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Very slowly varying functions - MaRDI portal

Very slowly varying functions (Q2264872)

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Very slowly varying functions
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    Very slowly varying functions (English)
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    1974
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    Let \(\varphi\) be a positive non-decreasing real valued function defined on \([0, \infty)\), and let \(f\) be any real valued function defined on \([0, \infty)\). We say that \(f\) is \(\varphi\)-slowly varying if \(\varphi (x)[f(x+ \alpha)-f(x)] \to 0\) as \(x \to \infty\) for each \(\alpha\). We say that \(f\) is uniformly \(\varphi\)-slowly varying if \(\sup \{\varphi (x) |f(x+ \alpha)-f(x)|: \alpha \in I \} \to 0\) as \(x \to \infty\) for every bounded interval \(I\). We state here five theorems that will be proved later in a longer communication. We also pose one question that seems to be difficult. Theorem 1. If \(f\) is \(\varphi\)-slowly varying and if \(\sum i/ \varphi (n)< \infty\), then \(f\) tends to a finite or infinite limit at \(\infty\). Theorem 2. If \(f\) is \(\varphi\)-slowly varying and measurable, then \(f\) is uniformly \(\varphi\)-slowly vayring. Theorem 3. Let \(f\) be \(\varphi\)-slowly varying and let \(\beta (x)= \sum ^\infty_{j=0} 1/ \varphi (x+j)\). If \(\varphi (x) \beta (x)\) is bounded, then \(f\) must be uniformly \(\varphi\)-slowly varying. Theorem 4. Suppose that \(\sum 1/ \varphi (n)< \infty\) and that \(\varphi (x+1)/ \varphi (x) \to 1\) as \(x \to \infty\). Then there exists a function \(f\) that is \(\varphi\)-slowly varying but not uniformly \(\varphi\)-slowly varying. Theorem 5. Let \(\beta (x)\) be the function of Theorem 4, and suppose that \(\varphi (x) \beta (x)\) is unbounded, but that \(\varphi (x) \beta (x)=o(x)\) as \(x \to \infty\). Then there exists a function \(f\) that is \(\varphi\)-slowly varying but not uniformly \(\varphi\)-slowly varying. Question. Does there exist a function \(f\) such that \(x[f(x+ \alpha)-f(x)] \to 0\) as \(x \to \infty\) for each \(\alpha\) but \(\sup \{|f(x+ \alpha)-f(x)|: \alpha \in [0,1] \} \nrightarrow 0\) as \(x \to \infty\)?
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