Is there a bounded function that is always greater than $M (t) = max_s in [0, t] left| zeta left( frac12 + i s right) right|$?

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Is there a bounded function that is always greater than $M (t) = max_s in [0, t] left| zeta left( frac12 + i s right) right|$ ?




riemann-zeta upper-lower-bounds




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  • 4




    This is simply equivalent to asking whether $zeta(z)$ is bounded on the line $Re z=1/2$.
    – Cave Johnson
    Jul 22 at 21:38






  • 1




    Probably not. That's stronger than the Lindelöf hypothesis.
    – Daniel Fischer♦
    Jul 22 at 21:39














up vote
2
down vote

favorite












Is there a bounded function that is always greater than $M (t) = max_s in [0, t] left| zeta left( frac12 + i s right) right|$ ?




riemann-zeta upper-lower-bounds




share|cite|improve this question















  • 4




    This is simply equivalent to asking whether $zeta(z)$ is bounded on the line $Re z=1/2$.
    – Cave Johnson
    Jul 22 at 21:38






  • 1




    Probably not. That's stronger than the Lindelöf hypothesis.
    – Daniel Fischer♦
    Jul 22 at 21:39












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Is there a bounded function that is always greater than $M (t) = max_s in [0, t] left| zeta left( frac12 + i s right) right|$ ?




riemann-zeta upper-lower-bounds




share|cite|improve this question











Is there a bounded function that is always greater than $M (t) = max_s in [0, t] left| zeta left( frac12 + i s right) right|$ ?





riemann-zeta upper-lower-bounds





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 22 at 21:16









crow

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  • 4




    This is simply equivalent to asking whether $zeta(z)$ is bounded on the line $Re z=1/2$.
    – Cave Johnson
    Jul 22 at 21:38






  • 1




    Probably not. That's stronger than the Lindelöf hypothesis.
    – Daniel Fischer♦
    Jul 22 at 21:39












  • 4




    This is simply equivalent to asking whether $zeta(z)$ is bounded on the line $Re z=1/2$.
    – Cave Johnson
    Jul 22 at 21:38






  • 1




    Probably not. That's stronger than the Lindelöf hypothesis.
    – Daniel Fischer♦
    Jul 22 at 21:39







4




4




This is simply equivalent to asking whether $zeta(z)$ is bounded on the line $Re z=1/2$.
– Cave Johnson
Jul 22 at 21:38




This is simply equivalent to asking whether $zeta(z)$ is bounded on the line $Re z=1/2$.
– Cave Johnson
Jul 22 at 21:38




1




1




Probably not. That's stronger than the Lindelöf hypothesis.
– Daniel Fischer♦
Jul 22 at 21:39




Probably not. That's stronger than the Lindelöf hypothesis.
– Daniel Fischer♦
Jul 22 at 21:39















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