Timeline for Does every distribution define a Radon measure?
Current License: CC BY-SA 2.5
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| Feb 1 at 19:59 | comment | added | Deane Yang | The key assumption is being continuous with respect to the topology of $C^0_c(\Omega$. This is a strong assumption. It is satisfied by the delta function, but not the derivative of the delta function. Indeed, the latter relies on the smoothness of the function, but that in fact is why it is NOT continuous with respect to the topology of $C^0_c(\Omega)$. In general the value of a distribution depends on the derivatives of a function up to some (finite) order. If the order is zero, then it is a Radon measure. | |
| Feb 1 at 12:56 | comment | added | demim00nde | @deane-yang I had a quick question. In his book on distributions, Treves (1969) says that a distribution is a Radon measure if it can be shown to be continuous on $C_c^{\infty}(\Omega)$ in the topology induced by $C_c^0(\Omega)$ (Prop.21.2). He extends the continuity to $C_c^0(\Omega)$ based on the density of $C_c^{\infty}(\Omega)$ in $C_c^0(\Omega)$. However, I'm having trouble buying this, because the distribution may rely on the smoothness of a given function to 'work'. Am I missing something? | |
| Jun 21, 2011 at 20:49 | comment | added | LSpice | I love the sentence “The condition that the distribution be non-negative for non-negative functions is non-trivial.” It seems that it should be simplifiable by some sort of elimination of double negations, but any such ‘simplification’ radically changes its meaning. | |
| Mar 12, 2010 at 8:49 | comment | added | Tom Ellis | ncatlab.org/nlab/show/distribution says "For an example of a distribution .. which does not arise from a measure, consider the derivative of the Dirac distribution. (As a functional, it maps a test function f to −f′(0).)" -- so my understanding of signed measures is not deep enough: there's something magic about measures that makes every signed measure the difference of two measures. The equivalent result is clearly not true for distributions! | |
| Mar 12, 2010 at 8:13 | comment | added | Tom Ellis | Perhaps I should have been clearer: does every distribution correspond to a signed measure? | |
| Mar 10, 2010 at 15:41 | history | answered | Deane Yang | CC BY-SA 2.5 |