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Whilst reading the article about restrictions of distributions (generalized functions) on Wikipedia (here)

I had trouble understanding the example of a distribution defined on the subset $V = (0,2) \subset \mathbb{R}$ that admits no extension to the space of Distributions on $U = \mathbb{R}$.

The example I am referring to is the distribution

\begin{equation} S(x) = \sum_{n = 1}^\infty n \delta \left(x - \frac{1}{n}\right) \end{equation}

Now, my question is how does this example act on test functions ? If I take a smooth function $\psi$ whith supp $\psi \subset V$, how do I apply $S$ to it ? I understand that $S$ is a modification of the Dirac delta distribution

\begin{equation} \langle \delta , \psi \rangle = \psi (0) \end{equation}

Below is my guess :

\begin{equation} \langle S , \psi \rangle = \sum^{\infty}_{n = 1} n \psi (\frac{1}{n}) \end{equation}

Is that correct ? If yes, I am still not sure how this is well defined, given only the restriction that $\psi$ is zero outside $(0,2)$. Tanks a lot for your help!

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The compact support of $\psi$ within $V$ forces the sum to have only finitely many non-zero summands! –  Vobo Feb 25 '12 at 22:17
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up vote 1 down vote accepted

I think the key is that $\psi$ is implicitly supposed (by the definition of $D(V)$ earlier in the article) to be smooth in addition to being zero outside of $(0,2)$.

Being smooth forces $\psi$ to be small enough close to $0$ to counteract the growth of the $n$ factors and make the entire sum finite. However, if we extend $V$ to $U$ -- or another open domain that contains $0$ -- then we have to deal with test functions that are not zero (and flat) at $0$, and then the sum doesn't necessarily converge.

Whether this actually implies that there is no possible extension of $S$ is not immediately clear to me. Of course, such an extension could not be writen as a sum of deltas, but there seems to be no requirement of that.

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Sorry, but smoothness is not the key point. It is the compact support within $V$, which forces the above sum to converge (in fact to have only finitely many non-zero summands for each test function). –  Vobo Feb 25 '12 at 22:01
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I think the key here is to use the fact that the map $\rho_{VU}\colon \mathcal D'(V)\to \mathcal D'(U)$ defined by $\langle \rho_{VU}T,\varphi\rangle=\langle T,E_{VU}\varphi\rangle$ is continuous. You can write $S=\lim_n\sum_{k=1}^nk\delta_{1/k}$ and write $S=\lim_n S_n$. So now we observe that $\rho_{VU} S_n$ and $S_n$ have the same expression, and we can see that we can't have convergence of $\{\rho_{VU} S_n\}$ in $\mathcal D'(U)$ taking a test function $\chi\in\mathcal D(\mathbb R)$ constant to $1$ in a neighborhood of $0$.

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