# Is space of Dirac measures Banach?

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.

• 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. – Michael Hardy Sep 10 '13 at 15:39
• You really should find something better to do. – BigUser Sep 10 '13 at 15:41
• 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? – Prahlad Vaidyanathan Sep 10 '13 at 15:47
• @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. – Cameron Williams Sep 10 '13 at 15:47
• @BigUSer, right, such measures are linear span od Dirac measures – Norbert Sep 11 '13 at 8:54

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

$\textit{Proof:}$

($\Leftarrow$)

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

($\Rightarrow$)

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.