On Hardy's integral inequality (Q2767900)

Let \(a=0\) and \(b=+\infty\). The classical Hardy inequality asserts that if \(p > 1\) and \(f\) is a non-negative measurable function on \((a,b)\), then NEWLINE\[NEWLINE \quad \int^b_a \left(\frac {1} {x} \int^x_a f (t) d t\right)^p d t < \left(\frac {p} {p-1}\right)^p \int^b_a f^p (t) d t NEWLINE\]NEWLINE unless \(f=0\) a.e. in \((a,b)\). By \textit{B. Yang, Z. Zeng} and \textit{L. Debnath} [J. Math. Anal. Appl. 217, No.~1, 321-327 (1998; Zbl 0893.26008)], this inequality remains true provided that \(0 < a < b < +\infty\). The author of the paper under review investigates the inequality mentioned above (and their modifications) with \(0 < a < b \leq + \infty\) in the case when the operator \(x^{-1} \int^x_a f(t) dt\) is replaced by \(x^{-1/r} \int^x_a f(t) dt\), \(r \geq 1\). However, his main results (Theorems 2.1 and 2.2) cannot hold if \(r > 1\). To see it, take \(b = +\infty\) and apply the Muckenhoupt criterion for the Hardy-type inequality. (Note also that all the fractions appearing in the paper are written without fraction lines).NEWLINENEWLINEFor the entire collection see [Zbl 0972.00006].