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I am working on exercises from Friedlander's Introduction to the Theory of Distributions and I am stuck in a particular problem.

The question is: "Show that $\langle u, \phi\rangle = \sum_\limits{k \geq 1} \partial^k \phi(1/k)$ is a distribution on $(0, \infty)$, but that there is no $v\in \mathcal{D}'(\mathbb{R})$ whose restriction to $(0, \infty)$ is equal to $u$."

I believe I have managed to prove the first part: Given any compact $K \subset (0,\infty)$, take a test function $\phi\in C^{\infty}_c(0, \infty)$ with $\operatorname{supp} \phi \subset K$. Take then $N$ such that $\frac{1}{N+1} < \min \operatorname{supp} \phi$. We have that $\langle u, \phi\rangle = \sum_\limits{k = 1}^N \partial^k \phi(1/k)$, since $\phi(1/k) = 0, \forall k>N+1$. And so it is clear that $\exists C$ and $\exists N$ such that $u$ satisfies the seminorm estimates $|\langle u, \phi\rangle| \leq \sum\limits_{k=1}^N |\sup\partial^k \phi|$ for any $\phi$.

Now, the second part is troubling me. I believe the way is to suppose there is a distribution $v\in \mathcal{D}'(\mathbb{R})$ with $v|_{(0,\infty)} = u$, and show that it would not satisfy the seminorm estimate because of the restriction. However, I am struggling to see how that should be done.

Note: I have recognized this distribution to be equivalent to $\sum\limits_{k\geq 1} \delta^{(k)}(x-1/k)$ but I am not sure how this helps!

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    $\begingroup$ If it were a distribution on $\mathbb{R}$, what would you get when you apply it to a smooth, compactly supported function that equals $e^x$ when $|x| \leq 1$? $\endgroup$ Feb 6 '17 at 4:19
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    $\begingroup$ Ah, that makes a lot of sense. I would get that $\langle u, \phi\rangle = \sum\limits_{k\geq 1} \exp(1/k) > \sum\limits_{k\geq 1} \exp(0)$ which is unbounded as $k\to \infty$. So $\exists \phi$ such that $|\langle u, \phi\rangle|$ cannot be bound by seminorm estimates. Is that the correct argument? Cheers! $\endgroup$
    – cako
    Feb 6 '17 at 11:29
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    $\begingroup$ that's basically it. $\endgroup$ Feb 6 '17 at 13:54
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    $\begingroup$ Cheers! Much simpler than I thought! $\endgroup$
    – cako
    Feb 6 '17 at 21:03
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To close the question, Willie Wong's suggestion was to choose a test function which is equal to $\exp(x)$ within $\{x:|x|<1\}$. Then $\langle u, \phi\rangle = \sum\limits_{k\geq 1} \exp(1/k) > \sum\limits_{k\geq 1} \exp(0)$ which is unbounded as $k\to\infty$. And so $|\langle u, \phi\rangle|$ cannot be bound by seminorm estimates for our chosen $\phi$.

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