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第一节

1. 证明\(C_c^\infty( {\mathbb{ R } }^n)\)在\(L^p({ \mathbb{ R } }^n)\)和\(C^0(\mathbb{R}^n)\)中稠密.
证明. 先证明\(L^p\)的情形, 设\(u\in L^p\). 对任何\(\varepsilon>0\), 取\(R\)充分大, 使得\(\|u\|_{L^p(B_R(0)^c)}<\varepsilon/2\). 再取\(\rho>0\), 使得\(\|(u\chi_{B_R(0)})_\rho-u\chi_{B_R(0)}\|_{L^p}<\varepsilon/2\). 有了这些, 立刻可以验证\(\|u-(u\chi_{B_R(0)})_\rho\|_{L^p}<\varepsilon\), 而\((u\chi_{B_R(0)})_\rho\in C_c^\infty\), 由此即得结论.
再来证\(C^0(\mathbb{R}^n)=\{u\in C(\mathbb{R}^n)|\lim_{|x|\rightarrow\infty}u(x)=0\}\)的情形. 和上面差不多, 不过直接截断换为光滑截断比较好. 即取\(R\)充分大使得\(\|u-u\chi_R\|_{L^\infty}<\varepsilon/2\), 这里\(\chi_R=(\chi_{B_R(0)})_{1}\)是球\(B_R(0)\)的示性函数做半径为1(1是瞎选的, 10也行)的正则化. 再注意\(\|(u\chi_R)_\rho-u\chi_R\|_{L^\infty}\le\rho\|\nabla(u\chi_R)\|_{L^\infty}\), 我们可以取\(\rho\)充分小, 使得\(\|(u\chi_R)_\rho-u\chi_R\|_{L^\infty}<\varepsilon/2\). 现在立刻可以验证\(\|u-(u\chi_R)_\rho\|_{L^\infty}<\varepsilon\), 由此即得结论. \(\blacksquare\)

2. 证明: 若\(u\in C^\infty(\mathbb{R}^n)\), 则\(J_\varepsilon u\rightarrow u(C^\infty(\mathbb{R}^n))\); 若\(u\in C_c^\infty({\mathbb{R}}^n)\), 则\(J_\varepsilon u\rightarrow u(C_c^\infty( { \mathbb{ R } }^n))\), 其中\(J_\varepsilon u\)是\(u\)的正则化.
证明. 我们只需要证明\(C^\infty( {\mathbb{R}}^n)\)的情形, 因为如果\(u\in C_c^\infty\), 那么正则化\(u_\rho\)在\(\rho<1\)时当然是具有一致的紧支集的, 例如\(\overline{\operatorname{supp}u+B_1(0)}\).
下面我们设\(u\in C^\infty(\mathbb{R}^n)\), 我们需要证\(\forall\alpha\in\mathbb{N}^n\), \(\forall R>0\), 有\(\|\partial^\alpha u_\rho-\partial^\alpha u\|_{L^\infty(B_R(0))}\rightarrow0\). 而简单的计算表明\(\|\partial^\alpha u_\rho-\partial^\alpha u\|_{L^\infty(B_R(0))}\le\rho\|\nabla\partial^\alpha u\|_{L^\infty(B_{R+\rho}(0))}\), 由此即得结论. \(\blacksquare\)

3. 证明\(C_c^\infty( \mathbb{ R }^n )\)是完备的.
证明. 设\(\{u_m\}\)是\(C_c^\infty\)基本列, 即存在紧集\(K\), 使得\(\forall m\ge1\), \(\operatorname{supp}u_m\subset K\), 且\(\forall\alpha\in\mathbb{N}^n\), \(\|\partial^\alpha u_k-\partial^\alpha u_l\|_{L^\infty(K)}\rightarrow0\). 由条件, 存在\(u^\alpha\in C(\mathbb{R}^n)\)使得在\(\mathbb{R}^n\)上\(\partial^\alpha u_m\rightrightarrows u^\alpha\), 且\(\operatorname{supp}u^\alpha\subset K\).
记\(u=u^0\). 现在我们来验证\(\partial^\alpha u=u^\alpha\). 事实上由归纳法, 我们只需要验证\(\partial_ju=u^{e_j}\). 注意\(u_m(x+te_j)=u_m(x)+\int_0^t\partial_ju_m(x+se_j)ds\), 由\(u_m,\partial_ju_m\)的一致收敛性得到\(u(x+te_j)=u(x)+\int_0^tu^{e_j}(x+se_j)ds\), 于是可以利用导数定义验证\(\partial_ju(x)=u^{e_j}(x)\). 这样我们就验证好了\(\partial^\alpha u=u^\alpha\), 所以\(u\in C_c^\infty(\mathbb{R}^n)\), 且\(\operatorname{supp}u\subset K\).
最后我们需要验证\(u_m\rightarrow u(C_c^\infty(\mathbb{R}^n))\). 共同的支集已经验证好了, 现在我们需要说明\(\|\partial^\alpha u_m-\partial^\alpha u\|_{L^\infty(K)}\rightarrow0\), 而这就是之前已经说明了的\(\partial^\alpha u_m\rightrightarrows u^\alpha=\partial^\alpha u\). \(\blacksquare\)

4. 证明速降函数关于卷积运算封闭.
证明. 设\(u,v\in\mathscr{S}(\mathbb{R}^n)\), 我们来验证\(u\ast v\in{ \mathscr{S}}(\mathbb{R}^n)\).
\(u\ast v\in C^\infty( \mathbb{ R } ^n )\)是容易的, 我们只需要验证\(\forall m\ge1\), \(\forall\alpha\in\mathbb{N}^n\), 有\(\|(1+|x|)^m\partial^\alpha(u\ast v)\|_{L^\infty}<\infty\)即可.
注意\(1+|x|\le(1+|x-y|)(1+|y|)\), 我们可以直接计算:

\[\left|(1+|x|)^m\partial^\alpha(u\ast v)(x)\right|=(1+|x|)^m\left|\int u(x-y)\partial^\alpha v(y)dy\right| \]

\[\le\int (1+|x-y|)^m|u(x-y)|\cdot(1+|y|)^m|\partial^\alpha v(y)|dy \]

\[\le\|(1+|x|)^mu\|_{L^2}\|(1+|x|)^m\partial^\alpha v\|_{L^2}<\infty \]

这就是我们想要的. \(\blacksquare\)

注. \(\{\|(1+|x|)^m\partial^\alpha\varphi\|_{L^\infty}\}_{m,\alpha}\), \(\{\|(1+|x|^2)^m\partial^\alpha\varphi\|_{L^\infty}\}_{m,\alpha}\), \(\{\|(1+|x|^2)^{m/2}\partial^\alpha\varphi\|_{L^\infty}\}_{m,\alpha}\), \(\{\|x^\alpha\partial^\beta\varphi\|_{L^\infty}\}_{\alpha,\beta}\), \(\{\|x^\alpha\partial^\beta\varphi\|_{L^p}\}_{\alpha,\beta}\)(\(1\le p\le\infty\))等等均是\(\mathscr{S}(\mathbb{R}^n)\)的等价半范数族, 它们诱导出同样的拓扑. 所以平时哪个方便用哪个.

5. 问: 是否存在\(0\not\equiv\varphi\in C_c^{\infty}({\mathbb{R}}^n)\), 使得存在一常数\(C\)与\(\alpha\)无关, \(\|\partial^\alpha\varphi\|_{L^\infty}\leq C\).
解. 不存在. 反证法, 假设存在这样的\(\varphi\), 我们来导出矛盾.
由条件, 我们可以得知\(\varphi\in C^\omega(\mathbb{R}^n)\), 这只消对Taylor定理里的余项简单估计一下即得. 取\(x_0\in(\operatorname{supp}\varphi)^c\), 由于\(\varphi\)在\(x_0\)处的各阶导数都是0, 所以把\(\varphi\)在\(x_0\)处展开即可得到\(\varphi\equiv0\), 矛盾. \(\blacksquare\)

6. 证明任一具有\(C^\infty( {\mathbb{R}}^n)\)系数的微分算子\(P(x,\partial)\)都是\(C_c^\infty( \mathbb{ R } ^n)\rightarrow C_c^\infty( \mathbb{R}^n )\)或\(C^\infty(\mathbb{R}^n)\rightarrow C^\infty(\mathbb{R}^n)\)的线性连续映照.
证明. 注意到\(P(x,\partial)=\sum_{|\alpha|\le\deg P}c_\alpha(x)\partial^\alpha\), 我们只需要证明两件事情: ①乘一个\(C^\infty\)函数是连续线性映射; ②求导是连续线性映射. 无需赘言, 加法对拓扑线性空间来说自然是连续的.
由于乘光滑函数和求导保持支集不变, 我们只需要证明这两个操作在\(C^\infty({\mathbb{R}}^n)\)上是连续的就好了. 对于线性映射来说, 验证它连续, 只需要验证它在原点处连续.
①设\(c\in C^\infty(\mathbb{R}^n)\), 我们来证\(\varphi\mapsto c\varphi\)是\(C^\infty({\mathbb{R}}^n)\)到\(C^\infty({\mathbb{R}}^n)\)的连续映射(线性是明显的). 取\(\varphi_m\rightarrow0(C^\infty( \mathbb{R}^n))\), 我们只需证明\(c\varphi_m\rightarrow0(C^\infty( \mathbb{R}^n))\). 或者说, 我们只需证明对任何紧集\(K\), \(\forall\alpha\in\mathbb{N}^n\), 有\(\|\partial^\alpha(c\varphi)\|_{L^\infty(K)}\rightarrow0\). 而\(\|\partial^\alpha(c\varphi)\|_{L^\infty(K)}\le\sum_{\beta\le\alpha}\binom{\alpha}{\beta}\|\partial^{\alpha-\beta}c\|_{L^\infty(K)}\|\partial^\beta\varphi\|_{L^\infty(K)}\), 即\(\|\partial^\alpha(c\varphi)\|_{L^\infty(K)}\lesssim_{c,\alpha}\sum_{\beta\le\alpha}\|\partial^\beta\varphi\|_{L^\infty(K)}\), 由于\(\|\partial^\beta\varphi\|_{L^\infty(K)}\)们趋于0, 故\(\|\partial^\alpha(c\varphi)\|_{L^\infty(K)}\rightarrow0\).
②我们来证\(\partial^\alpha:C^\infty(\mathbb{R}^n)\rightarrow C^\infty(\mathbb{R}^n)\)是连续线性映射. 线性是明显的. 为证连续, 任取\(\varphi_m\rightarrow0(C^\infty( \mathbb{R}^n))\), 来证\(\partial^\alpha\varphi_m\rightarrow0(C^\infty( \mathbb{R}^n))\). 这按定义验证是非常直接的. \(\blacksquare\)

7. 证明空间\(\mathscr{S}(\mathbb{R}^n)\)按其自身拓扑是完备的.
证明. 取\(\mathscr{S}(\mathbb{R}^n)\)中的Cauchy列\(\{u_m\}\), 即\(\forall\alpha,\beta\in\mathbb{N}^n\), \(p_{\alpha\beta}(u_k-u_l)=\|x^\alpha\partial^\beta(u_k-u_l)\|_{L^\infty}\rightarrow0\).
取\(\alpha=0\)即可知\(\{u_m\}\)亦是\(C^\infty({\mathbb{R}}^n)\)中的Cauchy列, 故存在\(u\in C^\infty(\mathbb{R}^n)\)使得\(u_m\rightarrow u(C^\infty({\mathbb{R}}^n))\).
我们来验证\(u\in\mathscr{S}(\mathbb{R}^n)\). 取\(N\ge1\), 使得当\(k,l\ge N\)时\(p_{\alpha\beta}(u_k-u_l)\le1\). 则固定\(x\in\mathbb{R}^n\), 有\(|x^\alpha\partial^\beta(u_k-u_l)(x)|\le1\), 由于\(\partial^\beta u_l(x)\rightarrow\partial^\beta(x)\), 我们令\(l\rightarrow\infty\), 有\(|x^\alpha\partial^\beta(u_k-u)(x)|\le1\). 故\(|x^\alpha\partial^\beta u(x)|\le1+|x^\alpha\partial^\beta u_k(x)|\le1+p_{\alpha\beta}(u_k)\), 考虑到\(x\)的任意性, 我们即有\(p_{\alpha\beta}(u)\le1+p_{\alpha\beta}(u_k)<\infty\). 故\(u\in\mathscr{S}(\mathbb{R}^n)\).
最后我们验证\(u_m\rightarrow u(\mathscr{S}(\mathbb{R}^n))\), 即\(\forall\alpha\forall\beta\), \(p_{\alpha\beta}(u_m-u)\rightarrow0\). 和上一段差不多, 对任何\(\varepsilon>0\), 取\(N\ge1\)使得\(\forall k,l\ge N\), 有\(p_{\alpha\beta}(u_k-u_l)<\varepsilon\), 然后用和上一段类似的论证得到\(p_{\alpha\beta}(u_k-u)\le\varepsilon\). \(\blacksquare\)

8. 设\(u\in\mathscr{S}(\mathbb{R}^n)\), 证明\(u_\varepsilon\rightarrow u(\mathscr{S}(\mathbb{R}^n))\).
证明. 我们要证明\(p_{m\beta}(u_\varepsilon-u)=\|(1+|x|)^m\partial^\alpha(u_\varepsilon-u)\|_{L^\infty}\rightarrow0\). 事实上我们只需要对\(\beta=0\)证明就够了, 因为\(p_{m\beta}(u_\varepsilon-u)=p_{m,0}((\partial^\beta u)_\varepsilon-\partial^\beta u)\), 而\(\partial^\beta u\)也是速降函数.
我们直接计算:

\[\left|(1+|x|)^m(u_\varepsilon-u)(x)\right|=(1+|x|)^m\left|\int (u(x-y)-u(x))\eta_\varepsilon(y)dy\right| \]

\[\le\varepsilon(1+|x|)^m\|\nabla u\|_{L^\infty(B_\varepsilon(x))}\overset{\text{设}\varepsilon<1}{\le}\varepsilon(1+|x|)^m\|\nabla u\|_{L^\infty(B_1(x))} \]

对任何\(y\in B_1(x)\), 利用\(1+|x|\le(1+|y|)(1+|x-y|)\), 有

\[|\nabla u(y)|=(1+|y|)^{-k}\cdot(1+|y|^{k})|\nabla u(y)| \]

\[\le(1+|x|)^{-k}(1+|x-y|)^k\|(1+|y|^k)\nabla u(y)\|_{L^\infty_y} \]

\[\le2^k(1+|x|)^k\sum_{j=1}^np_{k,e_j}(u) \]

这里\(k\)是随意的正整数, 现在我们取\(k=m\), 代入之前的估计, 即得

\[p_{m,0}(u_\varepsilon-u)\le2^m\varepsilon\sum_{j=1}^np_{m,e_j}(u)\overset{\varepsilon\rightarrow0}{\rightarrow}0 \]

这正是我们想要的. \(\blacksquare\)

9. 证明\(\varphi_m\rightarrow0(\mathscr{S}(\mathbb{R}^n))\)与下面两条件之一等价:
(1) 对任一重指标\(\alpha\), \(\|\partial^\alpha\varphi_m\|_{L^\infty}\rightarrow0\), 且对任何重指标\(\alpha,\beta\), \(\|x^\alpha\partial^\beta\varphi_m\|_{L^\infty}\le C_{\alpha,\beta}\).
(2) 对任一正整数\(\nu\), 在\(m\rightarrow\infty\)时, 有\(\sup_{|\alpha|\le\nu,x\in\mathbb{R}^n}(1+|x|^2)^\nu|\partial^\alpha\varphi_m|\rightarrow0\).
证明. 由第4题下面的注记, \(\varphi_m\rightarrow0(\mathscr{S}(\mathbb{R}^n))\)推出(1)或(2)是容易的(当然前提是证明了那些确实是等价范数, 至于等价范数的证明, 可以利用\((1+|x|^2)^\nu\approx_{\nu}\sum_{|\alpha|\le\nu}|x^\alpha|\)). 所以我们只需要分别证明(1)推出收敛, 以及(2)推出收敛即可. 但是(2)推收敛也是因为等价范数, 所以我们现在只需要从(1)推收敛.
假设(1)成立, 则\(\|x^\alpha\partial^\beta\varphi\|_{L^\infty}^2=\|x^{2\alpha}|\partial^\beta\varphi|^2\|_{L^\infty}\) \(\le\|x^{2\alpha}\partial^\beta\varphi\|_{L^\infty}\|\partial^\beta\varphi\|_{L^\infty}\) \(\le C_{2\alpha,\beta}\|\partial^\beta\varphi\|_{L^\infty}\rightarrow0\), 这就是我们需要的. \(\blacksquare\)

10. 设\(\varphi\in C_c^\infty(B_1(0))\), \(x_j\)为趋于\(\infty\)的一个点列, \(|x_{j+1}|>|x_j|+2\), 试证

\[\gamma(x):=\sum_{j=1}^\infty\frac{\varphi(x-x_j)}{(1+|x_j|^2)^j }\in\mathscr{S}(\mathbb{R}^n). \]

证明. 由于上边这个求和是局部有限的, 所以很容易验证\(\gamma\in C^\infty(\mathbb{R}^n)\), 且\(\gamma(x):=\sum_{j=1}^\infty\frac{{\partial^\alpha}\varphi(x-x_j)}{(1+|x_j|^2)^j }\). 现在我们只需要验证对任何\(m\ge0\), \(\forall\alpha\in\mathbb{N}^n\), 有\(\|(1+|x|^2)^m\partial^\alpha\gamma\|_{L^\infty}<\infty\)即可.
注意\(\varphi(x-x_j)\)的支集两两不交, 所以我们有:

\[|(1+|x|^2)^m\partial^\alpha\gamma(x)|=(1+|x|^2)^m \sum_{j=1}^\infty\frac{|\partial^\alpha\varphi(x-x_j)|}{(1+|x_j|^2)^j } \]

\[\le\sup_{j\ge1}\left\|(1+|x|^2)^m\frac{ |\partial^\alpha\varphi(x-x_j)|}{(1+|x_j|^2)^j }\right\|_{L^\infty_x(B_1(x_j))} \]

\[\le\sup_{j\ge1}(1+(|x_j|+1)^2)^m(1+|x_j|^2)^{-j}\|\partial^\alpha\varphi\|_{L^\infty} \]

\[\le3^m\sup_{j\ge1}(1+|x_j|^2)^{m-j}\|\partial^\alpha\varphi\|_{L^\infty}\lesssim_{m,\{x_j\}}\|\partial^\alpha\varphi\|_{L^\infty} \]

所以\(\|(1+|x|^2)^m\partial^\alpha\gamma\|_{L^\infty}\lesssim_{m,\{x_j\}}\|\partial^\alpha\varphi\|_{L^\infty}<\infty\). 这就证好了. \(\blacksquare\)

第二节

例 2.5. 设\(f_\nu(x)=\frac{1}{\pi}\frac{\sin\nu x}{x}\), 证明\(\nu\rightarrow\infty\)时, \(f_\nu\rightarrow\delta(\mathscr{D}'(\mathbb{R}))\).
证明. 只需任取\(\varphi\in C_c^\infty(\mathbb{R})\), 证明\(\langle f_\nu,\varphi\rangle\rightarrow\varphi(0)\)即可.
直接计算, 设\(\operatorname{supp}\varphi\subset[-R,R]\), 并且任取\(\varepsilon>0\), 有:

\[\langle f_\nu,\varphi\rangle=\frac{1}{\pi}\int_{|x|\le R}\frac{\sin\nu x}{x}\varphi(x)dx \]

\[=\frac{1}{\pi}\int_{\varepsilon\le|x|\le R}\frac{\sin\nu x}{x}\varphi(x)dx+\frac{1}{\pi}\int_{|x|\le\varepsilon}\frac{\sin\nu x}{x}\varphi(x)dx \]

\[=A+B \]

\(A\)在\(\nu\rightarrow\infty\)由Riemann-Lebesgue定理是趋于零的, 现在我们来处理\(B\).

\[|B-\varphi(0)|=\left|\frac{1}{\pi}\int_{|x|\le\varepsilon}\frac{\sin\nu x}{x}\varphi(x)dx-\varphi(0)\right| \]

\[\overset{y=\nu x}{=}\left|\frac{2}{\pi}\int_0^{\nu\varepsilon}\frac{\sin y}{y}\varphi(y/\nu)dy-\frac{2}{\pi}\int_0^{+\infty}\frac{\sin y}{y}\varphi(0)dy\right| \]

\[\le\frac{2}{\pi}\left|\int_0^{\nu\varepsilon}\frac{\sin y}{y}(\varphi(y/\nu)-\varphi(0))dy\right|+\left|\frac{2}{\pi}\int_{\nu\varepsilon}^{+\infty}\frac{\sin y}{y}\varphi(0)dy\right| \]

\[\le\frac{2}{\pi}\int_0^{\nu\varepsilon}\frac{1}{y}\|\varphi'\|_{L^\infty}\frac{y}{\nu}dy+o_{\nu\rightarrow\infty}(1) \]

\[=\frac{2 \varepsilon\|\varphi'\|_{L^\infty}}{\pi}+o_{\nu\rightarrow\infty}(1) \]

故\(\limsup_{\nu\rightarrow\infty}|B-\varphi(0)|\lesssim_\varphi\varepsilon\), 可是\(\varepsilon\)是任意的, 所以只能有\(\lim_{\nu\rightarrow\infty}B=\varphi(0)\). 所以\(\langle f_\nu,\varphi\rangle\rightarrow\varphi(0)\). \(\blacksquare\)

定理2.9 证明的修正:
首先重新陈述一遍定理: 设\(\Omega\)是\(\mathbb{R}^n\)中的连通开集(即开区域), \(T\in\mathscr{D}'(\Omega)\), 且\(T\)的各阶导数为零, 则\(T\)是常数.
书上为了证明\(T\)是常数, 要先证明\(T_\varepsilon\)是常数. 而要得到\(T_\varepsilon\)是常数, 我们需要\(\Omega_\varepsilon=\{x\in\Omega|\mathrm{dist}(x,\partial\Omega)>\varepsilon\}\)是连通的, 但这是不一定对的. 事实上我们可以构造出一个开区域\(\Omega\), 使得\(\forall\varepsilon>0\), \(\Omega_\varepsilon\)都不是连通的.
这里试图给出一个修正了的证明.
对任何\(x\in\Omega\), 令\(B_x=B(x,\mathrm{dist}(x,\partial\Omega)/2)\). 我们来证明, 对任何\(x\in\Omega\), 存在\(\lambda_x\in\mathbb{C}\), 使得对任何\(\varphi\in C_c^\infty(B_x)\), 有\(\langle T,\varphi\rangle=\lambda_x\int_\Omega\varphi dx\).
这是容易的, 因为我们可以取\(\rho<\mathrm{dist}(x,\partial\Omega)/2\), 考虑正则化\(T_\rho\in C^\infty(B_x)\), 由于\(T_\rho\)各阶导数为零, 故\(T_\rho\)在\(B_x\)上是常数, 比方说是\(\lambda_{x,\rho}\). 由于\(\rho\rightarrow0\)时\(T_{\rho}\rightarrow T(\mathscr{D'}(B_x))\), 故\(\lim_{\rho\rightarrow0}\lambda_{x,\rho}\)必然存在, 记为\(\lambda_x\). 可以验证\(\lambda_x\)具有我们所说的性质.
注意到一个简单的事实: 如果\(B_x\cap B_y\ne\emptyset\), 则\(\lambda_x=\lambda_y\). 这任取\(\varphi\in C_c^\infty(B_x\cap B_y)\)作用上\(T\)即可验证.
接下来, 我们证明对任何\(x,y\in\Omega\), 均有\(\lambda_x=\lambda_y\). 由于\(\Omega\)是连通的(自然也是道路连通的), 故存在连续映射\(\gamma:[0,1]\rightarrow\Omega\), 使得\(\gamma(0)=x\), \(\gamma(1)=y\). 由于\(\gamma([0,1])\)是紧的, 所以存在\(\{x_i\}_{i=1}^N\subset\Omega\), 使得\(\gamma([0,1])\subset\cup_i B_{x_i}\). 注意\(\{\gamma^{-1}(B_{x_i})\}\)是\([0,1]\)的开覆盖, 取这个开覆盖的Lebesgue数\(L\), 则当\(1/M<L\)时, \([0,1]\)的\(M\)等分\(0=t_0<t_1<\cdots<t_M=1\)满足: 相邻的每两个点\(t_j\)和\(t_{j+1}\)会落在同一个某个\(\gamma^{-1}(B_{x_i})\)里. 显然\(B_{\gamma(t_j)}\cap B_{x_i}\ne\emptyset\), \(B_{\gamma(t_{j+1})}\cap B_{x_i}\ne\emptyset\), 于是根据上一段的讨论有\(\lambda_{\gamma(t_j)}=\lambda_{x_i}=\lambda_{\gamma(t_{j+1})}\). 最后我们得到\(\lambda_x=\lambda_{\gamma(t_0)}=\cdots=\lambda_{\gamma(t_M)}=\lambda_y\).
现在我们知道存在\(\lambda\in\mathbb{C}\), 使得\(\forall x\in\Omega\), 有\(\lambda_x=\lambda\). 我们现在来证明\(\forall\varphi\in C_c^\infty(\Omega)\), 有\(\langle T,\varphi\rangle=\lambda\int_\Omega\eta dx\).
注意\(\{B_x\}_{x\in\Omega}\)是紧集\(\operatorname{supp}\varphi\)的开覆盖, 故存在有限子覆盖\(\{B_{x_i}\}\), 然后取附属于该子覆盖的单位分解\(\eta_i\in C_c^\infty(B_{x_i})\), 满足在\(\operatorname{supp}\varphi\)上\(\sum_i\eta_i=1\). 则\(\langle T,\varphi\rangle\) \(=\sum_i\langle T,\varphi\eta_i\rangle=\sum_i\lambda_{x_i}\int_\Omega\varphi\eta_i dx\) \(=\lambda\sum_i\int_\Omega\varphi\eta_i dx=\lambda\int_\Omega\varphi dx\). \(\blacksquare\)

1. 问下列函数属于哪个广义函数空间:
(1) \(f(x)=e^x\cos x\);
(2) \(f(x)=\left\{\begin{aligned}&1,\ |x|\le1\\ &0,\ |x|>1\end{aligned}\right.\).
(3) \(\langle f,\varphi\rangle=\sum_{\nu=1}^m\varphi^{(\nu)}(0)\), \(\forall\varphi\in C^m(\mathbb{R}^1)\).
解. (1)\(\mathscr{D}'(\mathbb{R}^1)\); (2)\(\mathscr{E}'(\mathbb{R}^1)\); (3)\(\mathscr{E}'(\mathbb{R}^1)\). \(\blacksquare\)

2. 设\(u(x)=\left\{\begin{aligned}&1,\ |x|\le1\\ &0,\ |x|>1\end{aligned}\right.\), 试问\(u\)的支集是什么? 它的正则化\(u_\varepsilon\)的支集是什么?
解. \(\operatorname{supp}u=[-1,1]\), \(\operatorname{supp}u_\varepsilon=[-1-\varepsilon,1+\varepsilon]\). \(\blacksquare\)

3. 证明: 若\(f_m\in C({\mathbb{R}})\), \(f_m\ge0\), 且\(\forall r>0\), \(\forall R>r\), \(f_m\)在\([r,R]\)和\([-R,-r]\)上一致趋于零, 并使\(\int_{-r}^r f_m(x)dx\rightarrow1\), 则\(m\rightarrow\infty\)时\(f_m\rightarrow\delta(\mathscr{D}'(\mathbb{R}))\).
证明. 任取\(\varphi\in C_c^\infty(\mathbb{R})\), 我们来证明\(\langle f_m,\varphi\rangle\rightarrow\varphi(0)\). 固定\(\varepsilon>0\), 取\(r>0\)使得\(\forall x\in[-r,r]\)有\(|\varphi(x)-\varphi(0)|<\varepsilon\). 又取\(R\)充分大使得\(R>r\)并且\(\operatorname{supp}\varphi\subset[-R,R]\). 我们来计算:

\[\left|\int_{\mathbb{R}}f_m(x)\varphi(x)dx-\varphi(0)\right| \]

\[\le\left|\int_{|x|\le r}f_m(x)\varphi(x)dx-\varphi(0)\right|+\left|\int_{|x|\ge r}f_m(x)\varphi(x)dx\right| \]

\[=A+B \]

一个个算:

\[A\le\left|\int_{|x|\le r}f_m(x)(\varphi(x)-\varphi(0))dx\right|+\left|\int_{|x|\le r}f_m(x)\varphi(0)dx-\varphi(0)\right| \]

\[\le\varepsilon\int_{|x|\le r}f_m(x)dx+|\varphi(0)|\left|\int_{|x|\le r}f_m(x)dx-1\right| \]

\[=\varepsilon(1+o_{m\rightarrow\infty}(1))+o_{m\rightarrow\infty}(1)=\varepsilon+o_{m\rightarrow\infty}(1) \]

再来看\(B\):

\[B\le\|\varphi\|_{L^\infty}\left|\int_{r\le|x|\le R}f_m(x)dx\right|=o_{m\rightarrow\infty}(1) \]

故\(\limsup_{m\rightarrow\infty}\left|\int_{\mathbb{R} }f_m(x)\varphi(x)dx-\varphi(0)\right|\le\varepsilon\), 由\(\varepsilon\)的任意性即得结论. \(\blacksquare\)

4. 证明在广义函数意义下下列极限成立:
(1) \(\lim_{\varepsilon\rightarrow0}\frac{\varepsilon}{\varepsilon^2+x^2}=\pi\delta(x)\);
(2) \(\lim_{\varepsilon\rightarrow0}\frac{1}{\sqrt{\pi\varepsilon}}e^{-\frac{x^2}{\varepsilon}}=\delta(x)\)
证明. 我们统一处理这两个极限.
假设\(f\in C^\infty(\mathbb{R})\)满足: ①\(f\ge0\); ②\(\int f=1\). 令\(f_{\varepsilon}(x)=\frac{1}{\varepsilon}f(\frac{x}{\varepsilon})\), 我们来证明\(f_\varepsilon\rightarrow\delta(\mathscr{D}'({\mathbb{R}}))\).
任取\(\varphi\in C_c^\infty(\mathbb{R})\), 我们来证明\(\int f_\varepsilon\varphi\rightarrow\varphi(0)\).

\[\left|\int_{ \mathbb{R}}f_\varepsilon(x)\varphi(x)dx-\varphi(0)\right|\le\frac{1}{\varepsilon}\int_{\mathbb{R}}f\left(\frac{x}{\varepsilon}\right)|\varphi(x)-\varphi(0)|dx \]

\[\overset{y=x/\varepsilon}{=}\int_{\mathbb{R}}f(y)|\varphi(\varepsilon y)-\varphi(0)|dy\le\varepsilon\|\varphi'\|_{L^\infty}\operatorname{diam}(\operatorname{supp}\varphi) \]

从而\(\limsup_{\varepsilon\rightarrow0}\left|\int f_\varepsilon\varphi-\varphi(0)\right|=0\). \(\blacksquare\)

5. 设\(f_\varepsilon(x)=(x+i\varepsilon)^{-1}+(x-i\varepsilon)^{-1}\), 证明\(\lim_{\varepsilon\rightarrow0}f_\varepsilon\)在\(\mathscr{D}'(\mathbb{R})\)中存在
证明1. 本来我们最好是证明\(f_{\varepsilon}\rightarrow 2P.V.(\frac{1}{x})\), 但是我们偷个懒. 我们只需要任取\(\varphi\in C_c^\infty(\mathbb{R})\), 证明 \(\lim_{\varepsilon\rightarrow0}\langle f_\varepsilon,\varphi\rangle\)存在就行了. 这是因为根据Rudin的泛函分析书里的定理2.7, 如果\(\Lambda_k\)是某拓扑线性空间\(X\)上的连续线性泛函, 并且\(\forall x\in X\), \(\lim_k\Lambda_kx=\Lambda x\)存在, 那么\(\Lambda\)也是\(X\)上的连续线性泛函.
任取\(\varepsilon_1,\varepsilon_2>0\), 我们来估计\(|\langle f_{\varepsilon_1}-f_{\varepsilon_2},\varphi\rangle|\). 注意\(f_\varepsilon\)是奇函数, 并且\(f_{\varepsilon_1}-f_{\varepsilon_2}\in L^1(\mathbb{R})\), 所以\(\int(f_{\varepsilon_1}-f_{\varepsilon_2})=0\). 从而我们有

\[|\langle f_{\varepsilon_1}-f_{\varepsilon_2},\varphi\rangle|=\left|\int(f_{\varepsilon_1}-f_{\varepsilon_2})\varphi dx\right| \]

\[=\left|\int(f_{\varepsilon_1}-f_{\varepsilon_2})(\varphi-\varphi(0))dx\right|\le\|\varphi'\|_{L^\infty}\int|x||f_{\varepsilon_1}-f_{\varepsilon_2}|dx \]

\[\le 2\|\varphi'\|_{L^\infty}|\varepsilon_1^2-\varepsilon_2^2|\int_0^\infty\frac{2x^2}{(x^2+\varepsilon_1^2)(x^2+\varepsilon_2^2) }dx \]

\[=4\|\varphi'\|_{L^\infty}\int_0^\infty\left|\frac{\varepsilon_1^2}{x+\varepsilon_1^2}-\frac{\varepsilon_2^2}{x+\varepsilon_2^2}\right|dx \]

\[=4\|\varphi'\|_{L^\infty}\left| \int_0^\infty\frac{\varepsilon_1^2}{x+\varepsilon_1^2}dx-\int_0^\infty \frac{\varepsilon_2^2}{x+\varepsilon_2^2}dx\right| \]

\[=2\pi\|\varphi'\|_{L^\infty}|\varepsilon_1-\varepsilon_2| \]

由此即可得到\(\lim_{\varepsilon\rightarrow0}\langle f_\varepsilon,\varphi\rangle\)是存在的. \(\blacksquare\)
证明2. 我们直接来证明\(f_{\varepsilon}\rightarrow 2P.V.(\frac{1}{x})\). 取定\(\varphi\in C_c^\infty(\mathbb{R})\). 对任何\(h>0\), 我们取\(\delta\in(0,1)\)使得\(\operatorname{supp}\varphi\subset[-\delta^{-1},\delta^{-1}]\), 并且\(2\delta\|\varphi'\|_{L^\infty}<h/3\), 以及

\[\left|\langle 2P.V.\frac{1}{x},\varphi\rangle-\int_{\delta\le|x|\le\delta^{-1}}\frac{2\varphi(x)}{x}dx\right|<\frac{h}{3} \]

再取\(\varepsilon\)充分小, 使得\(\frac{2\varepsilon^2}{\delta^4}\|\varphi\|_{L^\infty}< h/3\). 在取定此\(\delta,\varepsilon\)之后, 我们来计算:

\[| \langle f_\varepsilon,\varphi\rangle-\langle 2P.V.\frac{1}{x},\varphi\rangle|\le\left|\langle f_\varepsilon,\varphi\rangle-\int_{\delta\le|x|\le\delta^{-1}}\frac{2\varphi(x)}{x}dx\right| \]

\[+ \left|\langle 2P.V.\frac{1}{x},\varphi\rangle-\int_{\delta\le|x|\le\delta^{-1}}\frac{2\varphi(x)}{x}dx\right| \]

\[\le\left|\int_{\delta\le|x|\le\delta^{-1}}\left(\frac{2x}{x^2+\varepsilon^2}-\frac{2}{x}\right)\varphi(x)dx\right|+\left|\int_{|x|\le\delta}\frac{2x}{x^2+\varepsilon^2}\varphi(x)dx\right|+\frac{h}{3} \]

\[\le\|\varphi\|_{L^\infty}\int_{\delta\le|x|\le\delta^{-1}}\frac{2\varepsilon^2}{|x|( x^2+\varepsilon^2)}dx \]

\[+\left|\int_{|x|\le\delta}\frac{2x}{x^2+\varepsilon^2}(\varphi(x)-\varphi(0))dx\right|+\frac{h}{3} \]

\[\le\|\varphi\|_{L^\infty}\frac{2\varepsilon^2}{\delta^4}+\|\varphi'\|_{L^\infty}\int_{|x|\le\delta}\frac{2x^2}{x^2+\varepsilon^2}dx+\frac{h}{3} \]

\[\le\frac{2h}{3}+2\delta\|\varphi'\|_{L^\infty}<h \]

所以\(\lim_{\varepsilon\rightarrow0}\langle f_\varepsilon,\varphi\rangle=\langle 2P.V.(1/x),\varphi\rangle\). \(\blacksquare\)

标签：mathbb,infty,partial,varepsilon,陈恕行,varphi,alpha,习题,参考答案
来源： https://www.cnblogs.com/dxww/p/14075907.html