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Forgot the \pi in the Hilbert transform
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Phil Isett
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Every distribution is, locally, a finite number of derivatives of measures. You can prove this with Hahn Banach: restricted to an open set $\Omega$ with compact closure, your distribution belongs to the dual of $C^k(\Omega)$ for some $k$. Note that $C^k$ embeds into $(C^0)^N$ for some large $N$ by taking $f \mapsto (f, D f, D^2 f, \ldots, D^k f)$. By Hahn Banach your distribution is the restriction of some linear function on the dual of $(C^0)^N$, which has the form $u(f) = \sum_{|\alpha| \leq k} \int \partial^\alpha f(x) d\mu_\alpha$. This characterization can then be used to deduce the one mentioned in other responses where you replace $\mu_\alpha$ with some continuous functions, but you have to take more than $k$ derivatives so the latter characterization can be a bit misleading.

So your intuition is right, all a distribution does is take a few derivatives and integrate. In this sense, distribution theory is the natural setting to combine differential calculus with measure theory. Once you understand how crazy measures can be (e.g. measures on hypersurfaces; the derivative of the Cantor function is a measure supported on the Cantor set), you basically have the extent of the pathologies of distributions. But a lot of distributions don't come given to you as derivatives of measures ($p.v. \frac{1}{x}$ is a good exampleexample; it requires one derivative to define, but the Hilbert transform $f \mapsto \frac{1}{\pi} p.v. \int \frac{f(x-y)}{y} dy$ is a bounded operator on $L^2({\mathbb R})$!).

Every distribution is, locally, a finite number of derivatives of measures. You can prove this with Hahn Banach: restricted to an open set $\Omega$ with compact closure, your distribution belongs to the dual of $C^k(\Omega)$ for some $k$. Note that $C^k$ embeds into $(C^0)^N$ for some large $N$ by taking $f \mapsto (f, D f, D^2 f, \ldots, D^k f)$. By Hahn Banach your distribution is the restriction of some linear function on the dual of $(C^0)^N$, which has the form $u(f) = \sum_{|\alpha| \leq k} \int \partial^\alpha f(x) d\mu_\alpha$. This characterization can then be used to deduce the one mentioned in other responses where you replace $\mu_\alpha$ with some continuous functions, but you have to take more than $k$ derivatives so the latter characterization can be a bit misleading.

So your intuition is right, all a distribution does is take a few derivatives and integrate. In this sense, distribution theory is the natural setting to combine differential calculus with measure theory. Once you understand how crazy measures can be (e.g. measures on hypersurfaces; the derivative of the Cantor function is a measure supported on the Cantor set), you basically have the extent of the pathologies of distributions. But a lot of distributions don't come given to you as derivatives of measures ($p.v. \frac{1}{x}$ is a good example).

Every distribution is, locally, a finite number of derivatives of measures. You can prove this with Hahn Banach: restricted to an open set $\Omega$ with compact closure, your distribution belongs to the dual of $C^k(\Omega)$ for some $k$. Note that $C^k$ embeds into $(C^0)^N$ for some large $N$ by taking $f \mapsto (f, D f, D^2 f, \ldots, D^k f)$. By Hahn Banach your distribution is the restriction of some linear function on the dual of $(C^0)^N$, which has the form $u(f) = \sum_{|\alpha| \leq k} \int \partial^\alpha f(x) d\mu_\alpha$. This characterization can then be used to deduce the one mentioned in other responses where you replace $\mu_\alpha$ with some continuous functions, but you have to take more than $k$ derivatives so the latter characterization can be a bit misleading.

So your intuition is right, all a distribution does is take a few derivatives and integrate. In this sense, distribution theory is the natural setting to combine differential calculus with measure theory. Once you understand how crazy measures can be (e.g. measures on hypersurfaces; the derivative of the Cantor function is a measure supported on the Cantor set), you basically have the extent of the pathologies of distributions. But a lot of distributions don't come given to you as derivatives of measures ($p.v. \frac{1}{x}$ is a good example; it requires one derivative to define, but the Hilbert transform $f \mapsto \frac{1}{\pi} p.v. \int \frac{f(x-y)}{y} dy$ is a bounded operator on $L^2({\mathbb R})$!).

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Phil Isett
  • 2.2k
  • 1
  • 24
  • 27

Every distribution is, locally, a finite number of derivatives of measures. You can prove this with Hahn Banach: restricted to an open set $\Omega$ with compact closure, your distribution belongs to the dual of $C^k(\Omega)$ for some $k$. Note that $C^k$ embeds into $(C^0)^N$ for some large $N$ by taking $f \mapsto (f, D f, D^2 f, \ldots, D^k f)$. By Hahn Banach your distribution is the restriction of some linear function on the dual of $(C^0)^N$, which has the form $u(f) = \sum_{|\alpha| \leq k} \int \partial^\alpha f(x) d\mu_\alpha$. This characterization can then be used to deduce the one mentioned in other responses where you replace $\mu_\alpha$ with some continuous functions, but you have to take more than $k$ derivatives so the latter characterization can be a bit misleading.

So your intuition is right, all a distribution does is take a few derivatives and integrate. In this sense, distribution theory is the natural setting to combine differential calculus with measure theory. Once you understand how crazy measures can be (e.g. measures on hypersurfaces; the derivative of the Cantor function is a measure supported on the Cantor set), you basically have the extent of the pathologies of distributions. But a lot of distributions don't come given to you as derivatives of measures ($p.v. \frac{1}{x}$ is a good example).

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