The group $C^*$-algebra $C^*(G)$
Let $G$ be a unimodular Lie group endowed with a left and right invariant Haar measure dg, and let $f,g \in C_c^{\infty}(G)$ be two compactly supported smooth functions. Define the involution $$f^*(t):=\overline{f(t^{-1})}, $$ and the convolution product $$(f*g)(t):=\int_G f(s)g(s^{-1}t)ds.$$ The $C^*(G)$ norm is given by $$ ||f||_{C^*(G)}:=\displaystyle\sup_{(\rho,H)\in\hat{G}}||\rho(f)||_{B(H)}, $$ where $(\rho,H)$ denotes an unitary representation $\rho:G\to B(H)$ on the Hilbert space $H$ and $$\rho(f)v:=\int_G f(g)\rho(g)dg\in B(H). $$ The (full) group $C^*$-algebra $C^*(G)$ is given by the vector space completion of $C_c^{\infty}(G)$ with respect to the norm $||\cdot||_{C^*(G)}$.
The question
Let $f\in C_c^{\infty}(G)\subset C^*(G)$ be a positive element of $C^*(G)$. This means that $f$ is hermitian with respect to the involution and that the spectrum $\sigma_{C^*(G)}(f)\subset \mathbb{R}_+$. I would like to prove that $$\int_G f(g)dg = \int_G |f(g)|dg. $$
We cannot simply say that a positive (in $C^*(G)$) function is also point-wise positive, since being hermitian here actually means $\overline{f(g)}=f(g^{-1})$. If that was the case, the result would be immediate. I have come up with examples where the equation is actually satisfied. All in all, it is not hard to see that the integral on the left-hand side above is actually a positive real number (using unimodularity and the fact that $f$ is hermitian).
Any help with characterizing these functions $f\in C_c^{\infty}(G)\cap C^*(G)_+$ would be appreciated.
Consider the two-element cyclic group $G= \{1,a\}$, where $a^2=1$, and let $f\in C_c(G)$ be given by $f(1)=1/2$, and $f(a)=-1/2$. It is easy to show that $f=f^*=f*f$, so $f$ is positive. Now $$\int_G f(g)dg = 0 \neq 1 = \int_G |f(g)|dg, $$ hence a counter-example to the claim.