<ruby id="d9npn"></ruby>

<sub id="d9npn"><progress id="d9npn"></progress></sub>

<nobr id="d9npn"></nobr>

<rp id="d9npn"><big id="d9npn"><th id="d9npn"></th></big></rp>

<th id="d9npn"><meter id="d9npn"></meter></th>

 2 added 671 characters in body edited Feb 22 at 13:38 Abdelmalek Abdesselam 11.8k11 gold badge2929 silver badges7777 bronze badges It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $$\omega|_{\Gamma}$$ of $$\omega$$ to some subset $$\Gamma\subset\mathbb{R}^n$$. When one succeeds the result is called a trace theorem. This usually requires some quantitative regularity hypothesis on $$\omega$$, e.g., being in a Sobolev space $$H^s$$ with $$s>$$something. A particularly important case is when $$\Gamma$$ is an affine subspace, or say for simplicity a linear subspace like $$\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$$. A rather standard approach is to start with $$\omega\in\mathcal{D}'(\mathbb{R}^n)$$. The convolution $$\omega\ast \phi_j^0$$ is in the space of $$C^{\infty}$$ functions $$\mathcal{E}(\mathbb{R}^n)\subset \mathcal{D}'(\mathbb{R}^n)$$ and converges to $$\omega$$ in the topology of $$\mathcal{D}'(\mathbb{R}^n)$$ (the strong topology). The ordinary restriction $$\omega\ast \phi_j^0|_{\Gamma}$$ makes sense and you can ask if the limit $$\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$$ exists inside $$\mathcal{D}'(\mathbb{R}^m)$$. Your particular case $$p=0$$ corresponds to mine with $$m=0$$. Another problem of this kind is pointwise multiplication. If $$\omega_1(x)$$ and $$\omega_2(x)$$ are two distributions, then there is no problem defining $$\omega_1(x_1)\omega_2(x_2)$$ (tensor product), but the issue is how to restrict to the diagonal $$\Gamma=\{x_1=x_2\}$$. Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can sometimes do the (deterministically) impossible. Small addendum: Suppose that for some reason one has a trace theorem but only for large enough $$m$$ and one cannot do the $$m=0$$ or the point restriction case. Then one can still do the following "stabilization" trick: change $$\omega$$ to $$\omega\otimes 1$$ where one tensors with the constant function equal to one seen as a distribution in say $$p$$ new variables. If you can restrict it from $$\mathbb{R}^{n+p}$$ to a subspace of dimension $$p$$, then you will have your point evaluation after factoring out the $$\otimes 1$$. The last step of course needs your restriction construction to be invariant/covariant by translation along $$\Gamma$$. It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $$\omega|_{\Gamma}$$ of $$\omega$$ to some subset $$\Gamma\subset\mathbb{R}^n$$. When one succeeds the result is called a trace theorem. This usually requires some quantitative regularity hypothesis on $$\omega$$, e.g., being in a Sobolev space $$H^s$$ with $$s>$$something. A particularly important case is when $$\Gamma$$ is an affine subspace, or say for simplicity a linear subspace like $$\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$$. A rather standard approach is to start with $$\omega\in\mathcal{D}'(\mathbb{R}^n)$$. The convolution $$\omega\ast \phi_j^0$$ is in the space of $$C^{\infty}$$ functions $$\mathcal{E}(\mathbb{R}^n)\subset \mathcal{D}'(\mathbb{R}^n)$$ and converges to $$\omega$$ in the topology of $$\mathcal{D}'(\mathbb{R}^n)$$ (the strong topology). The ordinary restriction $$\omega\ast \phi_j^0|_{\Gamma}$$ makes sense and you can ask if the limit $$\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$$ exists inside $$\mathcal{D}'(\mathbb{R}^m)$$. Your particular case $$p=0$$ corresponds to mine with $$m=0$$. Another problem of this kind is pointwise multiplication. If $$\omega_1(x)$$ and $$\omega_2(x)$$ are two distributions, then there is no problem defining $$\omega_1(x_1)\omega_2(x_2)$$ (tensor product), but the issue is how to restrict to the diagonal $$\Gamma=\{x_1=x_2\}$$. Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can do the (deterministically) impossible. It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $$\omega|_{\Gamma}$$ of $$\omega$$ to some subset $$\Gamma\subset\mathbb{R}^n$$. When one succeeds the result is called a trace theorem. This usually requires some quantitative regularity hypothesis on $$\omega$$, e.g., being in a Sobolev space $$H^s$$ with $$s>$$something. A particularly important case is when $$\Gamma$$ is an affine subspace, or say for simplicity a linear subspace like $$\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$$. A rather standard approach is to start with $$\omega\in\mathcal{D}'(\mathbb{R}^n)$$. The convolution $$\omega\ast \phi_j^0$$ is in the space of $$C^{\infty}$$ functions $$\mathcal{E}(\mathbb{R}^n)\subset \mathcal{D}'(\mathbb{R}^n)$$ and converges to $$\omega$$ in the topology of $$\mathcal{D}'(\mathbb{R}^n)$$ (the strong topology). The ordinary restriction $$\omega\ast \phi_j^0|_{\Gamma}$$ makes sense and you can ask if the limit $$\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$$ exists inside $$\mathcal{D}'(\mathbb{R}^m)$$. Your particular case $$p=0$$ corresponds to mine with $$m=0$$. Another problem of this kind is pointwise multiplication. If $$\omega_1(x)$$ and $$\omega_2(x)$$ are two distributions, then there is no problem defining $$\omega_1(x_1)\omega_2(x_2)$$ (tensor product), but the issue is how to restrict to the diagonal $$\Gamma=\{x_1=x_2\}$$. Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can sometimes do the (deterministically) impossible. Small addendum: Suppose that for some reason one has a trace theorem but only for large enough $$m$$ and one cannot do the $$m=0$$ or the point restriction case. Then one can still do the following "stabilization" trick: change $$\omega$$ to $$\omega\otimes 1$$ where one tensors with the constant function equal to one seen as a distribution in say $$p$$ new variables. If you can restrict it from $$\mathbb{R}^{n+p}$$ to a subspace of dimension $$p$$, then you will have your point evaluation after factoring out the $$\otimes 1$$. The last step of course needs your restriction construction to be invariant/covariant by translation along $$\Gamma$$. 1 answered Feb 21 at 18:37 Abdelmalek Abdesselam 11.8k11 gold badge2929 silver badges7777 bronze badges It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $$\omega|_{\Gamma}$$ of $$\omega$$ to some subset $$\Gamma\subset\mathbb{R}^n$$. When one succeeds the result is called a trace theorem. This usually requires some quantitative regularity hypothesis on $$\omega$$, e.g., being in a Sobolev space $$H^s$$ with $$s>$$something. A particularly important case is when $$\Gamma$$ is an affine subspace, or say for simplicity a linear subspace like $$\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$$. A rather standard approach is to start with $$\omega\in\mathcal{D}'(\mathbb{R}^n)$$. The convolution $$\omega\ast \phi_j^0$$ is in the space of $$C^{\infty}$$ functions $$\mathcal{E}(\mathbb{R}^n)\subset \mathcal{D}'(\mathbb{R}^n)$$ and converges to $$\omega$$ in the topology of $$\mathcal{D}'(\mathbb{R}^n)$$ (the strong topology). The ordinary restriction $$\omega\ast \phi_j^0|_{\Gamma}$$ makes sense and you can ask if the limit $$\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$$ exists inside $$\mathcal{D}'(\mathbb{R}^m)$$. Your particular case $$p=0$$ corresponds to mine with $$m=0$$. Another problem of this kind is pointwise multiplication. If $$\omega_1(x)$$ and $$\omega_2(x)$$ are two distributions, then there is no problem defining $$\omega_1(x_1)\omega_2(x_2)$$ (tensor product), but the issue is how to restrict to the diagonal $$\Gamma=\{x_1=x_2\}$$. Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can do the (deterministically) impossible.
特码生肖图
<ruby id="d9npn"></ruby>

<sub id="d9npn"><progress id="d9npn"></progress></sub>

<nobr id="d9npn"></nobr>

<rp id="d9npn"><big id="d9npn"><th id="d9npn"></th></big></rp>

<th id="d9npn"><meter id="d9npn"></meter></th>

<ruby id="d9npn"></ruby>

<sub id="d9npn"><progress id="d9npn"></progress></sub>

<nobr id="d9npn"></nobr>

<rp id="d9npn"><big id="d9npn"><th id="d9npn"></th></big></rp>

<th id="d9npn"><meter id="d9npn"></meter></th>

挑战者游戏完整版 斗鸡怎么玩 辽宁11选5开奖视频 彩金水果拉霸游戏有什么套路 逆战燃烧箭头 摩纳哥vs卡昂直播 王者荣耀直播下载什么 阿森纳对伯恩茅斯比分 辽宁十一选五走势图吗 jdp夺宝电子假吗