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• Thread: 2071 Goods vs 891 No Goods

1. Economist
   86e8
   

   @5716, to be clear, are you saying that it doesn't follow from (18) that alpha_{0}(s) has a holomorphic CONTINUATION to Re(sigma) > Theta ?

   There are two separate assertions one can consider here, so let's be careful about it.

   1) The function {Re(s)>1} -> R \subset C defined by formula (18) is not holomorphic.

   2) Let U \subset R^2 be the subset of pairs (x,y) with x>1 and let V \subset C \times R be the open of pairs (s,y) where Re(s)>1. The function U -> C defined by (x,y) -> \alpha_0(x+iy) does admit an extension to a function V -> C that is holomorphic in the first variable; let me call this extension \alpha_1(s,y). In both of these, the paper convincingly argues that you can perfectly well analytically continue to x>\Theta or Re(s)>\Theta.

   However, \alpha_1(s,y) from 2) has essentially nothing to do with formula (18), though they coincide with \sigma is real.

   Now, the argument suggested in the paper to show f(\sigma) has an analytic extension to Re(\sigma)>\Theta needs to show that the integral \int_{t \in R} |\alpha_1(\sigma,t)|dt converges uniformly on compact sets K \subset {Re(\sigma)>\Theta}.

   To do this, you need to estimate the size of the function \alpha_1(\sigma,t). The analysis in the paper applies when \sigma is real, but this alone is not sufficient to deduce the assertion on analytic continuation of f(\sigma). Things would be easy if you had to estimate the function from 1) (i.e., defined by equation (18)), and you have written posts correctly saying this is trivial from what's written. But the problem is you need to estimate the function \alpha_1 from 2).

   6 months ago # QUOTE 0 Volod 0 Vlad !

2. Economist
   cfcf
   

   @5716, to be clear, are you saying that it doesn't follow from (18) that alpha_{0}(s) has a holomorphic CONTINUATION to Re(sigma) > Theta ?

   There are two separate assertions one can consider here, so let's be careful about it.
   1) The function {Re(s)>1} -> R \subset C defined by formula (18) is not holomorphic.
   2) Let U \subset R^2 be the subset of pairs (x,y) with x>1 and let V \subset C \times R be the open of pairs (s,y) where Re(s)>1. The function U -> C defined by (x,y) -> \alpha_0(x+iy) does admit an extension to a function V -> C that is holomorphic in the first variable; let me call this extension \alpha_1(s,y). In both of these, the paper convincingly argues that you can perfectly well analytically continue to x>\Theta or Re(s)>\Theta.
   However, \alpha_1(s,y) from 2) has essentially nothing to do with formula (18), though they coincide with \sigma is real.
   Now, the argument suggested in the paper to show f(\sigma) has an analytic extension to Re(\sigma)>\Theta needs to show that the integral \int_{t \in R} |\alpha_1(\sigma,t)|dt converges uniformly on compact sets K \subset {Re(\sigma)>\Theta}.
   To do this, you need to estimate the size of the function \alpha_1(\sigma,t). The analysis in the paper applies when \sigma is real, but this alone is not sufficient to deduce the assertion on analytic continuation of f(\sigma). Things would be easy if you had to estimate the function from 1) (i.e., defined by equation (18)), and you have written posts correctly saying this is trivial from what's written. But the problem is you need to estimate the function \alpha_1 from 2).

   It seems to me that when sigma is complex and Re(sigma) > Theta, the estimate for alpha_{1}(sigma + it) is exactly that of alpha_{0}(sigma + it), since the latter is a holomorphic continuation of the former.

   I mean, if sigma is fixed and complex so that sigma = sigma' + it', then

   s = sigma + it = sigma' + i(t+t')

   thus

   alpha_{1}(sigma + i(t + t')).

   Thus, estimating alpha_{1}(sigma + it) as t tends to infinity for fixed sigma in C, is essentially similar to estimating the quantity for real sigma.

   6 months ago # QUOTE 0 Volod 0 Vlad !

3. Economist
   cfcf
   

   Alternatively, you may forget about everything I said, and try to estimate alpha_{1}(sigma + it) yourself, for sigma in C. I don't think this should be too hard for you.

   It should help to recall that alpha_{1}(sigma + it) is the holomorphic continuation of alpha_{0}(sigma + it) to Re(sis) > Theta. So you just want to estimate alpha_{0}(sigma + it) for sigma in C.

   6 months ago # QUOTE 0 Volod 0 Vlad !

4. Economist
   86e8
   

   I mean, if sigma is fixed and complex so that sigma = sigma' + it', then
   s = sigma + it = sigma' + i(t+t')
   thus
   alpha_{1}(sigma + i(t + t')).
   Thus, estimating alpha_{1}(sigma + it) as t tends to infinity for fixed sigma in C, is essentially similar to estimating the quantity for real sigma.

   This is my objection. It's not true that \alpha_1(\sigma+it,t') = \alpha_1(\sigma,t+t') -- replace \alpha_0 by a function like z -> |z|^2 to see this (as I wrote explicitly earlier).

   6 months ago # QUOTE 0 Volod 0 Vlad !

5. Economist
   cfcf
   

   I mean, if sigma is fixed and complex so that sigma = sigma' + it', then
   s = sigma + it = sigma' + i(t+t')
   thus
   alpha_{1}(sigma + i(t + t')).
   Thus, estimating alpha_{1}(sigma + it) as t tends to infinity for fixed sigma in C, is essentially similar to estimating the quantity for real sigma.

   This is my objection. It's not true that \alpha_1(\sigma+it,t') = \alpha_1(\sigma,t+t') -- replace \alpha_0 by a function like z -> |z|^2 to see this (as I wrote explicitly earlier).

   Except that in this case, we indeed have

   alpha_1}(sigma, t) = alpha(sigma, t + t')

   since alpha_1 is holomorphic. It's bound is already what we have for alpha_0.

   6 months ago # QUOTE 0 Volod 0 Vlad !

6. Economist
   86e8
   

   Is that supposed to be \alpha_1 in place of \alpha on the right hand side?

   6 months ago # QUOTE 0 Volod 0 Vlad !

7. Economist
   cfcf
   

   Is that supposed to be \alpha_1 in place of \alpha on the right hand side?

   Oh, yes...typo.

   6 months ago # QUOTE 0 Volod 0 Vlad !

8. Economist
   cfcf
   

   In fact, we could actually bound alpha/{1}(sigma + it) for complex sigma, directly from (18), by the usual methods in the literature (of bounding alpha).

   6 months ago # QUOTE 0 Volod 0 Vlad !

9. Economist
   86e8
   

   Then I'm not sure my message got across. If your post should say:

   alpha_1(sigma+it', t) = alpha_1(sigma, t + t') = alpha_0(sigma+i(t+t'))

   Then this is false (for \alpha_1 defined as in my 2) from the post at the top of this page). Very simply, for t fixed, this formula would imply that the non-constant holomorphic function z \mapsto alpha_1(z,t) takes only real values, and this is impossible.

   In fact, we could actually bound alpha/{1}(sigma + it) for complex sigma, directly from (18), by the usual methods in the literature (of bounding alpha).

   Note that formula (18) describes a real-valued function, not \alpha_1 as considered in my point 2) above. So I don't understand your proposal.

   6 months ago # QUOTE 0 Volod 0 Vlad !

10. Economist
    86e8
    

    So are we settled that this is a gap?

    6 months ago # QUOTE 0 Volod 0 Vlad !

11. Economist
    2adc
    

    So are we settled that this is a gap?

    No. The main mistake you're making is that, it seems you'rethinking that the expression in (18) is real-valued, but it's actually not. That is, it is not a C \mapsto R function, but it is complex-valued and indeed analytic for sigma in C, Re(sigma) > Theta.

    It seems you're being mislead by the Re(...) and Im(...) in the brackets to thinking that the expression is real-valued. However, the correct way to see it is that,

    Re(x^{-sigma-1-it}) = x^{-sigma-1}cos(t log x),

    and

    Im(x^{-_sigma-1-it}) = -x^{-sigma - 1}sin(t log x).

    6 months ago # QUOTE 0 Volod 0 Vlad !

12. Economist
    7644
    

    Alessandro is a visiting postdoc working with Michael Tschantz in the Usable Security and Privacy group. He earned a MASt from the University of Cambridge, UK, and a Ph.D. from Stanford. His math research is in geometry and number theory and he has been progressively becoming interested in applied aspects of mathematics. His current research while at the Institute is in applications of game theory to differential privacy.

    4 months ago # QUOTE 0 Volod 0 Vlad !

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