Can the Cesaro limit of a positive definite function be negative?

Question

Let $G$ be a countable amenable group and $\gamma:G\to\mathbb{C}$ a positive (semi)definite function (i.e. such that $\gamma(g^{-1})=\overline{\gamma(g)}$ and $$\sum_{g,h\in G}f(g)\overline{f(h)}\gamma(h^{-1}g)\geq0$$ whenever $f:G\to\mathbb{C}$ is finitely supported). Let $(F_N)_{N\in\mathbb{N}}$ be a Folner sequence in $G$ and assume the Cesaro limit $$L:=\lim_{N\to\infty}\frac1{|F_N|}\sum_{g\in F_N}\gamma(g)$$ exists. Can $L$ be negative?

If $G$ is abelian then one can apply Bochner-Herglotz theorem to represent $\gamma$ as the Fourier transform of a positive measure, decompose it into atomic and continuous components, and then apply Wiener's lemma to the continuous component. It follows that $L\geq0$. Can this proof be adapted to the amenable case?

Answer 1

I hope it's not frowned upon to answer one's own question, but since I just figured out the answer it doesn't make sense to keep it unanswered and there are a few upvotes so the answer may interest other people as well.

The answer is no, i.e. $L$ must be non-negative. The proof is quite short if one uses the right theorems: A version of Naimark's dilation theorem (Theorem 5.20 in these notes of V. Paulsen; the version stated on wikipedia is significantly heavier) states that there exists a Hilbert space $H$, a unitary representation $(U_g)_{g\in G}$ of $G$ on $H$ and a vector $v\in H$ such that $$\gamma(g)=\langle U_gv,v\rangle.$$ Then the mean ergodic theorem (which holds for any Folner sequence in an amenable group) gives $$L=\lim_{N\to\infty}\frac1{|F_N|}\sum_{g\in F_N}\langle U_gv,v\rangle=\langle Pv,v\rangle=\|Pv\|^2\geq0$$ where $P:H\to H$ is the orthogonal projection onto the space of vectors fixed under each $U_g$.