Abstract
We prove that if { a n } n = 1 ∞ \{ {a_n}\} _{n = 1}^\infty is such that a n ↘ 0 {a_n} \searrow 0 and \[ lim n → ∞ ( a n + 1 / a n ) = 1 , \lim \limits _{n \to \infty } ({a_n}_{ + 1}/{a_n}) = 1, \] then for the typical continuous function f f we have \[ S n 0 := ∑ n = n 0 ∞ | f ( x n + 1 ) − f ( x n ) | = + ∞ {S_{{n_0}}}: = \sum \limits _{n = {n_0}}^\infty | f({x_n}_{ + 1}) - f({x_n})| = + \infty \] whenever x ∈ [ 0 , 1 − a n 0 ] x \in [0,1 - {a_{{n_0}}}] and x n ∈ [ x + a n + 1 , x + a n ] {x_n} \in [x + {a_{n + 1}},x + {a_n}] . Based on our result in a previous paper, we know that the above theorem fails to hold if a n + 1 / a n = λ > 1 {a_{n + 1}}/{a_n} = \lambda > 1 . We also prove that if { a n } n = 1 ∞ \{ {a_n}\} _{n = 1}^\infty is such that a n ↘ 0 {a_n} \searrow 0 , then for the typical continuous function f f we have S n 0 = + ∞ if x n = x + a n {S_{{n_0}}} = + \infty {\text { if }}{x_n} = x + {a_n} and x ∈ [ 0 , 1 − a n 0 ] x \in [0,1 - {a_{{n_0}}}] .
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.