Let $\Gamma$ be a countable discrete group and let $C_\lambda^*(\Gamma)$ denote the corresponding reduced group $C^*$-algebra.
In [Theorem 2.6.8., BO08] the authors prove the following equivalence:
(1) $\Gamma$ is amenable (2) There exist positive definite functions $(\varphi_n)_n\subset c_c(\Gamma)$ converging pointwise to 1.
In [Remark 2.3., BG13] the authors claim the following: For $\Gamma$ to be amenable it is sufficient that there exist positive definite functions $(h_n)_n \subset \ell^p(\Gamma)$ converging pointwise to 1.
Their reasoning is along these lines:
Let $k>p$. Then $h_n^k\in \ell^1(\Gamma)\subseteq C_\lambda^*(\Gamma)$ and these functions are positive definite and these functions also converge pointwise to one. Now for every $h_n^k$ consider compactly supported functions ${(f_n)}_m$ that approximate the square root of the $h_n^k$ in the norm of $C_\lambda^*(\Gamma)$. We get finitely supported positive definite functions $f_n^* \ast f_n$ that approximate $h_n^k$ and converge pointwise to 1.
I don't know in what sense square root is meant here. I would be very grateful if someone could give me an explenation, a hint or a reference, anything really as I am quite lost.
Edit: What I found so far is the notion of a "convolution square root" that seems to have been introduced by Godement in 1948 (for the general case of locally compact groups): Every positive-definite, square-integrable function over $\Gamma$ has a square root: It is of the form $g \ast g$, $g^* = g \in \ell^2(\Gamma)$. I don't speak french and seem to be unable to find any reproduction of that result in newer literature. Is that what should be applied here? How could one prove this for the countable discrete case?
I'm very sorry that I just have questions and don't have any input myself. Thank you very much in advance.
[BG13] Brown and Guentner, "New $C^*$-Completions of discrete groups and related spaces"
[BO08] Brown and Ozawa, "$C^*$-Algebras and Fnite-Dimensional Approximations"
Recall first that any positive definite function $\varphi$ is given by $\varphi(g) = \langle \xi, \pi(g) \xi \rangle$ for some unitary representation $\pi: \Gamma \to U(H_\varphi)$. To see this just notice that the following quadratic form over $\ell^1(\Gamma)$ is positive $$ \langle f, f \rangle_\varphi = \langle \varphi, f^\ast \ast f \rangle = \sum_{h} \sum_{g} \varphi(g^{-1} h) \overline{f(g)} f(h) $$ Then take completion of $\ell^1(\Gamma)$ with respect to the square root of that quadratic form to obtain a Hilbert space $H_\varphi$ and denote by $\pi$ the representation given by left translation. It is easy to check that $$ \varphi(g) = \langle \delta_e, \delta_g \rangle_\varphi = \langle \delta_e, \pi(g) \delta_e \rangle_\varphi. $$ For the details of this construction just check Folland [Fo].
Now, we have to see that if $\varphi \in \ell^1(\Gamma)$, then you can take the representation to be $\lambda$ and $H_\varphi$ to be $\ell_2(\Gamma)$ without loss of generality. Here is an sketch. See that $$ \langle f, f \rangle_\varphi = \sum_{h} \sum_{g} \varphi(g^{-1} h) \overline{f(g)} f(h) \leq \| \varphi \|_1 \| f \|_2 \| f \|_2 = \| \varphi \|_1 \langle f, f \rangle. $$ By the Lax-Milgram theorem that implies that there is an operator $V$, with $\| V \| \leq \| \varphi \|^\frac12_1$ such that $\langle f, f \rangle_\varphi = \langle V f, V f \rangle$. The fact that $\pi$ is unitary give you that $V^\ast V$ commutes with left translations and you can take $V$ to commute too, therefore: $$ \varphi(g) = \langle \delta_e, \pi(g) \delta_e \rangle_\varphi = \big\langle V(\delta_e), \lambda(g)V(\delta_e) \big\rangle. $$ The last expression is $f^\ast \ast f$ for $f = V(\delta_e)$
[Fo] Folland, Gerald B., A course in abstract harmonic analysis, Studies in Advanced Mathematics. Boca Raton, FL: CRC Press. viii, 276 p. (1995). ZBL0857.43001.