On the best exponent in Markov inequality
Recall that a compact set \(E\subset\mathbb{C}^{N}\) is said to be a Markov set if there are two constants \(M>0\) and \(m>0\) such that the following inequality \[ \||\nabla P|\|\leq M(\deg P)^m\| P\|_E,\tag{(*)} \] for each polynomial \(P\) holds, where \(\nabla P=(\frac{\partial P}{\partial z_1},\dotsc, \frac{\partial P}{\partial z_N})\), \(|\nabla P|=\left(\sum_{j=1}^N\left| \frac{\partial P}{\partial z_j}\right|^2\right)^{1/2}\) and \(\|\cdot\|_E\) is the sup-norm on \(E\). If \(E\) is a Markov set, then the Markov exponent of \(E\) is defined by \[ m(E):=\inf\{m>0:\exists M>0\text{ such that for all } P \text{ inequality (*) holds}\}. \] (If \(E\) is not a Markov set, then \(m(E):=\infty\).) (see [\textit{M. Baran} and \textit{W. Pleśniak}, Proc. Am. Math. Soc. 123, No. 9, 2785--2791 (1995; Zbl 0841.41011)]). The present paper contains several new results on Markov exponents. The main results of them are (i) for each \(N\geq2\) and an arbitrary \(p\geq1\) a compact set \(E\subset\mathbb{C}^N\) is constructed such \(m(E)=p\) and the Markov inequality (*) does not hold with exponent \(m=p\); (ii) a compact set \(E\subset\mathbb{C}^2\) is constructed with the property that the Markov inequality holds pointwise for each point of \(E\), but \(E\) is not a Markov set; (iii) in the case \(N=1\), it is shown that holomorphic functions transform Markov sets into Markov sets and an estimate of the transformation of the Markov exponent is given.
scientific article; zbMATH DE number 6136710
scientific article