First we recall some special means:
$$\begin{aligned}& A(\delta _{1}, \delta _{2}) = \frac{\delta _{1} + \delta _{2}}{2} \quad ( \text{arithmetic mean}), \\& G(\delta _{1}, \delta _{2}) = \pm \sqrt{\delta _{1}\delta _{2}}\quad (\text{geometric mean}), \\& L(\delta _{1}, \delta _{2})= \textstyle\begin{cases} \delta _{2} & \mbox{if } \delta _{1} = \delta _{2}, \\ \frac{\delta _{2} - \delta _{1}}{\ln \delta _{2} - \ln \delta _{1}} & \mbox{if } \delta _{1} \neq \delta _{2}, \delta _{1}, \delta _{2} > 0 \end{cases}\displaystyle \quad (\text{logarithmic mean}), \end{aligned}$$
and
$$ I(\delta _{1}, \delta _{2})= \textstyle\begin{cases} \delta _{2} & \mbox{if } \delta _{1} = \delta _{2}; \\ \frac{1}{e} ( \frac{\delta ^{{\delta _{2}}}_{2}}{\delta ^{{\delta _{1}}}_{1}} )^{\frac{1}{\delta _{2}-\delta _{1}}} & \mbox{if } \delta _{1} \neq \delta _{2} \end{cases}\displaystyle \quad (\text{identric mean}). $$
4.1 Kullback–Leibler divergence on time scales
Let \(\psi : (0, \infty ) \rightarrow \mathbb{R}\) be the convex mapping \(\psi (t) = t\ln t\). Then
$$ I_{\psi }(\tilde{r}, \tilde{s}) = \int _{a}^{b} \tilde{r}(y) \ln \biggl[ \frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr]\Delta y = D(\tilde{r}, \tilde{s}), $$
where \(D(\tilde{r}, \tilde{s})\) is the Kullback–Leibler distance.
Proposition 1
If
$$ \zeta _{1} \leq \frac{\tilde{r}(y)}{\tilde{s}(y)} \leq \zeta _{2} \quad \textit{for all } y \in \mathbb{T}, $$
(23)
then
$$ D(\tilde{r}, \tilde{s}) = \int _{a}^{b} \tilde{r}(y) \ln \biggl[ \frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr]\Delta y \leq \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1, $$
where \(G(\cdot , \cdot )\) is the usual geometric mean, \(L(\cdot , \cdot )\) is the logarithmic mean, and \(I(\cdot , \cdot )\) is the identric mean.
Proof
Using Theorem 2 for \(\psi (t) = t\ln t \), we obtain
$$\begin{aligned} D(\tilde{r}, \tilde{s}) =& \int _{a}^{b} \tilde{r}(y) \ln \biggl[ \frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr]\Delta y \leq \frac{\zeta _{2}-1}{\zeta _{2}-\zeta _{1}} \zeta _{1} \ln \zeta _{1} + \frac{1-\zeta _{1}}{\zeta _{2}-\zeta _{1}} \zeta _{2} \ln \zeta _{2} \\ =& \frac{\zeta _{1}\zeta _{2} \ln \zeta _{1} - \zeta _{1} \ln \zeta _{1} + \zeta _{2} \ln \zeta _{2} - \zeta _{1}\zeta _{2} \ln \zeta _{2}}{\zeta _{2}-\zeta _{1}} \\ =& \frac{\zeta _{2} \ln \zeta _{2} - \zeta _{1} \ln \zeta _{1}}{\zeta _{2}-\zeta _{1}} - \zeta _{1}\zeta _{2} \frac{[\ln \zeta _{2} - \ln \zeta _{1}]}{\zeta _{2}-\zeta _{1}} \\ =& \ln \biggl(\frac{\zeta _{2}^{\zeta _{2}}}{\zeta _{1}^{\zeta _{1}}} \biggr)^{\frac{1}{\zeta _{2}-\zeta _{1}}} - \ln e + \ln e - ( \sqrt{ \zeta _{1}\zeta _{2}})^{2} \frac{[\ln \zeta _{2} - \ln \zeta _{1}]}{\zeta _{2}-\zeta _{1}} \\ =& \ln I(\zeta _{1}, \zeta _{2}) +1 - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})}. \end{aligned}$$
□
Example 11
Choosing \(\mathbb{T} = \mathbb{R}\) in Proposition 1, we get [7, Proposition 1 on p. 6].
Example 12
Choosing \(\mathbb{T} = h\mathbb{Z}\), \(h > 0\), in Proposition 1, we obtain
$$ \sum_{k=\frac{a}{h}}^{\frac{b}{h} - 1} \tilde{r}(kh)h\ln \biggl[ \frac{\tilde{r}(kh)}{\tilde{s}(kh)} \biggr] \leq \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1. $$
(24)
Remark 5
Equation (24) is an extension of the specific bound for the Kullback–Leibler divergence obtained by Lovričević et al. [15, Corollary 4.4].
Example 13
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 1, we have
$$ \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1)\tilde{r} \bigl(q^{k} \bigr) \ln \biggl[ \frac{\tilde{r}(q^{k})}{\tilde{s}(q^{k})} \biggr] \leq \ln I( \zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1. $$
(25)
Remark 6
Equation (25) shows an upper bound for the Kullback–Leibler divergence, which is new in quantum calculus.
Proposition 2
Under the conditions of Proposition 1, we get
$$\begin{aligned} 0 \leq & \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1 - \int _{a}^{b} \tilde{r}(y) \ln \biggl[ \frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr]\Delta y \\ \leq & \frac{(\zeta _{2}-1)(1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s})}{L(\zeta _{1}, \zeta _{2})}, \end{aligned}$$
(26)
where \(D_{\chi ^{2}}(\tilde{r}, \tilde{s})\) is defined in Theorem 3.
Proof
Apply Theorem 3 for \(\psi (t) = t\ln t\):
$$ \frac{\psi ^{\prime }(\zeta _{2})-\psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{\ln \zeta _{2}-\ln \zeta _{1}}{\zeta _{2}-\zeta _{1}} = \frac{1}{L(\zeta _{1}, \zeta _{2})}. $$
□
Example 14
Putting \(\mathbb{T} = \mathbb{R}\) in Proposition 2, we get [7, Proposition 2 on p. 6].
Example 15
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)), in Proposition 2, we have
$$\begin{aligned} 0 \leq & \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1 - \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1)\tilde{r} \bigl(q^{k} \bigr) \ln \biggl[ \frac{\tilde{r}(q^{k})}{\tilde{s}(q^{k})} \biggr] \\ \leq & \frac{(\zeta _{2}-1)(1-\zeta _{1})- \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1)\tilde{r}(q^{k}) [ (\frac{\tilde{s}(q^{k})}{\tilde{r}(q^{k})} )^{2} - 1 ]}{L(\zeta _{1}, \zeta _{2})}. \end{aligned}$$
By using Theorem 4 we can improve (26) as follows.
Proposition 3
Let r̃, s̃ satisfy (23). Then we have
$$\begin{aligned}& \frac{1}{2\zeta _{2}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}( \tilde{r}, \tilde{s}) \bigr] \\& \quad \leq \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1 - D( \tilde{r}, \tilde{s}) \\& \quad \leq \frac{1}{2\zeta _{1}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}( \tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Consider \(\psi (t) = t\ln t\) in Theorem 4. In this case, \(\psi ^{\prime \prime }(t) = \frac{1}{t}\), \(t \in [\zeta _{1}, \zeta _{2}]\), and then
$$ \frac{1}{\zeta _{2}}\leq \psi ^{\prime \prime }(t) \leq \frac{1}{\zeta _{1}},\quad t \in [ \zeta _{1}, \zeta _{2}], $$
which gives the desired result. □
Remark 7
For \(\mathbb{T} = \mathbb{R}\) in Proposition 3, we get [7, Proposition 3 on p. 7].
Example 16
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 3, we have
$$\begin{aligned}& \frac{1}{2\zeta _{2}} \Biggl((\zeta _{2}-1) (1-\zeta _{1})- \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1) \tilde{r} \bigl(q^{k} \bigr) \biggl[ \biggl( \frac{\tilde{s}(q^{k})}{\tilde{r}(q^{k})} \biggr)^{2} - 1 \biggr] \Biggr) \\& \quad \leq \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2})}{L(\zeta _{1}, \zeta _{2})} + 1 - \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1) \tilde{r} \bigl(q^{k} \bigr) \ln \biggl[ \frac{\tilde{r}(q^{k})}{\tilde{s}(q^{k})} \biggr] \\& \quad \leq \frac{1}{2\zeta _{1}} \Biggl((\zeta _{2}-1) (1-\zeta _{1})- \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1) \tilde{r} \bigl(q^{k} \bigr) \biggl[ \biggl( \frac{\tilde{s}(q^{k})}{\tilde{r}(q^{k})} \biggr)^{2} - 1 \biggr] \Biggr). \end{aligned}$$
Now consider the convex mapping \(\psi (t) = -\ln t\). We get
$$\begin{aligned} I_{\psi }(\tilde{r}, \tilde{s}) =& - \int _{a}^{b}\tilde{s}(y) \ln \biggl[ \frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr]\Delta y \\ =& \int _{a}^{b}\tilde{s}(y) \ln \biggl[ \frac{\tilde{s}(y)}{\tilde{r}(y)} \biggr]\Delta y = D(\tilde{s}, \tilde{r}). \end{aligned}$$
By using Theorem 2 we obtain the following result.
Proposition 4
Let r̃, s̃ satisfy (23). Then
$$ D(\tilde{s}, \tilde{r}) = \int _{a}^{b}\tilde{s}(y) \ln \biggl[ \frac{\tilde{s}(y)}{\tilde{r}(y)} \biggr]\Delta y \leq \ln I \biggl( \frac{1}{\zeta _{1}}, \frac{1}{\zeta _{2}} \biggr) - \frac{1}{L(\zeta _{1}, \zeta _{2})} + 1. $$
Proof
Using (1) for \(\psi (t) = -\ln t\), we get
$$\begin{aligned} D(\tilde{s}, \tilde{r}) =& \int _{a}^{b}\tilde{s}(y) \ln \biggl[ \frac{\tilde{s}(y)}{\tilde{r}(y)} \biggr]\Delta y \leq \frac{(\zeta _{2}-1)(-\ln \zeta _{1})+(1-\zeta _{1})(-\ln \zeta _{2})}{\zeta _{2}-\zeta _{1}} \\ =& \frac{(\zeta _{1} \ln \zeta _{2} - \zeta _{2} \ln \zeta _{1})}{\zeta _{2}-\zeta _{1}} - \frac{(\ln \zeta _{2} - \ln \zeta _{1})}{\zeta _{2}-\zeta _{1}} \\ =& \frac{\zeta _{1}\zeta _{2}(\frac{1}{\zeta _{2}}\ln \zeta _{2} - \frac{1}{\zeta _{1}}\ln \zeta _{1})}{\zeta _{2}-\zeta _{1}} - \frac{1}{L(\zeta _{1}, \zeta _{2})} \\ =& \frac{(\frac{1}{\zeta _{1}}\ln \frac{1}{\zeta _{1}} - \frac{1}{\zeta _{2}}\ln \frac{1}{\zeta _{2}})}{\frac{1}{\zeta _{1}}-\frac{1}{\zeta _{2}}} - \frac{1}{L(\zeta _{1}, \zeta _{2})} \\ =& \ln I \biggl(\frac{1}{\zeta _{1}},\frac{1}{\zeta _{2}} \biggr) - \frac{1}{L(\zeta _{1}, \zeta _{2})} + 1. \end{aligned}$$
□
Remark 8
For \(\mathbb{T} = \mathbb{R}\) in Proposition 4, we get [7, Proposition 4 on p. 7].
Proposition 5
Let r̃, s̃ satisfy (23). Then we have
$$\begin{aligned} 0 \leq & \ln I \biggl(\frac{1}{\zeta _{1}}, \frac{1}{\zeta _{2}} \biggr) - \frac{1}{L(\zeta _{1}, \zeta _{2})} + 1 - \int _{a}^{b}\tilde{s}(y) \ln \biggl[ \frac{\tilde{s}(y)}{\tilde{r}(y)} \biggr]\Delta y \\ =& \frac{1}{G^{2}(\zeta _{1}, \zeta _{2})} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr], \end{aligned}$$
(27)
where \(D_{\chi ^{2}}(\tilde{r}, \tilde{s})\) is defined in Theorem 3.
Proof
Apply Theorem 3 for \(\psi (t) = - \ln t\):
$$ \frac{\psi ^{\prime }(\zeta _{2})-\psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{1}{\zeta _{1}\zeta _{2}} = \frac{1}{G^{2}(\zeta _{1}, \zeta _{2})}. $$
□
Remark 9
For \(\mathbb{T} = \mathbb{R}\), Proposition 5 becomes [7, Proposition 5 on p. 7].
Further improvement of (27) is as follows.
Proposition 6
Under the assumptions of Theorem 4, we have
$$\begin{aligned}& \frac{1}{2\zeta _{2}^{2}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}( \tilde{r}, \tilde{s}) \bigr] \\& \quad \leq \ln I \biggl(\frac{1}{\zeta _{1}},\frac{1}{\zeta _{2}} \biggr) - \frac{1}{L(\zeta _{1}, \zeta _{2})} + 1 - D(\tilde{s}, \tilde{r}). \\& \quad \leq \frac{1}{2\zeta _{1}^{2}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{ \chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Apply Theorem 4, for which \(\psi ^{\prime \prime }(t) = \frac{1}{t^{2}}\) and
$$ \frac{1}{2\zeta _{2}^{2}} \leq \psi ^{\prime \prime }(t) \leq \frac{1}{2\zeta _{1}^{2}} $$
for all \(t \in [\zeta _{1}, \zeta _{2}]\). □
Remark 10
Choosing \(\mathbb{T} = \mathbb{R}\) in Proposition 6, we get [7, Proposition 6 on p. 8].
4.2 Triangular discrimination on time scales
Let \(\psi : [0, \infty ) \rightarrow \mathbb{R}\) be the convex mapping \(\psi (t) = \frac{(t - 1)^{2}}{t + 1}\). Then
$$ I_{\psi }(\tilde{r}, \tilde{s})= \int _{a}^{b} \frac{[\tilde{r}(y) - \tilde{s}(y)]^{2}}{\tilde{r}(y) + \tilde{s}(y)}\Delta y = D_{\Delta }( \tilde{r}, \tilde{s}), $$
where \(D_{\Delta }(\tilde{r}, \tilde{s})\) is the triangular discrimination.
Proposition 7
Under the assumptions of Theorem 2, we have
$$ D_{\Delta }(\tilde{r}, \tilde{s}) \leq \frac{4A(\zeta _{1}, \zeta _{2}) - 2G^{2}(\zeta _{1}, \zeta _{2}) - 2}{2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2}) + 1}. $$
Proof
Using Theorem 2 for \(\psi (t) = \frac{(t - 1)^{2}}{t + 1}\), we obtain
$$\begin{aligned}& D_{\Delta }(\tilde{r}, \tilde{s}) \\& \quad \leq \frac{(\zeta _{2}-1)(\zeta _{1} - 1)^{2}(\zeta _{2} + 1) + (1-\zeta _{1})(\zeta _{2} - 1)^{2}(\zeta _{1} + 1)}{(\zeta _{2}-\zeta _{1})(\zeta _{1} + 1)(\zeta _{2} + 1)} \\& \quad = \frac{\zeta ^{2}_{2}\zeta ^{2}_{1} + \zeta ^{2}_{2} - 2\zeta _{1}\zeta ^{2}_{2} - \zeta ^{2}_{1} - 1 + 2\zeta _{1} + \zeta ^{2}_{2} + 1 - 2\zeta _{2} - \zeta ^{2}_{2}\zeta ^{2}_{1} - \zeta ^{2}_{1} + 2\zeta _{2}\zeta ^{2}_{1}}{(\zeta _{2}-\zeta _{1})(\zeta _{1} + 1)(\zeta _{2} + 1)} \\& \quad = \frac{(2\zeta ^{2}_{2} - 2\zeta ^{2}_{1}) - (2\zeta _{1}\zeta ^{2}_{2} - 2\zeta _{2}\zeta ^{2}_{1}) - (2\zeta _{2} - 2\zeta _{1})}{(\zeta _{2}-\zeta _{1})(\zeta _{1} + 1)(\zeta _{2} + 1)} \\& \quad = \frac{2(\zeta _{2} + \zeta _{1})(\zeta _{2} - \zeta _{1}) - 2\zeta _{1}\zeta _{2}(\zeta _{2} - \zeta _{1}) - 2(\zeta _{2} - \zeta _{1})}{(\zeta _{2}-\zeta _{1})(\zeta _{1} + 1)(\zeta _{2} + 1)} \\& \quad = \frac{2(\zeta _{2} + \zeta _{1} - \zeta _{1}\zeta _{2} - 1)}{(\zeta _{1} + 1)(\zeta _{2} + 1)} = \frac{2(\zeta _{2} + \zeta _{1} - \zeta _{1}\zeta _{2} - 1)}{\zeta _{1} + \zeta _{2} + \zeta _{1}\zeta _{2} + 1} \\& \quad = \frac{4A(\zeta _{1}, \zeta _{2}) - 2G^{2}(\zeta _{1}, \zeta _{2}) - 2}{2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2}) + 1}. \end{aligned}$$
□
The following example gives an upper bound for triangular discrimination, which is new in quantum calculus.
Example 17
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 7, we have
$$ \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1) \frac{[\tilde{r}(q^{k}) - \tilde{s}(q^{k})]^{2}}{\tilde{r}(q^{k}) + \tilde{s}(q^{k})} \leq \frac{4A(\zeta _{1}, \zeta _{2}) - 2G^{2}(\zeta _{1}, \zeta _{2}) - 2}{2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2}) + 1}. $$
Proposition 8
Under the conditions of Theorem 3, we have
$$\begin{aligned}& 0 \leq \frac{4A(\zeta _{1}, \zeta _{2}) - 2G^{2}(\zeta _{1}, \zeta _{2}) - 2}{2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2}) + 1}-D_{ \Delta }(\tilde{r}, \tilde{s}), \\& 0 \leq \frac{8A(\zeta _{1}, \zeta _{2}) + 8}{[G^{2}(\zeta _{1}, \zeta _{2}) + 2A(\zeta _{1}, \zeta _{2}) + 1]^{2}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Apply Theorem 3 with \(\psi (t) = \frac{(t - 1)^{2}}{t+1}\), which gives \(\psi ^{\prime }(t) = 1- \frac{4}{(1 + t)^{2}}\) and
$$ \frac{\psi ^{\prime }(\zeta _{2})- \psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{4(\zeta _{2} + \zeta _{1} + 2)}{[(\zeta _{2}+ 1)(\zeta _{1}+ 1)]^{2}} = \frac{8A(\zeta _{1}, \zeta _{2}) + 8}{[G^{2}(\zeta _{1}, \zeta _{2}) + 2A(\zeta _{1}, \zeta _{2}) + 1]^{2}}. $$
□
Proposition 9
Under the assumptions of Theorem 4, we have
$$\begin{aligned} 0 \leq & \frac{8}{(1 + \zeta _{2})^{3}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr] \\ \leq & \frac{4A(\zeta _{1}, \zeta _{2}) - 2G^{2}(\zeta _{1}, \zeta _{2}) - 2}{2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2}) + 1}-D_{ \Delta }(\tilde{r}, \tilde{s}) \\ \leq & \frac{8}{(1 + \zeta _{1})^{3}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{ \chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Use Theorem 4 for \(\psi (t) = \frac{(t - 1)^{2}}{t+1}\), which implies that \(\psi ^{\prime \prime }(t) = \frac{8}{(1 + t)^{3}}\) and
$$ \frac{8}{(1 + \zeta _{2})^{3}} \leq \psi ^{\prime \prime }(t) \leq \frac{8}{(1 + \zeta _{1})^{3}} \quad \text{for all } t \in [ \zeta _{1}, \zeta _{2}]. $$
□
4.3 Hellinger discrimination on time scales
Let \(\psi : [0, \infty ) \rightarrow \mathbb{R}\) be the convex mapping \(\psi (t) = \frac{1}{2}(\sqrt{t} - 1)^{2}\). Then
$$\begin{aligned} I_{\psi }(\tilde{r}, \tilde{s}) =& \frac{1}{2} \int _{a}^{b} \tilde{s}(y) \biggl(\sqrt{ \frac{\tilde{r}(y)}{\tilde{s}(y)}} - 1 \biggr)^{2}\Delta y \\ =& \frac{1}{2} \int _{a}^{b} \bigl[\sqrt{\tilde{r}(y)} - \sqrt{ \tilde{s}(y)} \bigr]^{2}\Delta y = h^{2}( \tilde{r}, \tilde{s}), \end{aligned}$$
where \(h^{2}(\tilde{r}, \tilde{s})\) is the Hellinger discrimination.
Proposition 10
Under the assumptions of Theorem 2, we get
$$ h^{2}(\tilde{r}, \tilde{s}) \leq \frac{2A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}}) - G(\zeta _{1}, \zeta _{2}) - 1}{2A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})}. $$
Proof
Use Theorem 2 for \(\psi (t) = \frac{1}{2}(\sqrt{t} - 1)^{2}\) to obtain
$$\begin{aligned}& h^{2}(\tilde{r}, \tilde{s}) \\& \quad \leq \frac{(\zeta _{2}-1)\frac{1}{2}(\sqrt{\zeta _{1}} - 1)^{2} + (1-\zeta _{1}) \frac{1}{2}(\sqrt{\zeta _{2}} - 1)^{2}}{\zeta _{2}-\zeta _{1}} \\& \quad = \frac{(\frac{1}{2}\sqrt{\zeta _{2}}-1)(1-\sqrt{\zeta _{1}})}{\zeta _{2}-\zeta _{1}} \bigl[( \sqrt{\zeta _{2}} + 1) (1-\sqrt{ \zeta _{1}}) + (1+\sqrt{\zeta _{1}}) ( \sqrt{\zeta _{2}} - 1) \bigr] \\& \quad = \frac{(\sqrt{\zeta _{2}}-1)(1-\sqrt{\zeta _{1}}) (\sqrt{\zeta _{2}} - \sqrt{\zeta _{1}})}{\zeta _{2}-\zeta _{1}} = \frac{(\sqrt{\zeta _{2}}-1)(1-\sqrt{\zeta _{1}}) }{(\sqrt{\zeta _{2}} + \sqrt{\zeta _{1}})} \\& \quad = \frac{\sqrt{\zeta _{2}} + \sqrt{\zeta _{1}} - \sqrt{\zeta _{1}\zeta _{2}} - 1}{\sqrt{\zeta _{1}} + \sqrt{\zeta _{2}}} = \frac{2A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}}) - G(\zeta _{1}, \zeta _{2}) - 1}{2A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})}. \end{aligned}$$
□
Example 18
For \(\mathbb{T} = \mathbb{R}\) in Proposition 10, we get [7, Proposition 7 on p. 8].
The following example gives an upper bound for the Hellinger discrimination, which is new in quantum calculus.
Example 19
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 10, we have
$$ \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1) \bigl[ \sqrt{ \tilde{r} \bigl(q^{k} \bigr)} - \sqrt{\tilde{s} \bigl(q^{k} \bigr)} \bigr]^{2} \leq \frac{2A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}}) - G(\zeta _{1}, \zeta _{2}) - 1}{2A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})}. $$
Proposition 11
Under the assumptions of Theorem 3, we have
$$\begin{aligned}& 0 \leq \frac{(\sqrt{\zeta _{2}}- 1)(1 - \sqrt{\zeta _{1}})}{\sqrt{\zeta _{2}} + \sqrt{\zeta _{1}}} - h^{2}(\tilde{r}, \tilde{s}), \\& 0 \leq \frac{1}{4\sqrt{\zeta _{1}\zeta _{2}} A(\sqrt{\zeta _{1}},\sqrt{\zeta _{2}})} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr], \end{aligned}$$
where \(A(\cdot ,\cdot )\) is the arithmetic mean.
Proof
Apply Theorem 3 with \(\psi (t) = \frac{1}{2}(\sqrt{t} - 1)^{2}\), which implies that \(\psi ^{\prime }(t) = \frac{1}{2} - \frac{1}{2\sqrt{t}}\) and
$$ \frac{\psi ^{\prime }(\zeta _{2})- \psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{1}{2\sqrt{\zeta _{1}\zeta _{2}}(\sqrt{\zeta _{2}}+ \sqrt{\zeta _{1}})}. $$
□
Remark 11
For \(\mathbb{T} = \mathbb{R}\), Proposition 11 becomes [7, Proposition 8 on p. 9].
Proposition 12
Under the assumptions of Theorem 4, we have
$$\begin{aligned} 0 \leq & \frac{1}{8\sqrt{\zeta _{2}^{3}}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr] \\ \leq &\frac{(\sqrt{\zeta _{2}}- 1)(1 - \sqrt{\zeta _{1}})}{\sqrt{\zeta _{2}} + \sqrt{\zeta _{1}}} - h^{2}(\tilde{r}, \tilde{s}) \\ \leq &\frac{1}{8\sqrt{\zeta _{1}^{3}}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Use Theorem 4 for \(\psi (t) = \frac{1}{2}(\sqrt{t} - 1)^{2}\), which gives \(\psi ^{\prime \prime }(t) = \frac{1}{4t^{\frac{3}{2}}}\) and, obviously,
$$ \frac{1}{4\zeta _{2}^{\frac{3}{2}}} \leq \psi ^{\prime \prime }(t) \leq \frac{1}{4\zeta _{1}^{\frac{3}{2}}}\quad \text{for all } t \in [\zeta _{1}, \zeta _{2}]. $$
□
Remark 12
Choosing \(\mathbb{T} = \mathbb{R}\) in Proposition 12, we get [7, Proposition 9 on p. 9].
4.4 Jeffreys distance on time scales
Let \(\psi : (0, \infty ) \rightarrow \mathbb{R}\) be the convex mapping \(\psi (t) = (t - 1)\ln (t)\). Then
$$ I_{\psi }(\tilde{r}, \tilde{s}) = \int _{a}^{b} \bigl(\tilde{r}(y) - \tilde{s}(y) \bigr) \ln \biggl[\frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr] \Delta y = D_{J}( \tilde{r}, \tilde{s}), $$
where \(D_{J}(\tilde{r}, \tilde{s})\) is the Jeffreys distance.
Proposition 13
Let r̃, s̃ satisfy (23). Then we get
$$ D_{J}(\tilde{r}, \tilde{s}) = \int _{a}^{b} \bigl(\tilde{r}(y) - \tilde{s}(y) \bigr) \ln \biggl[\frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr]\Delta y \leq \ln I( \zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2}) + 1}{L(\zeta _{1}, \zeta _{2})} + \ln I \biggl( \frac{1}{\zeta _{1}},\frac{1}{\zeta _{2}} \biggr) + 2. $$
Proof
Use Theorem 2 for \(\psi (t) = (t - 1)\ln t \) to get
$$\begin{aligned} D_{J}(\tilde{r}, \tilde{s}) =& \int _{a}^{b} \bigl(\tilde{r}(y) - \tilde{s}(y) \bigr) \ln \biggl[\frac{\tilde{r}(y)}{\tilde{s}(y)} \biggr] \Delta y \\ \leq & \frac{\zeta _{2}-1}{\zeta _{2}-\zeta _{1}} (\zeta _{1}-1) \ln \zeta _{1} + \frac{1-\zeta _{1}}{\zeta _{2}-\zeta _{1}} (\zeta _{2}-1) \ln \zeta _{2} \\ =& \frac{(\zeta _{2}-1)(\zeta _{1}-1)\ln \zeta _{1} - (1 - \zeta _{1}) (\zeta _{2} - 1) \ln \zeta _{2}}{\zeta _{2}-\zeta _{1}} \\ =& \frac{\zeta _{2} \ln \zeta _{2} - \zeta _{1} \ln \zeta _{1}}{\zeta _{2}-\zeta _{1}} - \zeta _{1}\zeta _{2} \frac{[\ln \zeta _{2} - \ln \zeta _{1}]}{\zeta _{2}-\zeta _{1}} + \frac{\zeta _{1} \ln \zeta _{2} - \zeta _{2} \ln \zeta _{1}}{\zeta _{2}-\zeta _{1}} - \frac{[\ln \zeta _{2} - \ln \zeta _{1}]}{\zeta _{2}-\zeta _{1}} \\ =& \ln \biggl(\frac{\zeta _{2}^{\zeta _{2}}}{\zeta _{1}^{\zeta _{1}}} \biggr)^{\frac{1}{\zeta _{2}-\zeta _{1}}} - \ln e + \ln e - ( \zeta _{1} \zeta _{2} + 1) \frac{[\ln \zeta _{2} - \ln \zeta _{1}]}{\zeta _{2}-\zeta _{1}} + \frac{\zeta _{1}\zeta _{2}(\frac{1}{\zeta _{2}}\ln \zeta _{2} - \frac{1}{\zeta _{1}}\ln \zeta _{1})}{\zeta _{2}-\zeta _{1}} \\ =& \ln I(\zeta _{1}, \zeta _{2}) +1 - \frac{[(\sqrt{\zeta _{1}\zeta _{2}})^{2} + 1]}{L(\zeta _{1}, \zeta _{2})} + \frac{(\frac{1}{\zeta _{1}}\ln \frac{1}{\zeta _{1}} - \frac{1}{\zeta _{2}}\ln \frac{1}{\zeta _{2}})}{\frac{1}{\zeta _{1}}-\frac{1}{\zeta _{2}}} \\ =&\ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2}) + 1}{L(\zeta _{1}, \zeta _{2})} + \ln I \biggl(\frac{1}{\zeta _{1}},\frac{1}{\zeta _{2}} \biggr) + 2. \end{aligned}$$
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The following example gives an upper bound for the Jeffreys distance, which is new in quantum calculus.
Example 20
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 13, we have
$$\begin{aligned}& \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} q^{k}(q-1) \bigl( \tilde{r} \bigl(q^{k} \bigr) - \tilde{s} \bigl(q^{k} \bigr) \bigr) \ln \biggl[\frac{\tilde{r}(q^{k})}{\tilde{s}(q^{k})} \biggr] \\& \quad \leq \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2}) + 1}{L(\zeta _{1}, \zeta _{2})} + \ln I \biggl(\frac{1}{\zeta _{1}}, \frac{1}{\zeta _{2}} \biggr) + 2. \end{aligned}$$
Proposition 14
Under the assumptions of Theorem 3, we get
$$\begin{aligned} 0 \leq & \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2}) + 1}{L(\zeta _{1}, \zeta _{2})} + \ln I \biggl(\frac{1}{\zeta _{1}},\frac{1}{\zeta _{2}} \biggr) + 2 - D_{J}( \tilde{r}, \tilde{s}) \\ \leq & \biggl[\frac{1}{G^{2}(\zeta _{1}, \zeta _{2})} + \frac{1}{L(\zeta _{1}, \zeta _{2})} \biggr] \bigl[( \zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}( \tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Apply Theorem 3 for \(\psi (t) = (t - 1)\ln t\), for which
$$ \frac{\psi ^{\prime }(\zeta _{2})-\psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{1}{\zeta _{1}\zeta _{2}} + \frac{\ln \zeta _{2}-\ln \zeta _{1}}{\zeta _{2}-\zeta _{1}} = \frac{1}{G^{2}(\zeta _{1}, \zeta _{2})} + \frac{1}{L(\zeta _{1}, \zeta _{2})}. $$
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Proposition 15
Under the assumptions of Theorem 4, we have
$$\begin{aligned}& \frac{\zeta _{2} + 1}{\zeta ^{2}_{2}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr] \\& \quad \leq \ln I(\zeta _{1}, \zeta _{2}) - \frac{G^{2}(\zeta _{1}, \zeta _{2}) + 1}{L(\zeta _{1}, \zeta _{2})} + \ln I \biggl(\frac{1}{\zeta _{1}},\frac{1}{\zeta _{2}} \biggr) + 2 - D_{J}( \tilde{r}, \tilde{s}) \\& \quad \leq \frac{\zeta _{1} + 1}{\zeta ^{2}_{1}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Consider \(\psi (t) = (t - 1)\ln t\) in Theorem 4. In this case, \(\psi ^{\prime \prime }(t) = \frac{t + 1}{t^{2}}\), \(t \in [\zeta _{1}, \zeta _{2}]\), and then
$$ \frac{\zeta _{2} + 1}{\zeta ^{2}_{2}}\leq \psi ^{\prime \prime }(t) \leq \frac{\zeta _{1} + 1}{\zeta ^{2}_{1}},\quad t \in [\zeta _{1}, \zeta _{2}], $$
which gives the desired result. □
4.5 Bhattacharyya distance on time scales
Let \(\psi : [0, \infty ) \rightarrow \mathbb{R}\) be a convex mapping, \(\psi (t) = -\sqrt{t}\). Then
$$\begin{aligned} I_{\psi }(\tilde{r}, \tilde{s}) =& - \int _{a}^{b} \tilde{s}(y) \sqrt{ \frac{\tilde{r}(y)}{\tilde{s}(y)}}\Delta y \\ =& - \int _{a}^{b} \sqrt{\tilde{s}(y)\tilde{r}(y)} \Delta y = D_{B}( \tilde{r}, \tilde{s}), \end{aligned}$$
where \(D_{B}(\tilde{r}, \tilde{s})\) is the Bhattacharyya distance.
Proposition 16
Under the assumptions of Theorem 2, we obtain
$$ D_{B}(\tilde{r}, \tilde{s}) \leq \frac{-1 - G(\zeta _{1}, \zeta _{2})}{2 A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})}. $$
Proof
Use Theorem 2 for \(\psi (t) = -\sqrt{t}\) to obtain
$$\begin{aligned} D_{B}(\tilde{r}, \tilde{s}) \leq & \frac{(\zeta _{2}-1)(-\sqrt{\zeta _{1}}) + (1-\zeta _{1}) (-\sqrt{\zeta _{2}})}{\zeta _{2}-\zeta _{1}} \\ =& \frac{-\sqrt{\zeta _{2}}+\sqrt{\zeta _{1}} - \zeta _{2}\sqrt{\zeta _{1}} + \zeta _{1} \sqrt{\zeta _{2}}}{\zeta _{2}-\zeta _{1}} \\ =& \frac{-1(\sqrt{\zeta _{2}}-\sqrt{\zeta _{1}}) - \sqrt{\zeta _{1}}\sqrt{\zeta _{2}}(\sqrt{\zeta _{2}} - \sqrt{\zeta _{1}})}{(\sqrt{\zeta _{2}} - \sqrt{\zeta _{1}})(\sqrt{\zeta _{2}} + \sqrt{\zeta _{1}})} \\ =& \frac{(-1 - \sqrt{\zeta _{1}}\sqrt{\zeta _{2}})}{(\sqrt{\zeta _{2}} + \sqrt{\zeta _{1}})} = \frac{-1 - G(\zeta _{1}, \zeta _{2})}{2 A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})}. \end{aligned}$$
□
Example 21
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 16, we have
$$ \sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} -\sqrt{q^{k}(q-1) \tilde{r} \bigl(q^{k} \bigr) \tilde{s} \bigl(q^{k} \bigr)} \leq \frac{-1 - G(\zeta _{1}, \zeta _{2})}{2 A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})}. $$
Remark 13
In Example 21, we get an upper bound for the Bhattacharyya distance, which is new in quantum calculus.
Proposition 17
Under the assumptions of Theorem 3, we have
$$\begin{aligned}& 0 \leq \frac{-1 - G(\zeta _{1}, \zeta _{2})}{2 A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})} - D_{B}(\tilde{r}, \tilde{s}), \\& 0 \leq \frac{1}{4G(\zeta _{1}, \zeta _{2}) A(\sqrt{\zeta _{1}},\sqrt{\zeta _{2}})} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr], \end{aligned}$$
where \(A(\cdot ,\cdot )\) is the arithmetic mean.
Proof
Apply Theorem 3 with \(\psi (t) = -\sqrt{t}\), which implies that \(\psi ^{\prime }(t) = - \frac{1}{2\sqrt{t}}\) and
$$ \frac{\psi ^{\prime }(\zeta _{2})- \psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{1}{2\sqrt{\zeta _{1}\zeta _{2}}(\sqrt{\zeta _{2}}+ \sqrt{\zeta _{1}})}. $$
□
Proposition 18
Under the assumptions of Theorem 4, we have
$$\begin{aligned} 0 \leq & \frac{1}{8\sqrt{\zeta _{2}^{3}}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr] \\ \leq & \frac{-1 - G(\zeta _{1}, \zeta _{2})}{2 A(\sqrt{\zeta _{1}}, \sqrt{\zeta _{2}})} - D_{B}(\tilde{r}, \tilde{s}) \\ \leq & \frac{1}{8\sqrt{\zeta _{1}^{3}}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Use Theorem 4 for \(\psi (t) = - \sqrt{t}\), which implies that \(\psi ^{\prime \prime }(t) = \frac{1}{4t^{\frac{3}{2}}}\) and, obviously,
$$ \frac{1}{4\zeta _{2}^{\frac{3}{2}}} \leq \psi ^{\prime \prime }(t) \leq \frac{1}{4\zeta _{1}^{\frac{3}{2}}}\quad \text{for all } t \in [\zeta _{1}, \zeta _{2}]. $$
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4.6 Harmonic distance on time scales
Let \(\psi : (0, \infty ) \rightarrow \mathbb{R}\) be the convex mapping \(\psi (t) = \frac{-2t}{1 + t}\). Then
$$ I_{\psi }(\tilde{r}, \tilde{s})= - \int _{a}^{b} \frac{2\tilde{r}(y)\tilde{s}(y)}{\tilde{r}(y) + \tilde{s}(y)}\Delta y = D_{\mathrm{Ha}}(\tilde{r}, \tilde{s}), $$
where \(D_{\mathrm{Ha}}(\tilde{r}, \tilde{s})\) is the harmonic distance.
Proposition 19
Under the assumptions of Theorem 2, we have
$$ D_{\mathrm{Ha}}(\tilde{r}, \tilde{s}) \leq \frac{-2}{1 + 2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2})}. $$
Proof
Use Theorem 2 for \(\psi (t) = \frac{-2t}{1 + t}\) to get
$$\begin{aligned} D_{\mathrm{Ha}}(\tilde{r}, \tilde{s}) \leq & \frac{\zeta _{2}-1}{\zeta _{2}-\zeta _{1}} \frac{-2\zeta _{1}}{1 + \zeta _{1}} + \frac{1-\zeta _{1}}{\zeta _{2}-\zeta _{1}} \frac{-2\zeta _{2}}{1 + \zeta _{2}} \\ =& \frac{-2\zeta _{1}\zeta _{2} + 2\zeta _{1} - 2\zeta _{2} + 2\zeta _{1} \zeta _{2}}{(\zeta _{2}-\zeta _{1})(\zeta _{1} + 1)(\zeta _{2} + 1)} \\ =& \frac{-2(\zeta _{2} - \zeta _{1})}{(\zeta _{2}-\zeta _{1})(\zeta _{1} + 1)(\zeta _{2} + 1)} \\ =& \frac{-2}{(\zeta _{1} + 1)(\zeta _{2} + 1)} = \frac{-2}{1 + 2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2})}. \end{aligned}$$
□
The following example gives an upper bound for the harmonic distance, which is new in quantum calculus.
Example 22
Choosing \(\mathbb{T} = q^{\mathbb{N}_{0}}\) (\(q > 1\)) in Proposition 19, we obtain
$$ -\sum_{k=\log _{q}(a)}^{\log _{q}(b) - 1} \frac{2q^{k}(q-1)\tilde{r}(q^{k})\tilde{s} (q^{k})}{\tilde{r}(q^{k}) + \tilde{s}(q^{k})} \leq \frac{-2}{1 + 2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2})}. $$
Proposition 20
Under the conditions of Theorem 3, we have
$$\begin{aligned}& 0 \leq \frac{-2}{1 + 2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2})} - D_{\mathrm{Ha}}(\tilde{r}, \tilde{s}), \\& 0 \leq \frac{8A(\zeta _{1}, \zeta _{2}) + 8}{[G^{2}(\zeta _{1}, \zeta _{2}) + 2A(\zeta _{1}, \zeta _{2}) + 1]^{2}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Apply Theorem 3 with \(\psi (t) = \frac{-2t}{1 + t}\), which implies that \(\psi ^{\prime }(t) = \frac{-1}{(1 + t)^{2}}\) and
$$ \frac{\psi ^{\prime }(\zeta _{2})- \psi ^{\prime }(\zeta _{1})}{\zeta _{2}-\zeta _{1}} = \frac{(\zeta _{2} + \zeta _{1} + 2)}{[(\zeta _{2}+ 1)(\zeta _{1}+ 1)]^{2}} = \frac{2A(\zeta _{1}, \zeta _{2}) + 2}{[G^{2}(\zeta _{1}, \zeta _{2}) + 2A(\zeta _{1}, \zeta _{2}) + 1]^{2}}. $$
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Proposition 21
Under the assumptions of Theorem 4, we get
$$\begin{aligned} 0 \leq & \frac{2}{(1 + \zeta _{2})^{3}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{\chi ^{2}}(\tilde{r}, \tilde{s}) \bigr] \\ \leq & \frac{-2}{1 + 2A(\zeta _{1}, \zeta _{2}) + G^{2}(\zeta _{1}, \zeta _{2})} - D_{\mathrm{Ha}}(\tilde{r}, \tilde{s}) \\ \leq & \frac{2}{(1 + \zeta _{1})^{3}} \bigl[(\zeta _{2}-1) (1-\zeta _{1})- D_{ \chi ^{2}}(\tilde{r}, \tilde{s}) \bigr]. \end{aligned}$$
Proof
Use Theorem 4 for \(\psi (t) = \frac{-2t}{1 + t}\), which implies that \(\psi ^{\prime \prime }(t) = \frac{2}{(1 + t)^{3}}\) and
$$ \frac{2}{(1 + \zeta _{2})^{3}} \leq \psi ^{\prime \prime }(t) \leq \frac{2}{(1 + \zeta _{1})^{3}} \quad \text{for all } t \in [ \zeta _{1}, \zeta _{2}]. $$
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