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The following is a toy version of something I've been fiddling with and I thought it might be more efficient to post it as a question here.

Just to fix definitions: for me, a complex-valued function $f$ on a group $G$ is said to be positive-semi-definite if, for every choice of points $x_1, \dots, x_n\in G$, the $n\times n$ matrix $[f(x_i^{-1}x_j)]_{i,j=1}^n$ is PSD in the usual sense. Note that in this definition I do not assume $G$ is countable, nor do I assume $f$ is continuous.

Question 1. Let $f$ be a continuous PSD function on $\mathbb R$, and pick a finite subset $F\subset R$; then let $E=\{-x+y \mid x,y \in F\}$. Does there always exist a PSD function $h$ on $\mathbb R$, with finite or countably infinite support,such that $h\vert_E = f\vert_E$ ?

Obviously, if we knew that the ``truncation'' of $f$ to the set $E$ was itself a PSD function on $\mathbb R$, then the answer to Q1 would be positive. But I don't see why this truncated function would always be PSD.

Question 2. The analogue of Q1 with $\mathbb R$ replaced by $\mathbb R^d$ for $d\geq 2$.

My personal suspicion is that Q2 will have a negative answer but this is based on pessimism rather than any genuine intuition.

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    $\begingroup$ Just to clarify: $x^{-1}y$ in the definition of $E$ is done in the group $(\mathbb R,+)$, so it's a fancy way of writing $y-x$ ? $\endgroup$
    – Christian Remling
    Commented Sep 9, 2016 at 16:23
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    $\begingroup$ As to "countably infinite" support, why don't you just take the restriction of $f$ to $\text{span}_{\mathbb Q}F$? $\endgroup$
    – fedja
    Commented Sep 9, 2016 at 19:32
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    $\begingroup$ Because $0=x_3-x_3=x_4-x_4\in T$ and $x_2-x_3$ is either in $T$, in which case you are fine, or not in $T$, in which case the corresponding element is $0$. In general, the matrix splits into blocks corresponding to cosets of $\mathbb R/T$ and each coset is as good as $T$ itself. Am I missing something? $\endgroup$
    – fedja
    Commented Sep 9, 2016 at 20:57
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    $\begingroup$ @fedja you miss nothing. For $H<G$ subgroup, the extension by 0 of a PD function on $H$ is a PD on $G$. From representation theoretic pov this corresponds to induction. In particular the answer to Q3 (general $G$) is positive. $\endgroup$
    – Uri Bader
    Commented Sep 10, 2016 at 8:53
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    $\begingroup$ Yemon, the two operations $PD(G)\to PD(H)$ and $PD(H)\to PD(G)$ given by "restriction" and "extending by 0" correspond to the functors "Restriction" and "Induction" on (cyclic) unireps of the (abstract) groups $G$ and $H$. Thus, from RT pov, the answer to your question is "restrict to a countable group containing $F$ and induce back to $G$". Practically this is the same as "restrict $f$ to a countable group containing $F$ and extend by 0" which @fedja gave. I'd be happy to explain further in case the correspondence (or anything else) is still unclear. $\endgroup$
    – Uri Bader
    Commented Sep 10, 2016 at 12:42

1 Answer 1

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The answer for countably infinite has been given by fedja and Uri Bader in the comments and is yes: put $f$ to $0$ outside of the subgroup generated by $E$, and leave $f$ unchanged on this subgroup.

For finite, the answer is no. For an example, take $f=1$ and $E=\{-1,0,1\}$. Since the only PSD function of the form $\begin{pmatrix} 1 & 1 & a \\ 1&1&1\\ b &1 &1\end{pmatrix}$ is the matrix $\begin{pmatrix} 1 & 1 & 1 \\ 1&1&1\\ 1 &1 &1\end{pmatrix}$, we have that any positive-semidefinite function whose restriction to $E$ is $1$ has to be constant equal to $1$ on $\mathbf Z$.

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  • $\begingroup$ Nice! I wonder if this example can be seen as some kind of argument with multiplicative domains or the Schwarz inequality for cp maps... but this is purely idle speculation $\endgroup$
    – Yemon Choi
    Commented Sep 19, 2016 at 13:53
  • $\begingroup$ @YemonChoi: sure. $f \colon G \to \mathbf{C}$ being PSD means that the Fourier multiplier $M_f$ with symbol $f$ is completely positive. So if we have such PSD $f$ and $t \in G$ such that $f(t) = f(0)=1$, we have (Choi) that $t$ belongs to the multiplicative domain of $M_f$, and so the whole group generated by $f$ belongs to the multiplicative domain, ie that $f$ is constant equal to $1$ on the group generated by $t$. $\endgroup$
    – Mikael de la Salle
    Commented Sep 20, 2016 at 11:09
  • $\begingroup$ I often feel that Choi (Y) needs to pay closer attention to ideas introduced by Choi (M-D) $\endgroup$
    – Yemon Choi
    Commented Sep 20, 2016 at 14:24
  • $\begingroup$ I have often wondered: is there any connection between you and Choi (M-D)? $\endgroup$
    – Mikael de la Salle
    Commented Sep 20, 2016 at 14:26
  • $\begingroup$ Not that I know of... $\endgroup$
    – Yemon Choi
    Commented Sep 20, 2016 at 14:31

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