Let $G$ be a discrete commutative group, and $\phi:G \rightarrow \mathbb{T}$ be a group homomorphism where $\mathbb{T}$ is the circle i.e. $\mathbb{T} := \{z \in \mathbb{Z}:|z| = 1\}$. Consider $l^1(G)$ with involution and convolution, and the define the function $\varphi(f) := \sum_{x \in G}\phi(x) f(x)$. I have shown that this is a linear functional on $l^1(G)$. I am having trouble showing that this functional is positive i.e. $\varphi(f^* * f) \geq 0$ for all $f \in l^1(G)$. I know it's not much, but this is what I have written so far $$\varphi(f^* * f) = \sum_{x \in G} \phi(x) (f^* * f)(x) = \sum_{x \in G} \phi(x) \sum_{y \in G} \overline{f(y^{-1})} f(y^{-1}x)$$


1 Answer 1


We have $$\sum_{x \in G} \phi(x) \sum_{y \in G} \overline{f(y^{-1})} f(y^{-1}x)\\ = \sum_{y \in G}\overline{\phi(y^{-1})} \overline{f(y^{-1})}\sum_{x \in G} \phi(y^{-1}x) f(y^{-1}x)\\ =\sum_{y \in G}\overline{\phi(y^{-1})} \overline{f(y^{-1})} \sum_{x \in G}{\phi(x)} {f(x)} \\ =\left |\sum_{x\in G} \phi(x)f(x)\right |^2$$

Remark Commutativity of $G$ is not essential.

  • $\begingroup$ This makes complete sense thanks $\endgroup$
    – 3j iwiojr3
    Nov 21 at 1:29

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