Abstract
For any countable set $D \subset [0,1]$, we construct a bounded measurable function $f$ such that the Fourier series of $f$ with respect to the regular general Haar system is divergent on $D$ and convergent on $[0,1]\backslash D$.
Highlights
Let {fn(x)}∞ n=1 be a sequence of functions, fn : [0, 1] → R for all n. ∞ For a functional series fn(x), the set D ⊂ [0, 1] is called a divergence n=1 set if the series is divergent for any x ∈ D and is convergent when x ∈/ D
Prokhorenko [18] proved that for any countable set F ⊂ [0, 1], there exists a bounded function such that the Fourier-Haar series of that function is divergent on F and convergent on [0, 1]\F
Bugadze [2] proved that for any set with 0 measure, there exists a bounded function such that its Fourier-Haar series is divergent on that set
Summary
Let {fn(x)}∞ n=1 be a sequence of functions, fn : [0, 1] → R for all n. ∞ For a functional series fn(x), the set D ⊂ [0, 1] is called a divergence n=1 set if the series is divergent for any x ∈ D and is convergent when x ∈/ D. There are many results concerning divergence sets for the Fourier series with respect to classical systems. V. Prokhorenko [18] proved that for any countable set F ⊂ [0, 1], there exists a bounded function such that the Fourier-Haar series of that function is divergent on F and convergent on [0, 1]\F. V. Bugadze [2] proved that for any set with 0 measure, there exists a bounded function such that its Fourier-Haar series is divergent on that set. Theorem 1 For any countable set D ⊂ [0, 1], there exists a bounded measurable function f : [0, 1] → R such that D is a divergence set for the Fourier series of f with respect to the regular general Haar system. The following question remains open: Is Theorem 1 true for every general Haar system (not regular)?
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