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Is this a continuous linear form on $\mathcal{D}(\mathbb{R})$ ($C^\infty$ with compact support) such that for any sequence $\varphi_n\to0 \Rightarrow \langle T, \varphi_n \rangle \to 0$.

$\int_\mathbb{R}\sum_{m \in \mathbb{N}} e^{ -(x-m)^2 } \varphi_n(x) dx \to 0$ as $n \to \infty$

I am guessing that it is not, so I could try to find a function for which this does not work, this is how far I have got, am I on the right track, should I try to use the translation of $\varphi_n$ for the counter example?

$\int_\mathbb{R} e^{ -x^2 } \sum_{m \in \mathbb{N}} \varphi_n(x+m) dx $

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  • $\begingroup$ $\sum_{n\in\mathbb{N}}e^{-(x-n)^2}$ is obviously a non-negative function bounded by $1+2\sum_{n\geq 1}e^{-n^2}<\frac{16}{9}$. Do you really need anything else? $\endgroup$ – Jack D'Aurizio Jan 5 '16 at 13:51
  • $\begingroup$ Yes I do, I could have worked out that it was non-negative and bounded. I did not know that that implies it is a distribution. $\endgroup$ – shilov Jan 5 '16 at 14:38
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Your $f$ is bounded, hence locally integrable (so it is a distribution ). For any x,

$$\sum_{n\in \Bbb N} \exp( -(x-n)^2 ) \leq \sum_{n\in \Bbb Z} \exp( -(x-n)^2 ) $$

But $x-n = \{x\} + \lfloor x\rfloor - n = \{x\} + k$ (where $ \{x\}$ is the fractional part of $x$) $$\leq \sum_{k\in \Bbb Z} \exp( -(k+ \{x\})^2 ) $$

$$\leq \sum_{k\in \Bbb Z} \exp( -k^2 ) $$

$$\leq 2 \sum_{n \in \Bbb N} e^{-n^2} = C < +\infty $$

And every bounded function is a distribution :

Take a sequence $\varphi_n \to 0$ in $\mathcal{D}$, there exists a compact $K$ such that $\forall n, \text{Supp}(\phi_n) \subset K$, and we have

$$\left| \langle T_f, \varphi \rangle \right| = \left| \int_{\Bbb R} f(x) \varphi_n(x) dx \right| = \left| \int_{K} f(x) \varphi_n(x) dx \right| \leq \| \varphi_n \|_\infty \| f \|_\infty \mu( K ) \to 0 $$

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  • $\begingroup$ Is your answer $f$ is bounded so it is locally integrable and so it is a distribution - here is the proof it is bounded? If so - why is a locally integrable function a distribution? My definition is different to that. $\endgroup$ – shilov Jan 5 '16 at 12:06
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    $\begingroup$ @shilov : yes, it is my answer. See my edit for "why is a L1 loc function a distribution?" (very very very usefull result). Here I just have showed the weaker result for bounded functions, but you can use the same reasonning+Hölder to get the locally integrable case $\endgroup$ – Tryss Jan 5 '16 at 12:57
  • $\begingroup$ that is very interesting thank you, could you direct me to a source for some similar results? For example what if we have $\sum_{n \in \mathbb{Z}} g(x)$ where $g(x)$ defines a distribution, is there anything we can say then? $\endgroup$ – shilov Jan 5 '16 at 13:48
  • $\begingroup$ @shilov : Your notation is lacking something. You're asking if a serie of functions that define a distributions is a distribution? I guess it depend wildly of the exact function, and I have no result specificly for distributions (just the usual result for functions, and the property that $\varphi \mapsto \int_{\Bbb R} f(x) \varphi(x) dx$ is a distribution if and only if $f\in L^1_{loc}$ $\endgroup$ – Tryss Jan 5 '16 at 14:51
  • $\begingroup$ What if I define $\int_{\mathbb{R}} \sum_{m \in \mathbb{Z}} f(x) \varphi_n( x-m ) dx $, would that make sense? $\endgroup$ – shilov Jan 5 '16 at 21:46

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