Generalized inequalities for integral operators via several kinds of convex functions

1.   Introduction

Convex functions play a vital role in Mathematical analysis, Mathematical statistics and Optimization theory. These functions produce an elegant theory of convex analysis, see (^[1,2,3]).

Definition 1. A function $f:[a, b]\rightarrow \mathbb{R}$ is said to be convex if

holds for all $x, y\in [a, b]$ and $t\in \left[0, 1\right]$. If inequality $(1.1)$ holds in revers order, then $f$ will be concave on $[a, b]$.

Convex functions have been generalized theoretically extensively; these generalizations include $m$-convex function, $n$-convex function, $r$-convex function, $h$-convex function, $(h-m)$-convex function, $(\alpha, m)$-convex function, $s$-convex function and many others. We are interested in using $(h-m)$-convex function ^[4] and $(\alpha, m)$-convex function ^[5].

Definition 2. Let $J\subseteq \mathbb{R}$ be an interval containing $(0, 1)$ and let $h: J\to \mathbb{R}$ be a non-negative function. A function $f:[0, b]\to \mathbb{R}$ is called $(h-m)$-convex function, if $f$ is non-negative satisfying

for all $x, y \in [0, b]$, $m\in [0, 1]$, $\alpha \in (0, 1)$.

By choosing suitable values for function $h$ and $m$, the above definition produces the functions defined on non-negative real line which are comprised in the following remark:

Remark 1. (i) If $m = 1$, then $h$-convex function is obtained.

(ii) If $h(x) = x$, then $m$-convex function is obtained.

(iii) If $h(x) = x$ and $m = 1$, then convex function is obtained.

(iv) If $h(x) = 1$ and $m = 1$, then $p$-function is obtained.

(v) If $h(x) = x^{s}$ and $m = 1$, then $s$-convex function is obtained.

(vi) If $h(x) = \dfrac{1}{x}$ and $m = 1$, then Godunova-Levin function is obtained.

(vii) If $h(x) = \dfrac{1}{x^{s}}$ and $m = 1$, then $s$-Godunova-Levin function of second kind is obtained.

For some recent citations and utilizations of $(h-m)$-convex functions one can see ^[4,6,7].

Definition 3. A function $f:[0, b]\rightarrow \mathbb{R}, \, b > 0$ is said to be $(\alpha, m)$-convex, where $(\alpha, m)\in[0, 1]^{2}$ if

holds for all $x, y\in[0, b]\, \, and\, \, t\in[0, 1].$

The functions which are deducible from above definition are given in the following remark:

Remark 2. (i) If $(\alpha, m)$ = $(1, m)$, then (1.3) produces the definition of $m$-convex function.

(ii) If $(\alpha, m)$ = $(1, 1)$, then (1.3) produces the definition of convex function.

(iii) If $(\alpha, m)$ = $(1, 0)$, then (1.3) produces the definition of star-shaped function.

For some recent citations and utilizations of $(\alpha, m)$-convex functions one can see ^{[8,9,10,11,12]}.

Recently, many mathematicians have used convex function to obtain fractional and conformable integral inequalities, see (^[13,14,15]). Our objective in this paper is to prove general inequalities for integral operators given in Definition 8 via $(h-m)$-convex and $(\alpha, m)$-convex functions. These results are interestingly associated with fractional and conformable integral operators. In the following we give some definitions of operators derivable from integral operators given in Definition 8.

Definition 4. Let $f\in L_{1}[a, b]$. Then the left-sided and right-sided Riemann-Liouville fractional integral operators of order $\mu\in\mathbb{C}, \, \, (\mathcal{R}(\mu) > 0)$ are defined by

A $k$-analogue of Riemann-Liouville fractional integral operator is given in ^[16].

Definition 5. Let $f\in L_{1}[a, b]$. Then the $k$-fractional integral operators of $f$ of order $\mu, \mathcal{R}(\mu) > 0$, $k > 0$ are defined by

A more general definition of the Riemann-Liouville fractional integral operators is given in ^[17].

Definition 6. Let $f:[a, b] \rightarrow \mathbb{R}$ be a integrable function. Also let $g$ be an increasing and positive function on $(a, b]$, having a continuous derivative $g'$ on $(a, b)$. The left-sided and right-sided fractional integrals of a function $f$ with respect to another function $g$ on $[a, b]$ of order $\mu\in\mathbb{C}, \, \, (\mathcal{R}(\mu) > 0)$ are defined by

where $\Gamma(.)$ is the gamma function.

A $k$-analogue of above definition is given in ^[18].

Definition 7. Let $f:[a, b] \rightarrow \mathbb{R}$ be an integrable function. Also let $g$ be an increasing and positive function on $(a, b]$, having a continuous derivative $g'$ on $(a, b)$. The left-sided and right-sided $k$-fractional integral operators of a function $f$ with respect to another function $g$ on $[a, b]$ of order $\mu, \mathcal{R}(\mu) > 0$ are defined by

where $\Gamma_{k}(.)$ is the $k$-analogue of gamma function.

The following integral operator is given in ^[19].

Definition 8. Let $f\, , \, g\, :[a, b]\longrightarrow \mathbb{R}\, \, , \, 0 < a < b$ be the functions such that $f$ be positive and $f\in L_{1}[a, b]$, and $g$ be differentiable and strictly increasing. Also let $\dfrac{\phi}{x}$ be an increasing function on $[a, \infty)$. Then for $x\in [a, b]$ the left and right integral operators are defined by

where $K_{g}(x, y;\phi) = \dfrac{\phi(g(x)-g(y))}{g(x)-g(y)}$.

Integral operators defined in (1.12) and (1.13) produce several fractional and conformable integral operators defined in ^{[17,20,21,22,23,24,25,26,27]}.

Remark 3. Integral operators given in (1.12) and (1.13) produce several known fractional and conformable integral operators corresponding to different settings of $\phi$ and $g$.

(i) If we put $\phi(t) = \frac{t^{\frac{\mu}{k}}}{k\Gamma_{k}(\mu)}$, then (1.12) and (1.13) integral operators coincide with (1.10) and (1.11) fractional integral operators.

(ii) If we put $\phi(t) = \frac{t^{\mu}}{\Gamma(\mu)}$, $\mu\, > \, 0$, then (1.12) and (1.13) integral operators coincide with (1.8) and (1.9) fractional integral operators.

(iii) If we put $\phi(t) = \frac{t^{\frac{\mu}{k}}}{k\Gamma_{k}(\mu)}, \mu\geq k$ and $g(x) = x$, then (1.12) and (1.13) integral operators coincide with (1.6) and (1.7) fractional integral operators.

(iv) If we put $\phi(t) = \frac{t^{\mu}}{\Gamma(\mu)}$, $\mu\, > \, 0$ and $g(x) = x$, then (1.12) and (1.13) integral operators coincide with (1.4) and (1.5) fractional integral operators.

(v) If we put $\phi(t) = \frac{t^{\mu}}{\Gamma(\mu)}$ and $g(x)\, = \, \frac{x^{\rho}}{\rho}, \, \rho\, > \, 0,$ then (1.12) and (1.13) produce Katugampola fractional integral operators defined by Chen et al. in ^[20].

(vi) If we put $\phi(t) = \frac{t^{\mu}}{\Gamma(\mu)}$ and $g(x)\, = \frac{x^{\tau+s}}{\tau+s}, \, s > 0$, then (1.12) and (1.13) produce generalized conformable integral operators defined by Khan et al. in ^[24].

(vii) If we put $\phi(t) = \frac{t^{\frac{\mu}{k}}}{k\Gamma_{k}(\mu)}$, $g(x) = \frac{(x-a)^s}{s}, \, s > 0$, in (1.12), and $\phi(t) = \frac{t^{\frac{\mu}{k}}}{k\Gamma_{k}(\mu)}$, $g(x)$ = $-\frac{(b-x)^s}{s}, \, s > 0$ in (1.13) respectively, then conformable $(k, s)$-fractional integrals are obtained as defined by Habib et al. in ^[22].

(viii) If we put $\phi(t) = \frac{t^{\frac{\mu}{k}}}{k\Gamma_{k}(\mu)}$ and $g(x)\, = \frac{x^{1+s}}{1+s}$, then (1.12) and (1.13) produce conformable fractional integrals defined by Sarıkaya et al. in ^[26].

(ix) If we put $\phi(t) = \frac{t^{\mu}}{\Gamma(\mu)}$, $g(x) = \frac{(x-a)^s}{s}, \, s > 0$ in (1.12) and $\phi(t) = \frac{t^{\mu}}{\Gamma(\mu)}$, $g(x) = -\frac{(b-x)^s}{s}, \, s > 0$ in (1.13) respectively, then conformable fractional integrals are obtained as defined by Jarad et al. in ^[23].

(x) If we put $\phi(t) = t^\frac{\mu}{k}\mathcal{F}_{\rho, \lambda}^{\sigma, k}(w(t)^{\rho})$, then (1.12) and (1.13) produce generalized $k$-fractional integral operators defined by Tunc et al. in ^[27].

(xi) If we put $\phi(t) = \dfrac{exp{(-At)}}{\mu}, A = \dfrac{1-\mu}{\mu}, \mu > 0,$ then following generalized fractional integral operators with exponential kernel defined in ^[21] are obtained and given as follows:

(xii) If we put $\phi(t) = \dfrac{t^{\mu}}{\Gamma(\mu)}$ and $g(t) = lnt$, then Hadamard fractional integral operators are obtained ^[17,25].

(xiii) If we put $\phi(t) = \dfrac{t^{\mu}}{\Gamma(\mu)}$ and $g(t) = -t^{-1}$, then Harmonic fractional integral operators defined in ^[17] are obtained and given as follows:

(xiv) If we put $\phi(t) = t^{\mu}lnt$, then left and right sided-logarithmic fractional integrals defined in ^[21] are obtained and given as follows:

Fractional integrals and derivatives play very vital role in advancement of almost all subjects of sciences and engineering. Now a days fractional and conformable integral operators have been used frequently in the advancements of classical inequalities, see ^{[7,10,18,20,21,22,28,29,30,31,32,33]}.

The aim of this paper is to derive general integral operator inequalities by using $(h-m)$ and $(\alpha, m)$-convex functions which will hold for all types of fractional and conformable integral operators and functions which are explained in the above discussions.

In Section 2 we derive the bounds of integral operators given in (1.12) and (1.13) by using $(h-m)$-convexity of function $f$ and $|f'|$. The bounds of various fractional integral operators and conformable integrals can be obtained by setting appropriate values of functions involved in (1.12) and (1.13). We also present results for $(\alpha, m)$ convex functions. The results of this paper hold for all kinds of functions explained in Remark 1 and Remark 2. In Section 3 we apply some of the results and get interesting consequences.

2.   Results and discussions

First we give results for $(h-m)$-convex functions.

2.1.   Inequalities for integral operators via $(h-m)$-convex function

Theorem 1. Let $f:[a, b]\longrightarrow \mathbb{R}, 0\leq a < b$, be a positive integrable $(h-m)$-convex function, $m\in(0, 1]$ and $h\in L_{\infty}[0, 1]$. Let $g:[a, b]\longrightarrow \mathbb{R}$ be differentiable and strictly increasing function, also let $\frac{\phi}{x}$ be an increasing function. Then for $x\in [a, b]$ the following inequalities for integral operators (1.12) and (1.13) hold:

Hence

Proof. Under given conditions, for $x \in (a, b]\, \, and\, \, t \in[a, x)$ the following inequality holds true:

Since $f$ is $(h-m)$-convex, we have

Inequalities (2.4) and (2.5) produce the following inequality:

This further takes form as follows:

Now on the other hand for $t \in (x, b]$ and $x \in [a, b)$, the following inequality holds true:

Again by using $(h-m)$-convexity of $f$ we have

From inequalities (2.8) and (2.9), the following integral operator inequality holds:

By adding (2.7) and (2.10), inequality $(2.3)$ can be obtained.

Remark 4. 1. By setting $h(x) = x$ and $m = 1$ in (2.3), [34], Theorem 1] can be reproduced.

2. By setting $h(x) = x$, $m = 1$ and $\phi(t) = \frac{t^{x}}{\Gamma(x)}, x > 0$ in (2.3), [33,Corollary 1] can be reproduced.

3. By setting $h(x) = x$, $m = 1$, $\phi(t) = \frac{t^{x}}{\Gamma(x)}, x > 0$ and $g(x) = x$ in (2.3), [28,Corollary 1] can be reproduced.

Theorem 2. Let $f:[a, b]\longrightarrow \mathbb{R}, 0\leq a < b$, be a differentiable function. Let $|f'|$ be $(h-m)$-convex, $m\in(0, 1]$, $h\in L_{\infty}[0, 1]$, $g:[a, b]\longrightarrow \mathbb{R}$ be differentiable and strictly increasing function, also let $\frac{\phi}{x}$ be an increasing function. Then for $x\in [a, b]$ the following inequality holds:

where

Proof. Let $x \in (a, b]$ and $t \in [a, x]$. Then $(h-m)$-convexity of $|f'|$ gives

From which we can write

Inequalities (2.4) and (2.13) produce the following integral inequality:

This further takes form as follows:

From (2.12) we can also write

Adopting the same procedure as we did for (2.13), the following inequality holds:

From (2.14) and (2.16), the following inequality is obtained:

Now using $(h-m)$-convexity of $|f'|$ on $(x, b]$ for $x\in(a, b)$ we have

From which we can write

Inequalities (2.8) and (2.19) produce the following integral inequality:

This further takes form as follows:

From (2.18) we can also write

Adopting the same procedure as we did for (2.19), the following inequality holds:

From (2.20) and (2.22), the following modulus inequality is obtained:

By adding (2.17) and (2.23), inequality (2.11) can be obtained.

Remark 5. By setting $h(x) = x$ and $m = 1$ in (2.11), [34], Theorem 1] can be reproduced.

To prove the next result we need the following lemma.

Lemma 1. ^[35] Let $f:[0, \infty)\rightarrow\mathbb{R}$ be a $(h-m)$-convex function, $m \in(0, 1]$. If $0\leq a < b$ and $f(x) = f\left(\dfrac{a+b-x}{m}\right)$, then the following inequality holds:

Theorem 3. Let $f: [a, b] \longrightarrow \mathbb{R}, 0\leq a < b$, be a positive $(h-m)$-convex with $m \in(0, 1]$, $f(x) = f\left(\dfrac{a+b-x}{m}\right)$, and $g:[a, b] \longrightarrow \mathbb{R}$ be a differentiable and strictly increasing function. Also let $\dfrac{\phi}{x}$ be an increasing function. Then the following Hadamard inequality is valid:

Proof. Under given conditions, the following inequality holds for $x\in(a, b]$:

Using $(h-m)$-convexity of $f$ for $x \in [a, b]$, we have

Inequalities (2.26) and (2.27) produce the following integral inequality:

This further takes form as follows:

On the other hand for $x\in[a, b)$, the following inequality holds true:

Adopting the same procedure as we did for (2.26) and (2.27), the following inequality can be obtained from $(2.27)$ and $(2.29)$:

By adding (2.28) and (2.30), the second inequality in (2.25) is obtained. Multiplying both sides of (2.24) by $K_{g}(x, a;\phi)g'(x)$, and integrating over $[a, b]$ we have

This further takes form as follows:

Similarly multiplying both sides of (2.24) by $K_{g}(b, x;\phi)g'(x)$, and integrating over $[a, b]$ we get

From (2.31) and (2.32), the first inequality in (2.25) is obtained.

Remark 6. By setting $h(x) = x$ and $m = 1$ in (2.25), [34], Theorem 3] can be reproduced.

2.2.   Inequalities for integral operators via $(\alpha, m)$-convex function

In this subsection we derive the bounds of integral operators defined in (1.12) and (1.13) by using $(\alpha, m)$-convexity of functions $f$ and $|f'|$.

Theorem 4. Let $f:[a, b]\longrightarrow \mathbb{R}, 0\leq a < b$, be a positive $(\alpha, m)$-convex function with $m\in(0, 1]$ and $g:[a, b]\longrightarrow \mathbb{R}$ be a differentiable and strictly increasing function. Also let $\dfrac{\phi}{x}$ be an increasing function. Then for $x\in [a, b]$ the following inequality for integral operators (1.12) and (1.13) holds:

Proof. By using $(\alpha, m)$-convexity of $f$ we get

Inequalities (2.4) and (2.34) produce the following integral inequality:

This takes the form as follows:

Again by using $(\alpha, m)$-convexity of $f$, we have

From inequalities (2.8) and (2.37), the following integral inequality is obtained:

This takes the form as follows:

By adding (2.36) and (2.39), (2.33) can be obtained.

Remark 7. 1. By setting $\phi(t) = \frac{t^\frac{\mu}{k}}{k\Gamma_{k}(\mu)}$ in (2.33), [10,Corollary 1] can be reproduced.

2. By setting $\phi(t) = \frac{t^{\alpha}}{\Gamma(\alpha)}, \alpha > 0$ and $(\alpha, m)$ = (1, 1) in (2.33), [33,Corollary 1] can be reproduced.

3. By setting $\phi(t) = \frac{t^{\alpha}}{\Gamma(\alpha)}, \alpha > 0$, $g(x) = x$ and $(\alpha, m)$ = (1, 1) in (2.33), [28,Corollary 1] can be reproduced.

Theorem 5. Let $f:[a, b]\longrightarrow \mathbb{R}, 0\leq a < b$, be a differentiable function. Let $|f'|$ be $(\alpha, m)$-convex, $m\in(0, 1]$, $g:[a, b]\longrightarrow \mathbb{R}$ be differentiable and strictly increasing function, also let $\frac{\phi}{x}$ be an increasing function. Then for $x\in [a, b]$ the following inequalities hold:

Proof. Since $|f'|$ is $(\alpha, m)$-convex, we have

From which we can write

Inequalities (2.4) and (2.43) produce the following integral inequality:

In compact form the following integral operator inequality holds:

From (2.42) we can write

Adopting the same procedure as we did for (2.43), the following inequality holds:

From (2.45) and (2.47), (2.40) can be obtained.

By using $(\alpha, m)$-convexity of $|f'|$, we have

From which we can write

Inequalities (2.49) and (2.8) produce the following integral inequality:

In compact form the following integral operator inequality holds:

From (2.48) we can write

Adopting the same procedure as we did for (2.49), the following inequality holds:

From (2.51) and (2.53), (2.41) can be obtained.

Lemma 2. ^[10] Let $f: [0, \infty] \longrightarrow \mathbb{R}$, be an $(\alpha, m)$-convex function. If $f(x) = f(\frac{a+b-x}{m})$, then the following inequality holds:

Theorem 6. Let $f: [a, b] \longrightarrow \mathbb{R}, 0\leq a < b$ be a positive $(\alpha, m)$-convex, $m\in(0, 1]$, $f(x) = f(\frac{a+b-x}{m})$ and $g:[a, b] \longrightarrow \mathbb{R}$ be a differentiable and strictly increasing function. Also let $\dfrac{\phi}{x}$ be an increasing function. Then for $(\alpha, m)\in[0, 1]^{2}$, the following Hadamard type inequality holds:

Proof. Under given conditions, the following inequality holds for $x\in(a, b]$:

By using $(\alpha, m)$-convexity of $f$, we have

Inequalities (2.56) and (2.57) produce the following integral inequality:

In compact form the following integral operator inequality holds:

On the other hand, under given conditions, the following inequality holds for $x\in[a, b)$:

Inequalities (2.57) and (2.60) produce the following integral inequality:

In compact form the following integral operator inequality holds:

From (2.59) and (2.61), the second inequality of (2.55) can be obtained.

Now using Lemma 2 and multiplying (2.54) with $K_{g}(x, a;\phi)g'(x)$, then integrating over $[a, b]$ we have

From which we get

Again by using Lemma 2 and multiplying (2.54) with $K_{g}(b, x;\phi)g'(x)$, then integrating over $[a, b]$ we get

From which we have

From (2.62) and (2.63), the first inequality of (2.55) can be obtained.

3.   Applications

In this section we will apply Theorem 4 for some particular functions and get upper bounds of several fractional and conformable integral operators. By applying Theorems 4, 5 & 6 we give bounds for $m$-convex functions. We also apply Theorem 1 to give the boundedness and continuity of all kinds of these operators. Further by applying other theorems reader can obtain specific results for these operators.

3.1.   Some special cases

Corollary 1. If we take $\alpha = 1$ in Theorem 4, then following inequality for $m$-convex functions holds:

Corollary 2. If we take $\alpha = 1$ in Theorem 5, then following inequality for $m$-convex functions holds:

Corollary 3. If we take $\alpha = 1$ in Theorem 6, then following inequality for $m$-convex functions holds:

3.2.   Consequences of Theorem 4

Proposition 1. Let $\phi(t) = \frac{t^{\alpha}}{\Gamma(\alpha)}$. Then (1.12) and (1.13) produce the fractional integral operators (1.8) and (1.9) as follows:

Further they satisfy the following bound for $\alpha\geq1$:

Proposition 2. Let $g(x) = I(x) = x$. Then (1.12) and (1.13) produce integral operators defined in ^[36] as follows:

Further they satisfy the following bound:

Corollary 4. If we take $\phi(t) = \frac{t^\frac{\alpha}{k}}{k\Gamma_k(\alpha)}$. Then (1.12) and (1.13) produce the fractional integral operators (1.10) and (1.11) as follows:

Further, the following bound holds for $\alpha\geq k$:

Corollary 5. If we take $\phi(t) = \frac{t^\alpha}{\Gamma(\alpha)}$, $\alpha > 0$ and $g(x) = I(x) = x$. Then (1.12) and (1.13) produce left and right Riemann-Liouville fractional integral operators (1.4) and (1.5) as follows:

Further, the following bound holds for $\alpha\geq 1$:

Corollary 6. If we take $\phi(t) = \frac{t^\frac{\alpha}{k}}{k\Gamma_k(\alpha)}$ and $g(x) = I(x) = x$. Then (1.12) and (1.13) produce the fractional integral operators (1.6) and (1.7) as follows:

Further, the following bound holds for $\alpha\geq k$:

Corollary 7. If we take $\phi(t) = \frac{t^\alpha}{\Gamma(\alpha)}, \, \, \alpha > 0$ and $g(x) = \frac{x^\rho}{\rho}, \ \rho > 0$. Then (1.12) and (1.13) produce the fractional integral operators defined in ^[20], as follows:

Further, they satisfy the following bound:

Corollary 8. If we take $\phi(t) = \frac{t^\alpha}{\Gamma(\alpha)}, \, \, \alpha > 0$ and $g(x) = \frac{x^{s+1}}{s+1}, \ s > 0$. Then (1.12) and (1.13) produce the fractional integral operators define as follows:

Further, they satisfy the following bound:

3.3.   Boundedness and continuity

We apply Theorem 1 and prove the boundedness and continuity of integral operators.

Theorem 7. Under the assumptions of Theorem 1, the following inequality holds for $m$-convex functions:

Proof. If we put $h(x) = x$ in (2.6), we have

This further takes form as follows:

Similarly, from (2.10) we have

From (3.9) and (3.10), (3.7) can be obtained.

Corollary 9. If $m = 1$, then the following inequality holds for convex functions:

Theorem 8. With assumptions of Theorem 7, if $f\in L_{\infty}[a, b]$, then integral operators (1.12) and (1.13) for $m$-convex functions are bounded and continuous.

Proof. From (3.9), we have

which further gives $\left|(F^{\phi, g}_{a^{+}}f)(x)\right|\leq K\|f\|_{\infty},$ where $K = (m+1)\phi(g(b)-g(a))$. Similarly, from (3.10) we get $\left|(F^{\phi, g}_{b^{-}}f)(x)\right|\leq K\|f\|_{\infty}$. Hence the integral operators (1.12) and (1.13) are bounded, also they are linear. Therefore continuity is followed.

Corollary 10. If $m = 1$ and $f\in L_{\infty}[a, b]$, then integral operators (1.12) and (1.13) for convex functions are bounded and continuous:

where $K = 2\phi(g(b)-g(a))$. Similarly we have $\left|(F^{\phi, g}_{b^{-}}f)(x)\right|\leq K \|f\|_{\infty}$.

4.   Conclusions

An integral operator has been studied for $(h-m)$-convex and $(\alpha, m)$-convex functions. The bounds of this operator are obtained in different forms. The Hadamard inequality is presented for $(h-m)$-convex as well as $(\alpha, m)$-convex functions. Further the results are deducible for various kinds of known functions given in Remarks 1 & 2. Some of the results are applied to obtain bounds of different kinds of conformable and fractional integral operators. The boundedness and continuity of integral operators (1.12) and (1.13) are given for $m$-convex and convex functions. Reader can deduce results for integral operators comprised in Remarks 3 for all functions given in Remarks 1 & 2.

Acknowledgments

We thank to the editor and referees for their careful reading and valuable suggestions to make the article friendly readable. The research work of Ghulam Farid is supported by the Higher Education Commission of Pakistan under NRPU 2016, Project No. 5421.

Conflict of interests

It is declared that authors have no competing interests.