🚀 go-pugleaf

It's pretty well-known that while the Gaussian "measure" with unit
variance on an infinite-dimensional real Hilbert space isn't really a 
measure, it is a positive linear functional on the algebra of
"exponential" functions, that is, finite linear combinations of functions
f(x) = exp(i<v,x>).  This is good enough to do the field theory of free
quantum fields - and that's how Segal, Zhou and I do it in our book.
The basic idea is that the algebra of exponential functions is a
*-algebra and has a natural C*-algebra completion, and the integral with
respect to the Gaussian "measure" extends to a positive linear
functional on this C*-algebra.  By general nonsense, the C*-algebra is
isomorphic the continuous functions on *some* space or other and the
positive linear functional corresponds to a measure on that space.

Of course, there are millions of ways to skin a cat, but this is the one
I want to consider.

Suppose we replace the inner product by a quadratic form that's not
positive definite.  I.e., I have a real vector space equipped with a
quadratic form and I want to study the "measure" exp(Q(x,x)) dx.  This
is rather nasty but we may still *formally* compute the integral of any
exponential function (defined as above) using analytic continuation, and
get a finite answer. 

Question 1: does the resulting linear functional on the algebra of
exponential functions extend to a positive (hence continuous) linear
functional on any C*-algebra completion of this algebra?  I don't think
so, since I don't think it's really positive, despite the formally
positive appearance of exp(Q(x,x)) dx.

Of course, if instead of Q(x,x) we work with iQ(x,x) things are a bit
more bounded looking and we are in a situation quite familiar in physics
(Feynman integrals).  There appears to exist an infinite amount of
literature on this, but I have a very specific question, namely:

Question 2: does the linear functional on the algebra of
exponential functions defined by (formal) integration against
exp(Q(x,x)) dx define a continuous linear functional on any C*-algebra
completion of this algebra?  I don't think so.  If not, I really believe
the claim that making these oscillatory "measures" rigorous is a
thankless task -- which is why people always analytically continue
to the Euclidean situation in QFT.

Note: I am posting to both sci.physics and sci.math.research.  Please
only post general divagations on field theory, references to vageuly
related physics literature, etc. to sci.physics, saving
sci.math.research for actual ANSWERS to my questions.  

I think I can answer my questions (which are really just calculus
exercises in disguise), but if they have been answered already I would
like references to spare me from reinventing the wheel and (worse)
claiming it was my idea.