Correction to: Results Math. 75:180 (2020) https://doi.org/10.1007/s00025-020-01308-y

There is an inadvertent mistake in the statement of Theorem 3.2 and its proof. The authors are indebted to Ioan Raşa for pointing out the mistake in the statement of Theorem 3.2, which has led us to revise the proof.

The theorem should be

FormalPara Theorem 3.2

Let \(n,q\in \mathbb {N}\). If f defined on \([0,\infty )\) and such that \(|f(x)|\le C(1+x)^{\gamma }\), for some \(C,\gamma >0\), is a q-monotone function there, then for all \(x,y\in [0,\infty )\),

$$\begin{aligned} \textrm{sgn}(x-y)^{q}\sum _{\nu _{1},\dots ,\nu _{q}=0}^{\infty }\sum _{j=0}^{q}(-1)^{q-j}\left( {\begin{array}{c}q\\ i\end{array}}\right)&\left( \prod _{i=1}^{j}m_{n,\nu _{i}}(x)\right) \left( \prod _{i=j+1}^{q}m_{n,\nu _{i}}(y)\right) \\&\times \int _0^1 f\left( \frac{\nu _{1}+\cdots +\nu _{q}+\alpha t}{n+\alpha }\right) dt \ge 0. \end{aligned}$$
FormalPara Proof of Theorem 3.2

Let \(n,m,q\in \mathbb {N}\), and denote \(\mathbf {\nu }:=(\nu _1,\dots ,\nu _q)\in ( \mathbb {N}_0)^q\). By (3.2) it follows that

$$\begin{aligned} \sum _{|\mathbf {\nu }|=m}\prod _{i=1}^{q}m_{n,\nu _i}(x_i)= & {} \frac{1}{m!}\sum _{| \mathbf {\nu }|=m}\left( {\begin{array}{c}m\\ \mathbf {\nu }\end{array}}\right) \prod _{i=1}^q\left. \left( \frac{\partial }{\partial z}\right) ^{\nu _i}(1-x_iz)^{-n}\right| _{z=-1}\nonumber \\= & {} \frac{1}{m!}\left. \left( \frac{\partial }{\partial z}\right) ^m \prod \limits _{i=1}^{q}\left( 1-x_iz\right) ^{-n}\right| _{z=-1}. \end{aligned}$$
(3.4)

Hence, as is done in (2.2), for \(0\le x,y<\infty \) and any sequence \( (a_k)_{k=0}^\infty \), we have

$$\begin{aligned}{} & {} \sum _{|\mathbf {\nu }|\ge 0}\sum _{j=0}^{q}(-1)^{q-j}\left( {\begin{array}{c}q\\ j\end{array}}\right) \left( \prod _{i=1}^{j}m_{n,\nu _{i}}(x)\right) \left( \prod _{i=j+1}^{q}m_{n,\nu _{i}}(y) \right) a_{|\mathbf {\nu }|}\nonumber \\{} & {} \quad =\sum \limits _{m=0}^{\infty }a_{m}\frac{1}{m!}\left. \left[ \left( \frac{ \partial }{\partial z}\right) ^m\left[ (1-xz)^{-n}-(1-yz)^{-n}\right] ^q \right] \right| _{z=-1}=:I.\nonumber \\ \end{aligned}$$
(3.5)

Let

$$\begin{aligned} h(z):=\bigl ((1-xz)^{-n}-(1-yz)^{-n}\bigr )^q. \end{aligned}$$

Then

$$\begin{aligned} h(z)= & {} ((1-xz)^{-1}-(1-yz)^{-1})^q\left( \frac{(1-xz)^{-n}-(1-yz)^{-n}}{(1-xz)^{-1}-(1-yz)^{-1}}\right) ^q\\= & {} \left( \frac{(1-yz)-(1-xz)}{(1-xz)(1-yz)}\right) ^q\left( \sum _{m=0}^{n-1}(1-xz)^{-m}(1-yz)^{-(n-1-m)}\right) ^q\\= & {} (x-y)^qz^q\left( \sum _{m=1}^n(1-xz)^{-m}(1-yz)^{-(n+1-m)}\right) ^q\\=: & {} (x-y)^qz^qg(z). \end{aligned}$$

Note that \(\left. \left( \partial /\partial z\right) ^j(1-uz)^{-m} \right| _{z=-1}=j!m_{m,j}(u)\ge 0\), for \(j,m=0,1,2,\ldots \), and \(0\le u<\infty \). Hence

$$\begin{aligned} g^{(k)}(-1)\ge 0,\quad k\ge 0. \end{aligned}$$
(3.6)

Finally, we may rewrite the last line in (3.5), as we did in (3.3),

$$\begin{aligned} I= & {} \sum _{m=0}^{\infty }a_{m}\frac{1}{m!}\left. \left[ \left( \frac{ \partial }{\partial z}\right) ^{m}\left[ (1-xz)^{-n}-(1-yz)^{-n}\right] ^{q}\right] \right| _{z=-1} \\= & {} (x-y)^q\sum _{m=0}^\infty a_m\frac{1}{m!}\left. \frac{d^m}{dz^m} \left( z^qg(z)\right) \right| _{z=-1} \\= & {} (x-y)^q\sum _{m=0}^\infty \frac{1}{m!}g^{(m)}(-1)\Delta ^qa_m, \end{aligned}$$

and Theorem 3.2 follows by taking \(a_m:=\int _0^1f\left( \frac{m+\alpha t}{n+\alpha }\right) dt\). \(\square \)