### 2.7 Uniqueness of analytic continuation

The uniqueness of the analytic continuation of to non-integer may not seem obvious,
especially if the field system in question is not relativistic so that there is no isometry in the polar angle
, which would allow us, without any trouble, to glue together pieces of the Euclidean space to form a
path integral over a conical space . However, some arguments can be given that the analytic
continuation to non-integer is in fact unique.
Consider a renormalized density matrix . The eigenvalues of lie in the interval
. If this matrix were a finite matrix we could use the triangle inequality to show
that

For infinite-size matrices the trace is usually infinite so that a regularization is needed. Suppose that is
the regularization parameter and is the regularized trace. Then

Thus is a bounded function in the complex half-plane, . Now suppose that we know
that for integer values of . Then, in the region , we
can represent in the form
where the function is analytic (for ). Since by condition (17) the function is
bounded, we obtain that, in order to compensate for the growth of the sine in Eq. (18) for complex values
of , the function should satisfy the condition
By Carlson’s theorem [36] an analytic function, which is bounded in the region and
which satisfies condition (19), vanishes identically. Thus, we conclude that and
there is only one analytic continuation to non-integer , namely the one given by function
.