Property (T) In Terms of the Spectrum of a Self-Adjoint Contraction Let $G$ be a finitely generated group.
Let $\mu$ be a non-degenerate, finitely supported symmetric probability measure on $G$.
For every unitary representation $(H,\pi)$ of $G$, we have a self-adjoint contraction
$$
T = \sum_{x \in G} \mu(x) \pi_x.
$$
If $G$ does not have Property (T), why does this mean that the spectrum $\sigma(T)$ of $T$ contains $1$, but not as an eigenvalue? (as claimed in Theorem A here)
 A: I think I need that $\mu$ is nonzero on the generating set of $G$.  If this isn't true I haven't thought about how to get around it. 
I'm also going to use the result from an answer of mine here.  In this answer, all that matters are that the $T_i$ are unitary operators.  The tricky part is figuring out how to adapt it to this weighted sum $\mu(x)$.  I'm sure it's possible to do it all at once, but it also follows from splitting it into the case where $\mu$ only takes rational values, and then the general case.  It gets a bit messy with approximations, but it's just because of this weight.  
Each $\pi(x)$ is a unitary, and let's suppose $y$ is a unit vector.  Since this $\mu$ has finite support, this sum will be finite.  Let's rename the group elements that we're summing over to $e=x_1,...,x_n$, and denote $T_i=\pi(x_i)$.  Furthermore, let's suppose that $\mu(x_i)=\frac{p_i}{q}$ is rational, and so we must have that the sum over all the $p_i$ must equal $q$.  Note that $T_1y=y$.
Now we have that $\sum_{i}T_i y=y$, and hence $q\sum_{i}T_i y=\sum_{i}p_iT_i y=qy$.  This element has norm $q$.  Also, instead of thinking it as a sum of $i$ operators, think of it as a sum over $q$ operators, with some repeated if $p_i>1$.  Essentially, we have $q$ unitary operators, and when we apply their sum to $y$, we get back $q*y$.  This happens if and only if $T_iy=y$ for all $i$.  The linked answer shows this, as it doesn't matter I assumed $C<n$ in that answer (here $q=n$).  So we get that $\pi(x)y=y$ on a generating set of $G$, and hence on all of $G$.
For the general case, you should be able to find rationals close enough to each $\mu(x_i)$, apply the above case, and get some estimate on the difference between $||T_iy-y||$ that depends on $\epsilon$, and since it holds for every $\epsilon$, you're done.  
