I will answer some of my questions in the negative.
3.
First consider the case of rescaling an operator A by some (positive) number λ. Then ζλA(s) = λ-sζA(s), and so TR λA = λ TR A. This is all well and good. How does the determinant behave? Define the "perceived dimension" DIM A to be logλ[ (DET λA)/(DET A) ]. Then it's easy to see that DIM A = ζA(0). What this means is that DET λA = λζA(0) DET A.
This is all well and good if the perceived dimension of a vector space does not depend on A. Unfortunately, it does. For example, the Hurwitz zeta functions ζ(s,μ) = Σ0∞(n+μ)-s (-μ not in N) naturally arise as the zeta functions of differential operators — e.g. as the operator x(d/dx) + μ on the space of (nice) functions on R. One can look up the values of this function, e.g. in Elizalde, et al. In particular, ζ(0,μ) = 1/2 - μ. Thus, let A and B be two such operators, with ζA = ζ(s,α) and ζB = ζ(s,β). For generic α and β, and provided A and B commute (e.g. for the suggested differential operators), then DET AB exists. But if DET were multiplicative, then:
DET(λAB) = DET(λA) DET(B) = λ1/2 - α DET A DET B
but a similar calculation would yield λ1/2 - β DET A DET B.
This proves that DET is not multiplicative.
1.
My negative answer to 1. is not quite as satisfying, but it's OK. Consider an operator A (e.g. x(d/dx)+1) with eigenvalues 1,2,..., and so zeta function the Reimann function ζ(s). Then TR A = ζ(-1) = -1/12. On the other hand, exp A has eigenvalues e, e2, etc., and so zeta function ζexp A(s) = Σ e-ns = e-s/(1 - e-s) = 1/(es-1). This has a pole at s=0, and so DET exp A = lims→0 es/(es-1)2 = ∞. So question 1. is hopeless in the sense that A might be zeta-function regularizable but exp A not. I don't have a counterexample when all the zeta functions give finite values.
5.
As in my answer to 3. above, I will continue to consider the Hurwitz function ζ(s,a) = Σn=0∞ (n+_a_)-s, which is the zeta function corresponding, for example, to the operator x(d/dx)+a, and we consider the case when a is not a nonpositive integer. One can look up various special values of (the analytic continuation) of the Hurwitz function, e.g. ζ(-m,a) = -Bm+1(a)/(m+1), where Br is the _r_th Bernoulli polynomial.
In particular,
TR(x(d/dx)+a) = -ζ(-1,a)/2 = -a2/2 + a/2 - 1/12
since, for example (from Wikipedia):
B2(a) = Σn=02 1/(n+1) Σk=0 n (-1)k {n \choose k} (a+_k_)2 = a2 - a + 1/6
Thus, consider the operator 2_x_(d/dx)+a+_b_. On the one hand:
TR(x(d/dx)+a) + TR(x(d/dx)+b) = -(a2+b2)/2 + (a+_b_)/2 - 1/6
On the other hand, TR is "linear" when it comes to multiplication by positive reals, and so:
TR(2_x_(d/dx)+a+_b_) = 2 TR(x(d/dx) + (a+_b_)/2) = -(a2+2_ab_+b2)/4 + (a+_b_)/2 - 1/6
In particular, we have TR(x(d/dx)+a) + TR(x(d/dx)+b) = TR( x(d/dx)+a + x(d/dx)+b ) if and only if a=_b_; otherwise 2_ab_ < a2+b2 is a strict inequality.
So the zeta-function regularized trace TR is not linear.
0./2.
My last comment is not so much to break number 2. above, but to suggest that it is limited in scope. In particular, for an operator A on an infinite-dimensional vector space, it is impossible for A-s to be trace-class for s in an open neighborhood of 0, and so if the zeta-function regularized DET makes sense, then det doesn't. (I.e. it's hopeless to say that det A = DET A.) Indeed, if the series converges for s=0, then it must be a finite sum.
Similarly, it is impossible for A to be trace class and also for A-s to be trace class for large s. If A is trace class, then its eigenvalues have finite sum, and in particular cluster near 0 (by the "divergence test" from freshman calculus). But then the eigenvalues of A-s tend to ∞ for positive s. I.e. it's hopeless to say that tr A = TR A.
My proof for 2. says the following. Suppose that dA/dt A-1 is trace class, and suppose that DET A makes sense as above. Then
d/dt [ DET A ] = (DET A)(tr dA/dt A-1)
I have no idea what happens, or even how to attack the problem, when dA/dt A-1 has a zeta-function-regularized trace.
Best Answer
OK, shameless cheating, as promised.
Part 1. Let's start with something.
We need a function bounded in $\Re z>1$ and growing not too fast on each vertical line whose zeroes are somewhere on the left. The first thing that comes to mind is $1$. No zeroes anywhere in sight, beautiful control on vertical lines. All that is lacking is the unboundedness on $\Re z=1$.
Part 2. Push it up!
We now want to add some bumps on the uneventful road $\Re z=1$. It is natural to add one bump a time. We have two options for bumping: addition and multiplication. Since we want to control the zeroes without trouble, we'll use multiplication. So, we'll be looking for an infinite product.
Part 3. A tiny little bump.
Take some entire function $g$ bounded by $1$ in the right half-plane, tending to $0$ at infinity in any right half-plane, and attaining its maximum of absolute value in $\Re z\ge 1$ at $1$, where it is real and positive. Denote $g(1)=a>0$. The exact choice doesn't matter. I'll take $g(z)=\frac{1-e^{-z}}{z}$. Put $F(z)=1+g(z)$. Now the ride along the line $\Re z=1$ is not that smooth anymore: you have to ascend to a small hill at $1$. However, at infinity everything levels to $1$ uniformly in any right half-plane. Also, if there are any zeroes, they all have non-positive real parts.
Part 4. Amplify the bump (being naive and fair)
Just raise $F$ to a high power $N$. You'll get as huge bump as you want. The problem is that it also becomes huge well to the right of $1$.
Part 5. Discriminate against numbers with the large real part.
Replace $g(z)$ by $g(z)e^{-n^2z}$. Of course the value at $1$ will suffer enormously, but everything with real part greater than $1$ will suffer much more (which is the whole point of any true discrimination).
Part 6: Amplify with discrimination.
Raise $F(z)=1+g(z)e^{-nz}$ to the power $N$. We'll get $(1+ae^{-n^2})^N$ at $1$ but only at most $(1+ae^{-n^2-2n})^N$ for $\Re z>1+\frac 2{n}$. Choose $N\approx a^{-1}e^{n^2+n}$. We'll get about $e^{n}$ at $1$ and at most $1+2e^{-n}$ to the right of $1+\frac 2n$. Now it is quite a bump, and it is next to invisible just a tiny bit to the right of $1$.
Part 7: Ship it up the line to satisfy the local regulations.
Replace $F(z)^N$ with $F(z-iy_n)^N$ with large $y_n$ to satisfy the polynomial growth restriction in $\Re z>-n$: let's even make $|F(z)^N-1|<2^{-n}(1+|z|)^{2^{-n}}$ in $\Re z>-n$. Remember that though our bump function is huge, it is still bounded in any right half plane and levels to $1$ at infinity there. We also have $|F(z)|^N\le 1+2e^{-n}$ when $\Re z>1+\frac 2n$ regardless of the shipment.
Part 8. Put the production and shipment of bumps on the conveyor belt with $n=1,2,3,...$, and enjoy the product.
Of course, this is as shameless, abominable, and mostly illegal as any manufacturing under loose government regulations. Every loophole that could be exploited in the formulation of the problem has been exploited. So, please, do not accept or upvote. Instead, think of how to tighten the regulations to force someone to do honest work. :)