Is the space of all Dirac measures on a set $\Omega$ Banach? With the total variation norm. I don't know what convergence means in this norm.. I mean how do I even think about it.

  • $\begingroup$ I wonder if I'm the only person in the universe who finds this grammatical form bizarre? No, the space is not Banach. It may be a Banach space (I don't know whether it is), but it is not Banach. A sequence may be a Cauchy sequence, but a sequence cannot be Cauchy. Cauchy was a person who died in the 19th century. $\endgroup$ – Michael Hardy Sep 10 '13 at 15:39
  • 2
    $\begingroup$ You really should find something better to do. $\endgroup$ – BigUser Sep 10 '13 at 15:41
  • $\begingroup$ By Dirac measure, you mean "point mass" measures? Why is this space even a vector space? If you add two such things, is it still a point mass measure? $\endgroup$ – Prahlad Vaidyanathan Sep 10 '13 at 15:47
  • 3
    $\begingroup$ @MichaelHardy I think it's pretty clear what OP means. I've often heard someone say something like "Is this space Frechet?" or the like. It's not grammatically incorrect. $\endgroup$ – Cameron Williams Sep 10 '13 at 15:47
  • 1
    $\begingroup$ @BigUSer, right, such measures are linear span od Dirac measures $\endgroup$ – Norbert Sep 11 '13 at 8:54

The space $V$ spanned by the Dirac measures on $\Omega$ is Banach $\Leftrightarrow$ $\Omega$ is finite.



If $|\Omega|=n$, then $V\simeq \mathbb{R}^n$ as a linear space. Every finite dimensional normed space is Banach.


Let $x_n\in\Omega$ be a countable sequence of distinct points. Then

$$\mu_n=\sum_{k=0}^n 2^{-k}\delta_{x_k}$$

is a Cauchy sequence in $V$ since

$$\| \mu_n-\mu_m\| = \left\|\sum_{k=m+1}^n 2^{-k}\delta_{x_k} \right\|\leq \sum_{k=m+1}^n 2^{-k} \|\delta_{x_k} \| =\sum_{k=m+1}^n 2^{-k}$$

which can be made arbitrarily small if one choose $n$ and $m$ sufficiently large. The limiting measure however is not in $V$, thus $V$ is not Banach.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.