<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>

 4 added 227 characters in body edited Feb 25 at 14:20 B K 64533 silver badges1212 bronze badges Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Additional thoughts after 1st edit: Some "consistency checks" for the definition would in my opinion be the following: If the value of $$\omega$$ exists at every point in some open set $$U\subset \mathbb R^n$$ and the function $$f$$ defined on $$U$$ by these values is continuous, then $$\omega|_U$$ is given by $$f$$. If the value of $$\omega$$ exists at Lebesgue-almost every point in some open set $$U\subset \mathbb R^n$$ and the values define a function $$f\in L^1_{\mathrm{loc}}(U)$$, then $$\omega|_U$$ is given by $$f$$. I believe that at least property 1 should be true and I'll check it once I find the time. 2nd edit: My question is related to this MO question which corresponds to the case $$f\equiv 0$$. Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Additional thoughts after 1st edit: Some "consistency checks" for the definition would in my opinion be the following: If the value of $$\omega$$ exists at every point in some open set $$U\subset \mathbb R^n$$ and the function $$f$$ defined on $$U$$ by these values is continuous, then $$\omega|_U$$ is given by $$f$$. If the value of $$\omega$$ exists at Lebesgue-almost every point in some open set $$U\subset \mathbb R^n$$ and the values define a function $$f\in L^1_{\mathrm{loc}}(U)$$, then $$\omega|_U$$ is given by $$f$$. I believe that at least property 1 should be true and I'll check it once I find the time. Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Additional thoughts after 1st edit: Some "consistency checks" for the definition would in my opinion be the following: If the value of $$\omega$$ exists at every point in some open set $$U\subset \mathbb R^n$$ and the function $$f$$ defined on $$U$$ by these values is continuous, then $$\omega|_U$$ is given by $$f$$. If the value of $$\omega$$ exists at Lebesgue-almost every point in some open set $$U\subset \mathbb R^n$$ and the values define a function $$f\in L^1_{\mathrm{loc}}(U)$$, then $$\omega|_U$$ is given by $$f$$. I believe that at least property 1 should be true and I'll check it once I find the time. 2nd edit: My question is related to this MO question which corresponds to the case $$f\equiv 0$$. 3 Added some additional thoughts edited Feb 22 at 10:49 B K 64533 silver badges1212 bronze badges Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Additional thoughts after 1st edit: Some "consistency checks" for the definition would in my opinion be the following: If the value of $$\omega$$ exists at every point in some open set $$U\subset \mathbb R^n$$ and the function $$f$$ defined on $$U$$ by these values is continuous, then $$\omega|_U$$ is given by $$f$$. If the value of $$\omega$$ exists at Lebesgue-almost every point in some open set $$U\subset \mathbb R^n$$ and the values define a function $$f\in L^1_{\mathrm{loc}}(U)$$, then $$\omega|_U$$ is given by $$f$$. I believe that at least property 1 should be true and I'll check it once I find the time. Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Additional thoughts after 1st edit: Some "consistency checks" for the definition would in my opinion be the following: If the value of $$\omega$$ exists at every point in some open set $$U\subset \mathbb R^n$$ and the function $$f$$ defined on $$U$$ by these values is continuous, then $$\omega|_U$$ is given by $$f$$. If the value of $$\omega$$ exists at Lebesgue-almost every point in some open set $$U\subset \mathbb R^n$$ and the values define a function $$f\in L^1_{\mathrm{loc}}(U)$$, then $$\omega|_U$$ is given by $$f$$. I believe that at least property 1 should be true and I'll check it once I find the time. 2 deleted 22 characters in body edited Feb 21 at 18:04 B K 64533 silver badges1212 bronze badges Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is a regular distribution given given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is a regular distribution given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". Let $$\omega\in D'(\mathbb R^n)$$ be a distribution and $$p\in \mathbb R^n$$. If there is an open set $$U\subset \mathbb R^n$$ containing $$p$$ such that $$\omega|_U$$ is given by a continuous function $$f\in C(U)$$, then for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ we can define a Dirac sequence $$\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$$ by $$\phi^p_j(x):=j^n\phi(j(x-p))$$ which fulfills $$\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.$$ This shows that we can recover the value $$\omega(p)\equiv f(p)$$ of the distribution $$\omega$$ at the point $$p$$ via a limit of such Dirac sequences. Now, suppose that for some $$\omega\in D'(\mathbb R^n)$$ and $$p\in \mathbb R^n$$ we just know that $$\lim_{j\to \infty}\omega(\phi^p_j)$$ exists for every $$\phi\in C^\infty_c(\mathbb R^n)$$ with $$\int_{\mathbb R^n}\phi(x)d x=1$$ and is independent of $$\phi$$. In view of the above it then seems reasonable to define $$\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$$ and to say that $$\omega$$ has a well-defined value at the point $$p$$. Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point". 1 asked Feb 21 at 15:59 B K 64533 silver badges1212 bronze badges
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>

北京赛车开奖结果记录 有你的校园闯关 河南快赢481开奖视频结果 p3试机号今天 网易电竞 跑跑卡丁车手游怎么断位漂移 棉农对波斯波利斯 云南快乐十分开奖详情 坐标新浪爱彩 欢乐捕鱼人赢话费是真的吗