$\newcommand{ \reals }{ \mathbb{R}} $ $\newcommand{ \distfns }{ \mathcal{D}'(\reals) } $ $\newcommand{ \testfns }{ \mathcal{D}( \reals ) } $ $\newcommand{ \ints }{ \mathbb{Z}} $ $\newcommand{ \x }{ \varphi }$

Let $\testfns$ denote $C^\infty$ functions with compact support.

Let $\distfns$ denote the space continuous linear forms on $\testfns$ such that if $\x_n \to \x$ in $\testfns$ then $\langle T, \x_n \rangle \to \langle T, \x \rangle$.

Given $\x \in \testfns$. How should one interpret the following, for $T \in \distfns$

$$ S=\sum_{m \in \ints} \tau_m T $$

where $\tau_m$ is the translation operator. Is it $ \langle S, \x \rangle = \sum_{m \in \ints} \langle \tau_m T, \x \rangle$?

  • 1
    $\begingroup$ What is $\mathcal D(\Bbb R)$ and $\mathcal D'(\Bbb R)$? $\endgroup$ Jan 6 '16 at 21:57
  • $\begingroup$ I added the definitions to the top of the question, is that clearer? $\endgroup$
    – shilov
    Jan 6 '16 at 22:12
  • $\begingroup$ Still a bit unclear the distinction between $\tau_m T$ and $T_m$, ... Especially since sometimes $\tau_m$ could be a translation operator... What is your intention? $\endgroup$ Jan 6 '16 at 23:01
  • $\begingroup$ My intention was to originally just say $T_m$ in place of $\tau_m T$ but I don't think I can do that. I do mean the translation operator yes. $\endgroup$
    – shilov
    Jan 6 '16 at 23:10
  • $\begingroup$ I would interpret this expression as $\langle S,\phi\rangle=\sum_{m\in\mathbb{Z}}\langle T,\tau_m\phi\rangle$. $\endgroup$
    – MaoWao
    Jan 7 '16 at 11:19

Special case. Suppose that $T$ is compactly supported.

Here we have : $$\langle S, \phi \rangle = \sum_{n \in \mathbb{Z}} \langle \tau_n T , \phi \rangle = \sum_{n \in \mathbb{Z}} \langle T, \tau_{-n} \phi \rangle = \sum_{n \in \mathbb{Z}} \langle T, \phi ( \cdot - n) \rangle $$ But you have to prove this is a distribution. In this way, set $K \subset \mathbb{R}$ a compact set. Then set $\phi \in \mathcal{D}( \mathbb{R})$ compactly supported in $K$ and note that : $$\langle S, \phi \rangle = \sum_{n= -N}^N \langle T, \phi ( \cdot - n) \rangle$$ With $N$ an integer. The sum is truncated because when $n$ is large inough, the suport of $\phi ( \cdot -n)$ has no intersection with the support of $T$.

But $\|\tau_{-n} \phi\|_{\mathcal{C}^p} = \|\phi\|_{ \mathcal{C}^p}$ for every $p$ so that since $T$ is a distribution there existe $C_K >0$ and $p_K \geqslant 0$ such that : $| \langle T, \phi \rangle | \leqslant C_K \|\phi\|_{\mathcal{C}^{p_K}}$ where $C_K,p_K$ only depend on $K$, not $\phi$. Thus we obtain : $$\langle S, \phi \rangle \leqslant (2N+1) C_K \|\phi\|_{\mathcal{C}^{p_K}}$$

This shows that $S$ is a distribution.

More generally I think that letting $\displaystyle S_N:= \sum_{n=-N}^N \tau_n T$ you have to figure out if $(S_N)_{N \geqslant 0}$ converges or not.

For example if you set $\displaystyle \langle T, \phi \rangle = \int_{\mathbb{R}} \phi(x) \mathrm{d}x$ then $S$ is not defined :

Let $\phi \in \mathcal{D}(\mathbb{R})$ such that $\displaystyle \int_{\mathbb{R}} f(x) \mathrm{d}x = 1$ then : [\tau_n T=T] for all $n \in \mathbb{Z}$ thus $\langle S_N, \phi \rangle = 2N+1$ and does not converge.

  • $\begingroup$ The argument of $\phi$ is shifted by $n$ for each term of the sum, so it seems to me that for each $\phi(\cdot - n)$ we have a function which has compact support which is definitely in $K \subset [-N-n,N-n]$, please could you explain how the sum is truncated? $\endgroup$
    – shilov
    Jan 8 '16 at 10:59
  • $\begingroup$ My answer is not correct, I was too quick. But do you have more informations over $T$ ? Is $T$ compactly supported ? If it's the case than my idea works (with little changes I will write), thus if $T$ is not compactly supported I don't think we can give a sens to $S$ $\endgroup$
    – M.LTA
    Jan 8 '16 at 11:20
  • $\begingroup$ $T$ is not necessarily compact supported, it is just a distribution and I don't think $S$ is a distribution. Could that be shown by contradiction? $\endgroup$
    – shilov
    Jan 8 '16 at 11:27
  • $\begingroup$ If $T$ is compactly supported yes $S$ is a distribution. If not, then you can have counter examples. For example with $T=\mathbf{1}$ the constant distribution, $S$ is not well defined. $\endgroup$
    – M.LTA
    Jan 8 '16 at 11:30
  • 1
    $\begingroup$ See the last edition. $\endgroup$
    – M.LTA
    Jan 8 '16 at 11:35

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