1 Correction to: Results Math (2018) 73:92 https://doi.org/10.1007/s00025-018-0852-3

Abstract. Corrigendum to Results Math. (73)(2018), no. 3, Art. 92, 23 pp. Concerning the examples in Remark 4.3 and Remark 5.4, I correct a few calculations, but which do not influence the conclusions in these remarks.

Mathematics Subject Classification. 41A36, 41A25, 28A10, 28A12, 28A25.

Keywords. Submodular set function, Nonlinear Choquet integral, Picard–Choquet operators, Gauss–Weierstrass–Choquet operators, Poisson–Cauchy–Choquet operators.

2 Introduction

I correct a few calculations in Remarks 4.3 and 5.4 in [1], but which do not influence the conclusions of these remarks, as follows.

On page 14, line 5 from below, the equality \(W_{t, \mu _{t, x}}(f)(x)=f(x+t^{2})\) must be changed to \(f(x)\le W_{t, \mu _{t, x}}(f)(x)\le f(x+t^2)\).

On page 14, line 3 from below (i.e. in relation (3)), the sign “=” must be changed to “\(\le \)”.

On page 15, the whole line 18 from above must be replaced by

$$\begin{aligned} W_{t, \mu _{t, x}}(f)(x)= & {} \int _{0}^{f(x+t^{2})}\sup \{e^{-(s-x)^{2}/t^{2}} ; s\in \mathbb {R}, f(s)e^{-(s-x)^{2}/t^2}\ge \alpha \}d \alpha \\\le & {} f(x+t^{2}). \end{aligned}$$

Then, after this line must be inserted the following two formulas

$$\begin{aligned} W_{t, \mu _{t, x}}(f)(x)= & {} \int _{0}^{f(x+t^{2})}\sup \{e^{-(s-x)^{2}/t^{2}} ; s\in \mathbb {R}, f(s)e^{-(s-x)^{2}/t^2}\ge \alpha \}d \alpha \\\ge & {} \int _{0}^{f(x)}\sup \{e^{-(s-x)^{2}/t^{2}} ; s\in \mathbb {R}, f(s)e^{-(s-x)^{2}/t^2}\ge \alpha \}d \alpha \\= & {} \int _{0}^{1}1 d\alpha =f(x). \end{aligned}$$

On page 20, line 5 from below, the equality \(Q_{t, \mu _{t, x}}(f)(x)=f(x+t^{2})\) must be changed to \(f(x)\le Q_{t, \mu _{t, x}}(f)(x)\le f(x+t^2)\).

On page 20, line 3 from below (i.e. in relation (6)), the sign “=” must be changed to “\(\le \)”.

On page 21, the whole line 10 from below must be replaced by

$$\begin{aligned} Q_{t, \mu _{t, x}}(f)(x)= & {} \int _{0}^{f(x+t^{2})}\sup \left\{ \frac{1}{|s-x|^{2}/t^2+1} ; s\in \mathbb {R}, f(s)\cdot \frac{1}{|s-x|^{2}/t^{2}+1}\ge \alpha \right\} d \alpha \\\le & {} f(x+t^{2}). \end{aligned}$$

Then, after this line must be inserted the following three lines

$$\begin{aligned} Q_{t, \mu _{t, x}}(f)(x)= & {} \int _{0}^{f(x+t^{2})}\sup \left\{ \frac{1}{|s-x|^{2}/t^2+1} ; s\in \mathbb {R}, f(s)\cdot \frac{1}{|s-x|^{2}/t^{2}+1}\ge \alpha \right\} d \alpha \\\ge & {} \int _{0}^{f(x)}\sup \left\{ \frac{1}{|s-x|^{2}/t^2+1} ; s\in \mathbb {R}, f(s)\cdot \frac{1}{|s-x|^{2}/t^{2}+1}\ge \alpha \right\} d \alpha \\= & {} \int _{0}^{f(x)}1 d\alpha =f(x). \end{aligned}$$