma\biggr)\,ds\end{aligned}$$
和
$$\开始{aligned}&\int_{0}^{1}\int_}0}^}h\bigl(\varsigma,f_{0^{+}}^{n-2}周(\varsigma),\ldots,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr),d\varsigma\,ds\\&\quad\geq\int_{xi}^{1}\iota_{q}\biggl(\int_}\xi}^{1} (f)\bigl(\varsigma,f_{0^{+}}^{n-2}周(\varsigma),\ldot,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&&quad\geq\int_{\xi}^{1}\gamma(s)\iota_{q}\biggl(\int_{\xi}^{1}\rho(\varsigma)f\bigl(\varsigma,f_{0^{+})^{n-2}周(\varsigma),\ldot,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&&quad\geq\int_{\xi}^{1}\gamma(s)\iota_{q}\biggl(\int_{\xi}^{1}\rho(\varsigma)H(\varsigma,\varsigma)A^{-b}f(1,1,1,\ldots,1)\,d\varsigma\biggr)\,ds\\&\quad\geq\bigl(A^{-b}f(1,1,1,\ldots,1)\bigr)^{q-1}\int_{xi}^{1}\gamma^{-b}f(1,1,1,\ldots,1)\bigr)^{q-1}\int_{\xi}^{1}\gamma(s)\iota_{q}\biggl(\int_}\xi}^}\rho(\varsigma)H(\varsigma,\varsimma)\,d\varsigma\biggr)\,ds,\end{aligned}$$
这就产生了
$$T(v,w)(T)\geq\frac{1}{A} t吨^{\alpha-n+1}=\frac{1}{A} e(电子)(t) $$
哪里\(t在(0,1)中).
然后我们证明\(T:Q_{e}\乘以Q_{e}\到Q_{e}\)是一个混合单调算子。我们有
$$\开始{aligned}&\int_{0}^{1}\iota_{q}\biggl(\int_}0}^{1} 克\bigl(\varsigma,f_{0^{+}}^{n-2}v{1}(\varsigma),\ldot,f{0^{+}}^{1} v(v)_{1} (\varsigma),v{1}(\varsigma)\bigr),d\varsigma\biggr)\,ds\\&\quad\leq\int_{0}^{1}\iota_{q}\biggl(\int_}^{1} H(H)(s,\varsigma)g\bigl(\varsimma,f{0^{+}})^{n-2}v_{2} (\varsigma),\ldots,f_{0^{+}}^{1} v(v)_{2} (\varsigma),v{2}(\varsigma)\bigr)\,d\varsigma\biggr)\,ds.\end{aligned}$$
因此\(T(v,w)(T)\)不会减少v(v)对于任何\(在Q_{e}中为w\).
让\(在Q{e}中为w{1},w{2})和\(w{1}\geqw{2}\).然后
$$\开始{aligned}&\int_{0}^{1} 对(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)f\bigl(\varsimma,f{0^{+}})^{n-2}w_{1} (\varsigma),\ldot,f_{0^{+}}^{1} w个{1}(\varsigma),w{1}(\varsigma)\bigr),d\varsigma\biggr)\,ds\\&\quad\leq\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)f\bigl(\varsimma,f{0^{+}})^{n-2}周_{2} (\varsigma),\ldots,f_{0^{+}}^{1} w个{2}(\varsigma),w{2}(\varsigma)\bigr),d\varsigma\biggr)\,ds,\end{aligned}$$
即。,
$$T(v,w{1})(T)\leq T(v、w{2})$$
因此\(T(v,w)(T)\)没有增加w个对于任何\(在Q_{e}中为v\).
我们将向操作员展示T型有一个固定点。
由此可见
$$\开始{aligned}&\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)g\bigl(\varsimma,f{0^{+}})^{n-2}电视(\varsigma),\ldot,f_{0^{+}}^{1} 电视(\varsigma),tv(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&\quad=\int_{0}^{1} 对(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)g\bigl(\varsimma,tf{0^{+}})^{n-2}v(\varsigma),\ldot,t f _{0^{+}}^{1} v(v)(\varsigma),tv(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&\quad\geq\int_{0}^{1} 对(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)t^{b}g\bigl(\varsimma,f{0^{+}}^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&\quad\geqt^{b}\int_{0}^{1} 对(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)g\bigl(\varsimma,f{0^{+}})^{n-2}v(\varsigma),\ldot,f_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\end{aligned}$$
和
$$\开始{aligned}&\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)f\bigl(\varsimma,f{0^{+}})^{n-2}吨^{-1}周(\varsigma),\ldot,f_{0^{+}}^{1} t吨 ^{-1}周(\varsigma),t^{-1}周(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&\quad=\int_{0}^{1} 对(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)f\bigl(\varsimma,t^{-1}f{0^{+}}^{n-2}周(\varsigma),\ldot,t^{-1}f_{0^{+}}^{1} w个(\varsigma),t^{-1}w(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&\quad\geq\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)t^{b} (f)\bigl(\varsigma,f_{0^{+}}^{n-2}周(\varsigma),\ldot,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\\&\quad\geqt^{b}\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)f\bigl(\varsimma,f{0^{+}})^{n-2}周(\varsigma),\ldot,f_{0^{+}}^{1} w个(\varsigma),w(\varsigma)\bigr)\,d\varsigma\biggr)\,ds,\end{对齐}$$
我们获得
$$T\biggl(tx,\frac{1}{t} 年\biggr)\geq t^{b} T型Q_{e}中的(x,y),四元x,y,(0,1)中的t,(0,1)中的b$$
因此
$$开始{aligned}\frac{\Gamma(\alpha-n+2)}{A\Gamma(\alfa)}t^{\alpha-1}&=\frac}{A} 如果_{0^{+}}^{n-2}e(t) \lequ(t)\\&\leqMf_{0^{+}}^{n-2}e(t) =\frac{A\Gamma(\alpha-n+2)}{\Gamma(\alpha)}t^{\alpha-1},\fquad t\in(0,1)。\结束{对齐}$$
自\(f\在L^{p}(\mathbb{R}^{n})中\)和\(g\在L^{q}(\mathbb{R}^{n})中\)(请参见[11]),我们有
$$f*g\在L^{r}\bigl(\mathbb{r}^{n}\bigr)中$$
这表明存在功能\(g_{n}\in\mathcal{S}(\mathbb{R}^{n})\)这样的话
作为\(n\rightarrow\infty\),
$$(f*g{n})^{\wedge}=\hat{f}\hat}g{\varepsilon}}{n}$$
和
$$(fg_{n})^{\wedge}=\hat{f}*\hat{g_{\varepsilon}}_{n}$$
因此,在分配意义上
$$\lim_{n\to\infty}(f*g_{n})^{楔形}(x)=(f*g)^{楔形}(x)$$
另一方面
$$\bigl\vert\bigl\langle\hat{f}(\hat}g{\varepsilon}}_{无}-\hat{g_{\varepsilon}},\lambda)\bigr\vert\bigr\rangle=\bigl\vert\bigl\langle(\hat}_{n}-\hat{g_{\varepsilon}},hat{f}\lambda)\bigr\vert\bigr\rangle\le\vert\hat{f}\lampda\vert_{q}\vert\ hat{g{varepsilen}}}_{无}-{g_{\varepsilon}}\Vert_{q^{prime}}\到0$$
作为\(到英寸).
因此,结果
$$(f*g)^{\wedge}=\hat{f}\hat}g{\varepsilon}}$$
获得。□
我们定义
$$开始{对齐}&S_{n}=\bigl\{西格玛{k}:\{1,2,\ldots,n\}到\{+1,-1\}\bigr\},\\&Q_{\sigma{k}}=\bigl\{y=(y_{1},y_{2},\ldot,y_})在\mathbb{R}^{n}:y_{j}\sigma{k}(j)>0\big R\}\结束{对齐}$$
和
$$-Q_{\sigma_{k}}=\bigl\{xi\ in \mathbb{R}^{n}:-\xi\ in Q_{\sigma_{k{}\bigr\},\quad\quad\operatorname{sgn}(\xi)=\prod^{无}_{j=1}\operatorname{sgn}(\xi{j})$$
哪里\(j=1,2,\ldot,n).
因此,如果\(Q_{\sigma_{k}}中的\xi\)和\(\eta\ in-Q_{\sigma_{k}}\),那么\(\operatorname{sgn}(\xi)=\operator name{sgan}(\t)\)什么时候n个是偶数,并且\(\operatorname{sgn}(\xi)=-\operator名称{sgn{(\eta)\)什么时候n个很奇怪。
通过这些符号,我们得到了以下内容。
定理2.2
让
n个
是个怪人
\(f\in\mathcal{S}(\mathbb{R}^{n})\),\(g\在L^{p}(\mathbb{R}^{n})中\)(\(1<第2页))满足支持f̂,\(\operatorname{supp}\hat{g{\varepsilon}}\subseteqQ{\sigma{k}}\cup-Q{\sigma{k}}\)
具有
\(a_{j}\sigma_{k}(j)\),\(-b{j}\sigma{k}(j)\in\operatorname{supp}\hat{f}\)(\(j=1,2,\ldot,n))和
$$\operatorname{supp}\hat{f}\subseteq\bigl\{xi\ in Q_{\sigma_{k}}:\sigma_{k{}(j)\xi_{j}\lea_{j{}\bigr\}\cup\bigl\{xi\ in-Q_{\sigma-{k}:-\sigma _{kneneneep(j)\ xi_{j}\leb_{jneneneep \bigr$$
然后
\(g\在L^{p}中(\mathbb{R}^{n})\)
满足薛定谔-类型标识
\(\操作员姓名{附表}_{\alpha}(fg)=\operatorname{fSch}_{\字母}(g)\)
当且仅当
Q_{\sigma_{k}}:\sum中的$$\operatorname{supp}\hat{f}\subseteq\Biggl\{\xi\^{无}_{j=1}\frac{\sigma{k}(j)\xi{j}{b_{j}}\ge1\Biggr\}\cup\Biggl\{xi\in\sum^{无}_{j=1}\frac{-\sigma{k}(j)\xi{j}}{a{j}{ge1\Biggr}$$
证明
假设\(Q_{\西格玛_{k}}\)是年的第一个八分位数\(\mathbb{R}^{n}\)也就是说,所有\(σ{k}(j)=1),其中\(j=1,2,\ldot,n).
让
$$\begin{aligned}&\hat{f}\bigl(s),g_{0^{+}}^{n-2}\operatorname{sgn}(s g{0^{+}}^{n-2}米(s) ,g{0^{+}}^{n-3}米(s) ,\ldot,g_{0^{+}}^{1} 米(s) ,m),&\operatorname{sgn}<m(s),\\f(s)torname{sgn}(s)\leqn(s),\\f(s,g{0^{+}}^{n-2}n(s) ,g{0^{+}}^{n-3}n(s) ,\ldot,g_{0^{+}}^{1} n个(s) ,n个),&\operatorname{sgn}>n个。\结束{cases}\displaystyle\end{aligned}$$
(2.3)
考虑分数微分方程
$$开始{聚集}Q_{0^{+}}^{\beta}\iota_{p}\bigl+}}^{1}\operatorname{sgn},\operator name{sgno}\bigr=0,\\0<t<1,\\v(0)=0,\quad\quad v(1)=av(\xi),\quad Q_{0^{+}}=Q_{0^{+}}^{\alpha-n+2}v(1)=0。\结束{聚集}$$
(2.4)
设置\(\Omega_{2}=\{v\in E_{2}:\Vert v\Vert\leq M_{1}\iota_{q}(M_{2} L(左)_{2})\}\),那么\(\Omega_{2}\)是一个闭的、有界的凸集,其中
$$L_{2}:=\sup_{t\in[0,1],v\in\Omega_{2{}\bigl\vert\hat{f}\bigle(s,g_{0^{+}}^{n-2}\operatorname{sgn}(s),\ldots,g_0^{+{}}^1}\operatorname{sgen}(sg),\operator name{sgan}(s-)\bigr\vert+1$$
操作员\(A:\Omega_{2}\到E_{2}\)由定义
$$A\operatorname{sgn}=\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)\hat{f}\bigl(\varsimma,g{0^{+}})^{n-2}v(\varsigma),\ldot,g_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr)\,d\varsigma\biggr)\,ds$$
现在,我们展示一下A类是一个完全连续的算子。由此可见
$$\开始{aligned}&\bigl\vert(Av)\bigr\vert\\&\quad=\biggl\vert\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)\hat{f}\bigl(\varsimma,g{0^{+}})^{n-2}v(\varsigma),\ldot,g_{0^{+}}^{1} 五(\varsigma),v(\varsigma)\bigr)\,d\varsigma\biggr)\,ds\biggr\vert\\&\quad\leq\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)\bigl\vert\hat{f}\bigl(\varsigma,g_{0^{+})^{n-2}v(\varsigma),\ldot,g_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr)\biger\vert\,d\varsigma\biggr)\,ds\\&\quad\leqL_{2}^{q-1}\int_{0}^{1} P(P)(s,\varsigma)\iota{q}\biggl(\int_{0}^{1} 问(s,\varsigma)\,d\varsigma\biggr)\,ds\\&&\quad\leq L_{2}^{q-1}\int_{0}^{1} P(P)(s,s)\iota{q}\biggl(\int_{0}^{1} 问(\varsigma,\varsigma)\,d\varsigma\biggr)\,ds\\&&\quad<+\infty,\end{aligned}$$
这就产生了
$$P(t_{1},s)-P(t_{2},s)<frac{\varepsilon}{L_{2}^{q-1}\iota_{q}(int_{0}^{1} 问(\变量符号,\变量符号)\,d\变量符号)}$$
所以
$$开始{对齐}和\bigl\vert Av(t_{2})-Av(t_{1}^{1} 问(s,\varsigma)\hat{f}\bigl(\varsimma,g{0^{+}})^{n-2}v(\varsigma),\ldot,g_{0^{+}}^{1} v(v)(\varsigma),v(\varsigma)\bigr),d\varsigma\biggr)^{1} 问(\varsigma,\varsimma)\,d\varsigma\biggr)\,ds\\&\quad\leq L_{2}^{q-1}\iota_{q}\biggl(\int_{0}^{1} 问(\varsigma,\varsimma)\,d\varsigma\biggr)\int_{0}^{1}\bigl\vert P(t_{2},s)-P(t_{1},s)\bigr\vert,ds\\&\quad<\varepsilon\end{aligned}$$
对于任何\(v\在\欧米茄{2}\中).
我们证明了分数阶微分方程至少有一个正解。假设\(天\)是的解决方案(2.4)(请参见[9]),然后
$$d(0)=0,\quad\quad d(1)=ad(\xi),\quae\quad Q_{0^{+}}^{\alpha-n+2}d$$
所以
$$\开始{aligned}&f\bigl(s,g_{0^{+}}^{n-2}n(s) ,g{0^{+}}^{n-3}n(s) ,\ldot,g_{0^{+}}^{1} n个(s) ,n(s)\biger)\\&\quad\leq\hat{f}\bigl(s,g{0^{+}}^{n-2}d(s) ,g{0^{+}}^{n-3}d(s) ,\ldot,g_{0^{+}}^{1} d日(s) ,d(s)\bigr)\\&&quad\leq f\bigl(s,g_{0^{+}}^{n-2}米(s) ,g{0^{+}}^{n-3}米(s) ,\ldots,g_{0^{+}}^{1} 米(s) ,米\更大)。\结束{对齐}$$
所以
$$\开始{aligned}&f\bigl(s,g_{0^{+}}^{n-2}q(s) ,g{0^{+}}^{n-3}q(s) ,\ldot,g_{0^{+}}^{1} q个(s) ,q(s)\bigr)\\&\quad\leq\hat{f}\bigl(s,g{0^{+}}^{n-2}d(s) ,g{0^{+}}^{n-3}d(s) ,\ldot,g_{0^{+}}^{1} d日(s) ,d(s)\bigr)\\&\quad\leq f \bigl(s),g_{0^{+}}^{n-2}p(s) ,g{0^{+}}^{n-3}p(s) ,\ldots,g_{0^{+}}^{1} 第页(s) ,p(s)\biger),\end{对齐}$$
这就产生了
$$开始{对齐}Q_{0^{+}}^{\beta}\iota_{p}\bigl}^{n-2}p(s) ,g{0^{+}}^{n-3}p(s) ,\ldots,g_{0^{+}}^{1} 第页(s) ,p(s)\更大)。\结束{对齐}$$
从上述讨论中,我们得出
$$开始{对齐}和Q_{0^{+}}^{\beta}\iota_{p}\bigl(Q_{0 ^{+{}}^}\alpha-n+2}n(s)\bigr)-Q_{0^{+}}^{\beta}\ieta_{p}\bigle(Q_ 0^{+}}^ \alpha-n+2}d(s)\ bigr^{n-2}p(s) ,g{0^{+}}^{n-3}p(s) ,\ldots,g_{0^{+}}^{1} 第页(s) ,p(s)\bigr)\\&\quad\quad{}-\hat{f}\bigl(s,g{0^{+}}^{n-2}d(s) ,g_{0^{+}}^{n-3}d(s) ,\ldot,g_{0^{+}}^{1} d日(s) ,d(s)\bigr)\\&&quad\geq 0,\end{aligned}$$
哪里\(在[0,1]\中).
如果我们允许\(z(s)=\iota{p}(Q_{0^{+}}^{\alpha-n+2}n(s))-\iota}p},那么\(z(0)=z(1)=0).莱玛2.1,我们有\(z(s)\leq 0\).
因此,
$$\iota{p}\bigl(Q_{0^{+}}^{\alpha-n+2}n(s)\biger)\leq\iota_{p}\ bigl$$
哪里\([0,1]\中的秒).
自\(\iota{p}\)单调增加,
$$Q_{0^{+}}^{\alpha-n+2}n(s)\leqQ_{0 ^{+{}}^}\alpha-n+2}d(s)$$
也就是说,
$$Q_{0^{+}}^{\alpha-n+2}(n-d)\leq 0$$
假设支持f̂,\(\operatorname{supp}\hat{g{\varepsilon}}\subseteq\sigma{k}(j)\cup-\sigma{k}(j)\),我们获得
$$\int_{Q_{\sigma_{k}}}\hat{f}(x-t)\hat}g{varepsilon}}(s)\,ds=0$$
哪里\(x\in-Q_{\sigma_{k}}\)、和
$$\int_{-Q_{\sigma_{k}}}(x-t)\hat{g_{varepsilon}},ds=0$$
哪里\(x\在Q_{\sigma_{k}}\中).
所以
$$\operatorname{supp}\hat{g_{\varepsilon}}\chi Q_{\sigma_{k}}\subseteq\Biggl\{\xi\在Q_{\sigma_{k{}}:\sum中^{无}_{j=1}\压裂{xi_{j}}{b_{j{}}\ge1\Biggr}$$
(2.5)
因为另一种情况可以以类似的方式获得。
让\(\lambda=\hat{f}\chi{-Q{\sigma{k}}\)和\(\varrho=\hat{g_{\varepsilon}}\chi Q_{\sigma_{k}}\).我们分解ϱ进入之内
$$\varrho=\varrho{1}+\varrho2}$$
具有\(\operatorname{supp}\varrho_{1}\subseteq\prod^{无}_{j=1}[0,b{j}]\)和\(\operatorname{supp}\varrho{2}\subseteq\上一行{Q{\sigma{k}}\setminus\prod^{n}_{j=1}(0,b_{j})}\).
由(2.5)我们获得
$$(\varrho_{1}*\lambda)(x)=-(\varrro_{2}*\lambda)$$
哪里\(x\in-Q_{\sigma_{k}}\).
与此同时
$$\operatorname{supp}(\varrho_{2}*\lambda)\subseteq\operator名称{supp}\varrho{2}+\ operatorname{supp}\slambda\subsetq\上划线{Q_{sigma_{k}}\setminus\prod^{无}_{j=1}(0,b_{j})}+产品^{无}_{j=1}[-b_{j},0]\substeq\mathbb{R}^{n}\setminus(-Q_{\sigma_{k}})$$
(2.6)
和
$$\operatorname{supp}(\varrho_{1}*\lambda)\substeq\operator名称{supp}\varrho{1}+\operatormame{supp}\slambda\substeq \prod^{无}_{j=1}[0,b{j}]+\prod^{无}_{j=1}[-b{j},0]\subseteq\prod^{无}_{j=1}[-b{j},b{j}]$$
(2.7)
由(2.6)很明显
$$\operatorname{supp}(\varrho_{1}*\lambda)\subsetq\mathbb{R}^{n}\setminus(-)Q_{\sigma_{k}}$$
这与(2.7)意味着
$$\operatorname{conv}\operator name{supp}(\varrho_{1}*\lambda)\substeq\Biggl\{xi\in\mathbb{R}^{n}:-b_{j}\le\xi_{j{}\leb_{jneneneep,\sum^{无}_{j=1}\压裂{xi{j}}{a{j}{ge1-n\Biggr}$$
(2.8)
我们声称,对于任何\(\xi\in\operatorname{supp}\varrho{1}\),
$$\总额^{无}_{j=1}\压裂{xi{j}}{a{j}{ge 1$$
持有。
如果无效,则\(\xi^{1}\in\operatorname{conv}\operator名称{supp}\varrho{1}\)令人满意的
$$\总额^{无}_{j=1}\压裂{xi{j}}{b{j}{<1$$
请注意\(\xi^{2}=b\in\operatorname{supp}\lambda\)满足
$$\总额^{无}_{j=1}\压裂{xi{j}}{b{j}=-n$$
自
$$\operatorname{conv}\operator name{supp}(\varrho_{1}*\lambda)=\operatoriname{conv}\operatorname{supp}\varrho{1}+\operatoraname{conv-}\opperator name{supp{lambda$$
存在一些问题\(\xi\in\operatorname{conv}\operator name{supp}(\varrho{1}*\lambda)\)这样的话
$$\总额^{无}_{j=1}\frac{xi_{j}}{a_{j}}<1-n$$
这与之相矛盾(2.8). 我们的结论是
$$\operatorname{supp}\varrho_{1}=\operator名称{supp}\hat{g{\varepsilon}}\chi Q_{\sigma_{k}}\subseteq\Biggl\{\xi\ in Q_{\sigma_{k{}}:\sum^{无}_{j=1}\frac{\xi_{j}}{b_{j{}}\ge1\Biggr\}$$
这就完成了证明。□