# Newman's Inequality for Muntz Polynomials on Positive Intervals

Borwein, Peter and Erdelyi, Tamas (1994) Newman's Inequality for Muntz Polynomials on Positive Intervals. [Preprint]

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The principal result of this paper is the following Markov-type inequality for lacunary polynomials. Theorem (Newman's Inequality on $[a,b] \subset [0, \infty)$): Let $\Lambda:=\{\lambda_i\}_{i=1}^{\infty}$ be a sequence of nonnegative real numbers. Assume that there exists a $\delta > 0$ so that $\lambda_i \geq \delta i$ for each $i$. Suppose that $[a,b] \subset [0, \infty)$. Then there exists a constant $c(a,b,\delta)$ depending only on $a$, $b$, and $\delta$ so that $$\|P^{\prime}\|_{[a,b]} \leq c(a,b,\delta) \left( \shave{\sum_{j=0}^n} {\lambda_j} \right)\|P\|_{[a,b]}$$ for every $P \in M_n(\Lambda)$, where $M_n(\Lambda)$ denotes the linear span of $\{x^{\lambda_0}, x^{\lambda_1}, \ldots, x^{\lambda_n}\}$ over ${\Bbb R}$. When $[a,b] = [0,1]$ this was proved by Newman. Note that the interval $[0,1]$ plays a special role in the study of M\"untz spaces $M_n(\Lambda)$. A linear transformation $y=ax+b$ does not preserve membership in $M_n(\Lambda)$ in general (unless $b=0$). So the analogue of Newman's Inequality on $[a,b]$, $a >0$ does not seem to be obtainable in any straightforward fashion from the $[0,b]$ case.