**3. Bounds of the Weighted Tsallis and Kaniadakis Divergences**

The proof of the following lemma is elementary and is omitted.

**Lemma 1.** *For any x* <sup>∈</sup> (0, <sup>∞</sup>) \ {1}*, let <sup>ϕ</sup><sup>x</sup>* : <sup>R</sup> \ {0} → <sup>R</sup>*,*

$$
\varphi\_{\mathbf{x}}(t) = \frac{\mathbf{x}^t - 1}{t}.
$$

*The function ϕ<sup>x</sup> is strictly increasing.*

The next two corollaries are very useful in this article.

**Corollary 1.** *Let x* > 0 *and k* ∈ (−1, ∞) \ {0}*. Then,*

$$\propto \log\_k^T \ge \ge -1$$

*and the equality is valid if and only if x* = 1*.*

\*\*Corollary 2.\*\*  $Let \ x > 0 \ and \ k \in \left(-\frac{1}{2}, 1\right)$  \\
\*\*1\*\*  $Then$  
$$\frac{2(\sqrt{x} - 1)}{\sqrt{x}} \le \log\_k^T x \le x - 1.$$

*We have equality in these inequalities if and only if x* = 1*.*

**Theorem 1.** *Let A* ∈ T *such that μi*(*A*) = 0 *for any i* = 1, 2*.* (*a*) *Assume that k* ∈ (−1, ∞) \ {0}*. Then,*

$$D\_k^{w,T}(\mu\_1|\mu\_2, A) \ge \log\_k^T \left(\frac{\int\_A w d\mu\_1}{\int\_A w d\mu\_2}\right).$$

(*b*) *Assume that k* ∈ (−1, 1) \ {0}*. Then,*

$$D\_k^{w,K}(\mu\_1|\mu\_2, A) \ge \log\_k^K \left(\frac{\int\_A w d\mu\_1}{\int\_A w d\mu\_2}\right).$$

$$\text{In both cases, the equality holds if and only if } \frac{f\_1(\omega)}{f\_2(\omega)} = \frac{\int\_A w d\mu\_1}{\int\_A w d\mu\_2} \text{ }\lambda - a.e. \text{ for } \omega \in A. \text{ }\lambda$$

**Proof.** (*a*) We will make the proof in two steps.

$$\text{Step 1. Assume that } \int\_A w d\mu\_1 = \int\_A w d\mu\_2. \text{ Because } \log\_k^T \left(\frac{\int\_A w d\mu\_1}{\int\_A w d\mu\_2}\right) = 0 \text{, we have to}$$

show that *Dw*,*<sup>T</sup> <sup>k</sup>* (*μ*1|*μ*2, *A*) ≥ 0.

According to Corollary 1, we have

$$\begin{split} D\_{k}^{w,T}(\mu\_{1}|\mu\_{2},A) &= \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \log\_{k}^{T} \left(\frac{f\_{1}}{f\_{2}}\right) d\mu\_{1} = \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{f\_{2}}{f\_{1}} \cdot \frac{f\_{1}}{f\_{2}} \log\_{k}^{T} \left(\frac{f\_{1}}{f\_{2}}\right) d\mu\_{1} \geq 0\\ &\quad \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{f\_{2}}{f\_{1}} \left(\frac{f\_{1}}{f\_{2}} - 1\right) d\mu\_{1} = \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \left(1 - \frac{f\_{2}}{f\_{1}}\right) d\mu\_{1} =\\ &\quad 1 - \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{f\_{2}}{f\_{1}} d\mu\_{1} = 1 - \frac{\int\_{A} w d\mu\_{2}}{\int\_{A} w d\mu\_{1}} = 0. \end{split}$$

The equality holds if and only if *<sup>f</sup>*1(*ω*) *<sup>f</sup>*2(*ω*) <sup>=</sup> <sup>1</sup> *<sup>μ</sup>*<sup>1</sup> <sup>−</sup> *<sup>a</sup>*.*e*, i.e., if and only if *<sup>f</sup>*1(*ω*) *<sup>f</sup>*2(*ω*) <sup>=</sup> <sup>1</sup> *λ* − *a*.*e*.

*Step 2.* Let *<sup>A</sup>* ∈ T with *<sup>μ</sup>i*(*A*) <sup>=</sup> 0 for any *<sup>i</sup>* <sup>=</sup> 1, 2. We define the measures *<sup>μ</sup>*\*<sup>1</sup> and *<sup>μ</sup>*\*<sup>2</sup> via

$$\widetilde{\mu}\_i(B) = \frac{\mu\_i(B)}{\int\_A w d\mu\_i} \text{ for any } B \in \mathcal{T} \text{ and any } i = 1, 2.$$

We remark that *A wdμ*\*<sup>1</sup> <sup>=</sup> *A wdμ*\*<sup>2</sup> <sup>=</sup> 1. Hence the weighted Tsallis divergence between *<sup>μ</sup>*\*<sup>1</sup> and *<sup>μ</sup>*\*<sup>2</sup> on *<sup>A</sup>* is

$$\begin{aligned} D\_k^T(\tilde{\mu}\_1 | \tilde{\mu}\_{2'} A) &= \int\_A w \cdot \frac{f\_1}{\int\_A w d\mu\_1} \cdot \log\_k^T \left( \frac{\frac{f\_1}{\int\_A w d\mu\_1}}{\frac{f\_2}{\int\_A w d\mu\_2}} \right) d\lambda = 0, \\ &\int\_A w \cdot \frac{f\_1}{\int\_A w d\mu\_1} \cdot \log\_k^T \left( \frac{f\_1}{f\_2} \int\_A w d\mu\_2}{f\_2} \right) d\lambda. \end{aligned}$$

We deduce from *Step 1* that *D<sup>T</sup> <sup>k</sup>* (*μ*\*1|*μ*\*2, *<sup>A</sup>*) <sup>≥</sup> 0. So,

0 ≤ *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *A wdμ*<sup>1</sup> · log*<sup>T</sup> k* ⎛ ⎜⎝ *f*1 *A wdμ*<sup>2</sup> *f*2 *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *dλ* = *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *A wdμ*<sup>1</sup> · ⎛ ⎜⎝ *f*1 *A wdμ*<sup>2</sup> *f*2 *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k* − 1 *<sup>k</sup> <sup>d</sup><sup>λ</sup>* <sup>=</sup> *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *A wdμ*<sup>1</sup> · ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k* · *f*<sup>1</sup> *f*2 *k* − 1 *<sup>k</sup> <sup>d</sup><sup>λ</sup>* <sup>=</sup> *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *A wdμ*<sup>1</sup> · ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k* · ( *<sup>f</sup>*<sup>1</sup> *f*2 *k* − 1 + 1 ) − 1 *<sup>k</sup> <sup>d</sup><sup>λ</sup>* <sup>=</sup> *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *A wdμ*<sup>1</sup> · ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k* · *f*<sup>1</sup> *f*2 *k* − 1 *<sup>k</sup>* <sup>+</sup> ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k* − 1 *k* ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ *dλ* = ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k Dw*,*<sup>T</sup> <sup>k</sup>* (*μ*1|*μ*2, *A*) + ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k* − 1 *<sup>k</sup>* . Hence, *Dw*,*<sup>T</sup> <sup>k</sup>* (*μ*1|*μ*2, *A*) ≥ ⎛ ⎜⎝ *A wdμ*<sup>1</sup> *A wdμ*<sup>2</sup> ⎞ ⎟⎠ *k* · 1 − ⎛ ⎜⎝ *A wdμ*<sup>2</sup> *A wdμ*<sup>1</sup> ⎞ ⎟⎠ *k <sup>k</sup>* <sup>=</sup> log*<sup>T</sup> k* ⎛ ⎜⎝ *A wdμ*<sup>1</sup> *A wdμ*<sup>2</sup> ⎞ ⎟⎠. The equality holds if and only if *<sup>f</sup>*1(*ω*) *A wdμ*<sup>1</sup> <sup>=</sup> *<sup>f</sup>*2(*ω*) *A wdμ*<sup>2</sup> *λ* − *a*.*e*. for *ω* ∈ *A*, i.e., if and only if *<sup>f</sup>*1(*ω*) *<sup>f</sup>*2(*ω*) <sup>=</sup> *A wdμ*<sup>1</sup> *A wdμ*<sup>2</sup> *λ* − *a*.*e*. for *ω* ∈ *A*. (*b*) Because log*<sup>K</sup> <sup>k</sup> <sup>x</sup>* <sup>=</sup> <sup>1</sup> 2 - log*<sup>T</sup> <sup>k</sup> <sup>x</sup>* <sup>+</sup> log*<sup>T</sup>* <sup>−</sup>*<sup>k</sup> <sup>x</sup>* , we obtain *Dw*,*<sup>K</sup> <sup>k</sup>* (*μ*1|*μ*2) = <sup>1</sup> 2 - *Dw*,*<sup>T</sup> <sup>k</sup>* (*μ*1|*μ*2) + *<sup>D</sup>w*,*<sup>T</sup>* <sup>−</sup>*<sup>k</sup>* (*μ*1|*μ*2) ≥ 1 2 ⎛ ⎜⎝log*<sup>T</sup> k* ⎛ ⎜⎝ *A wdμ*<sup>1</sup> *wdμ*<sup>2</sup> ⎞ ⎟⎠ <sup>+</sup> log*<sup>T</sup>* −*k* ⎛ ⎜⎝ *A wdμ*<sup>1</sup> *wdμ*<sup>2</sup> ⎞ ⎟⎠ ⎞ ⎟⎠ <sup>=</sup> log*<sup>K</sup> k* ⎛ ⎜⎝ *A wdμ*<sup>1</sup> *wdμ*<sup>2</sup> ⎞ ⎟⎠ (see (*a*)).

We have equality in the preceding inequality if and only if

*A*

*A*

*A*

$$\frac{1}{2}\left(D\_{k}^{w,T}(\mu\_{1}|\mu\_{2},A) + D\_{-k}^{w,T}(\mu\_{1}|\mu\_{2},A)\right) = \frac{1}{2}\left(\log\_{k}^{T}\left(\frac{\int\_{A}wd\mu\_{1}}{\int\_{A}wd\mu\_{2}}\right) + \log\_{-k}^{T}\left(\frac{\int\_{A}wd\mu\_{1}}{\int\_{A}wd\mu\_{2}}\right)\right).$$

$$\text{which is equivalent to } D\_{k}^{w,T}(\mu\_{1}|\mu\_{2},A) = \log\_{k}^{T}\left(\frac{\int\_{A}wd\mu\_{1}}{\int\_{A}wd\mu\_{2}}\right) \text{ and } D\_{-k}^{w,T}(\mu\_{1}|\mu\_{2},A) = \log\_{-k}^{T}\left(\frac{\int\_{A}wd\mu\_{1}}{\int\_{A}wd\mu\_{2}}\right) \text{ and therefore }$$

$$\log\_{-k}^{T}\left(\frac{\int\_{A}wd\mu\_{1}}{\int\_{A}wd\mu\_{2}}\right) \text{ and these are equivalent to } \frac{f\_{1}(\omega)}{f\_{2}(\omega)} = \frac{\int\_{A}wd\mu\_{1}}{\int\_{A}wd\mu\_{2}}\lambda - a.c. \text{ for } \omega \in A.$$

**Theorem 2.** *Let A* ∈ T *such that <sup>μ</sup>i*(*A*) <sup>=</sup> <sup>0</sup> *for any i* <sup>=</sup> 1, 2 *and A wdμ*<sup>1</sup> = *A wdμ*2*.* (*a*) *Assume that k* ∈ −1 2 , 1 \ {0}*. Then,*

$$\frac{1}{\int\_{A} w d\mu\_1} \int\_{A} w \left( \sqrt{\frac{d\mu\_1}{d\mu\_2}} - 1 \right)^2 d\mu\_2 \le D\_k^{w,T} (\mu\_1 | \mu\_2) \le \frac{1}{\int\_{A} w d\mu\_1} \int\_{A} w \left( \frac{d\mu\_1}{d\mu\_2} - 1 \right)^2 d\mu\_2.$$
 
$$(\mu\_1, \mu\_2)$$

(*b*) *Assume that k* ∈ −1 2 , 1 2 \ {0}*. Then,*

$$\frac{1}{\int\_A w d\mu\_1} \int\_A w \left( \sqrt{\frac{d\mu\_1}{d\mu\_2}} - 1 \right)^2 d\mu\_2 \le D\_k^{w,K}(\mu\_1 | \mu\_2) \le \frac{1}{\int\_A w d\mu\_1} \int\_A w \left( \frac{d\mu\_1}{d\mu\_2} - 1 \right)^2 d\mu\_2.$$

**Proof.** (*a*) We have (see Corollary 2)

*Dw*,*<sup>T</sup> <sup>k</sup>* (*μ*1|*μ*2) = <sup>1</sup> *A wdμ*<sup>1</sup> *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *f*2 · log*<sup>T</sup> k f*<sup>1</sup> *f*2 *dμ*<sup>2</sup> ≥ 1 *A wdμ*<sup>1</sup> *A <sup>w</sup>* · *<sup>f</sup>*<sup>1</sup> *f*2 · 2 *f*1 *f*2 − 1 *f*1 *f*2 *<sup>d</sup>μ*<sup>2</sup> <sup>=</sup> <sup>1</sup> *A wdμ*<sup>1</sup> *A* 2*w* · *f*1 *f*2 · *f*1 *f*2 − 1 *dμ*<sup>2</sup> = 1 *A wdμ*<sup>1</sup> *A* 2*w* · *dμ*<sup>1</sup> *dμ*<sup>2</sup> · *dμ*<sup>1</sup> *dμ*<sup>2</sup> − 1 *<sup>d</sup>μ*<sup>2</sup> <sup>=</sup> <sup>1</sup> *A wdμ*<sup>1</sup> *A w* · 2 · *dμ*<sup>1</sup> *dμ*<sup>2</sup> − 2 *dμ*<sup>1</sup> *dμ*<sup>2</sup> *dμ*<sup>2</sup> = 1 *A wdμ*<sup>1</sup> *A w* · *dμ*<sup>1</sup> *dμ*<sup>2</sup> *<sup>d</sup>μ*<sup>2</sup> <sup>−</sup> <sup>2</sup> · <sup>1</sup> *A wdμ*<sup>1</sup> *A w* · *dμ*<sup>1</sup> *dμ*<sup>2</sup> *dμ*<sup>2</sup> + 1 *A wdμ*<sup>1</sup> *A wdμ*<sup>2</sup> = 1 *A wdμ*<sup>1</sup> *A w* · *dμ*<sup>1</sup> *dμ*<sup>2</sup> − 1 2 *dμ*2.

On the other hand (see again Corollary 2),

$$\begin{split} D\_{k}^{w,T}(\mu\_{1}|\mu\_{2}) &= \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{f\_{1}}{f\_{2}} \cdot \log\_{k}^{T} \left(\frac{f\_{1}}{f\_{2}}\right) d\mu\_{2} \leq \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{f\_{1}}{f\_{2}} \cdot \left(\frac{f\_{1}}{f\_{2}} - 1\right) d\mu\_{2} = \\ &\frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{d\mu\_{1}}{d\mu\_{2}} \cdot \left(\frac{d\mu\_{1}}{d\mu\_{2}} - 1\right) d\mu\_{2} = \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \left(\left(\frac{d\mu\_{1}}{d\mu\_{2}}\right)^{2} - \frac{d\mu\_{1}}{d\mu\_{2}}\right) d\mu\_{2} = \end{split}$$

$$\begin{split} \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \left(\frac{d\mu\_{1}}{d\mu\_{2}}\right)^{2} d\mu\_{2} - 2 \cdot \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \cdot \frac{d\mu\_{1}}{d\mu\_{2}} d\mu\_{2} + \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w d\mu\_{2} &= 0\\ \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \left(\frac{d\mu\_{1}}{d\mu\_{2}} - 1\right)^{2} d\mu\_{2}. \end{split}$$

(*b*) Using (*a*) we obtain

$$\begin{split} \frac{1}{2} \cdot \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \left( \sqrt{\frac{d\mu\_{1}}{d\mu\_{2}}} - 1 \right)^{2} d\mu\_{2} + \frac{1}{2} \cdot \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \left( \sqrt{\frac{d\mu\_{1}}{d\mu\_{2}}} - 1 \right)^{2} d\mu\_{2} \leq \frac{1}{2} \\ \frac{1}{2} D\_{k}^{w,T}(\mu\_{1}|\mu\_{2}) + \frac{1}{2} D\_{-k}^{w,T}(\mu\_{1}|\mu\_{2}) \leq \\ \frac{1}{2} \cdot \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \left( \frac{d\mu\_{1}}{d\mu\_{2}} - 1 \right)^{2} d\mu\_{2} + \frac{1}{2} \cdot \frac{1}{\int\_{A} w d\mu\_{1}} \int\_{A} w \left( \frac{d\mu\_{1}}{d\mu\_{2}} - 1 \right)^{2} d\mu\_{2}. \end{split}$$

Hence,

$$\frac{1}{\int\_{A} w d\mu\_1} \int\_{A} w \left( \sqrt{\frac{d\mu\_1}{d\mu\_2}} - 1 \right)^2 d\mu\_2 \le D\_k^{w, K} (\mu\_1 | \mu\_2) \le \frac{1}{\int\_{A} w d\mu\_1} \int\_{A} w \left( \frac{d\mu\_1}{d\mu\_2} - 1 \right)^2 d\mu\_2.$$
