## 2 The \(\alpha\)-Divergences Induced by a Pair of Strictly Comparable Weighted Means
### Limit Cases of \(1\)-divergences and \(0\)-divergences  
We seek a closed-form formula of the limit divergence \(\lim_{\alpha\to 0}I_{\alpha}^{f,g}(p:q)\) when \(\alpha\to 0\). **Lemma 1**. : _A first-order [PERSON] approximation of the quasi-arithmetic mean [56]\(M_{\alpha}^{f}\) for a \(C_{1}\) strictly increasing generator \(f\) when \(\alpha\simeq 0\) yields_  
\[M_{\alpha}^{f}(p,q)=p+\frac{\alpha(f(q)-f(p))}{f^{\prime}(p)}+o(\alpha(f(q)-f( p))). \tag{47}\]  
Proof. : By taking the first-order [PERSON] expansion of \(f^{-1}(x)\) at \(x_{0}\) (i.e., Taylor polynomial of order \(1\)), we get:  
\[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})(f^{-1})^{\prime}(x_{0})+o(x-x_{0}). \tag{48}\]  
Using the property of the derivative of an inverse function  
\[(f^{-1})^{\prime}(x)=\frac{1}{(f^{\prime}(f^{-1})(x))}, \tag{49}\]it follows that the first-order Taylor expansion of \(f^{-1}(x)\) is:  
\[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})\frac{1}{(f^{\prime}(f^{-1})(x_{0}))}+o(x-x_{0}). \tag{50}\]  
Plugging \(x_{0}=f(p)\) and \(x=f(p)+\alpha(f(q)-f(p))\), we get a first-order approximation of the weighted quasi-arithmetic mean \(M_{\alpha}^{f}\) when \(\alpha\to 0\):  
\[M_{\alpha}^{f}(p,q)=p+\frac{\alpha(f(q)-f(p))}{f^{\prime}(p)}+o(\alpha(f(q)-f(p ))). \tag{51}\]  
Let us introduce the following bivariate function:  
\[E_{f}(p,q):=\frac{f(q)-f(p)}{f^{\prime}(p)}. \tag{52}\]  
**Remark 3**. : _Notice that \(E_{f}(p,q)=E_{-f}(p,q)\) matches the fact that \(M_{\alpha}^{f}(p,q)=M_{\alpha}^{-f}(p,q)\). That is, we may either consider a strictly increasing differentiable generator \(f\), or equivalently a strictly decreasing differentiable generator \(-f\)._  
Thus, we obtain closed-form formulas for the \(I_{1}\)-divergence and \(I_{0}\)-divergence:  
**Theorem 2** (Quasi-arithmetic \(I_{1}\)-divergence and reverse \(I_{0}\)-divergence). : _The quasi-arithmetic \(I_{1}\)-divergence induced by two strictly increasing and differentiable functions \(f\) and \(g\) such that \(f\circ g^{-1}\) is strictly convex is_  
\[I_{1}^{f,g}(p:q):=\lim_{\alpha\to 1}I_{\alpha}^{f,g}(p:q) = \int\Big{(}E_{f}(p,q)-E_{g}(p,q)\Big{)}\mathrm{d}\mu\geq 0, \tag{53}\] \[= \int\bigg{(}\frac{f(q)-f(p)}{f^{\prime}(p)}-\frac{g(q)-g(p)}{g^{ \prime}(p)}\bigg{)}\mathrm{d}\mu. \tag{54}\]  
_Furthermore, we have \(I_{0}^{f,g}(p:q)=I_{1}^{f,g}(q:p)=(I_{1}^{f,g})^{*}(p:q)\), the reverse divergence._  
Proof. : Let us prove that \(I_{1}^{f,g}\) is a proper divergence satisfying axioms D1 and D2. Note that a sufficient condition for \(I_{1}^{f,g}(p:q)\geq 0\) is to check that  
\[E_{f}(p,q) \geq E_{g}(p,q), \tag{55}\] \[\frac{f(q)-f(p)}{f^{\prime}(p)} \geq \frac{g(q)-g(p)}{g^{\prime}(p)}. \tag{56}\]  
If \(p=q\)\(\mu\)-almost everywhere then clearly \(I_{1}^{f,g}(p:q)=0\). Consider \(p\
eq q\) (i.e., at some observation \(x\): \(p(x)\
eq q(x)\)). We use the following property of a strictly convex and differentiable function \(h\) for \(x<y\) (sometimes called the chordal slope lemma, see [29]):  
\[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})(f^{-1})^{\prime}(x_{0})+o(x-x_{0}). \tag{48}\]0  
We consider \(h(x)=(f\circ g^{-1})(x)\) so that \(h^{\prime}(x)=\frac{f^{\prime}(g^{-1}(x))}{g^{\prime}(g^{-1}(x))}\). There are two cases to consider:* \(p<q\) and therefore \(g(p)<g(q)\). Let \(y=g(q)\) and \(x=g(p)\) in Equation (57). We have \(h^{\prime}(x)=\frac{f^{\prime}(p)}{g^{\prime}(p)}\) and \(h^{\prime}(y)=\frac{f^{\prime}(q)}{g^{\prime}(q)}\), and the double inequality of Equation (57) becomes \[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})(f^{-1})^{\prime}(x_{0})+o(x-x_{0}). \tag{48}\]1 Since \(g(q)-g(p)>0\), \(g^{\prime}(p)>0\), and \(f^{\prime}(p)>0\), we get \[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})(f^{-1})^{\prime}(x_{0})+o(x-x_{0}). \tag{48}\]2
* \(q<p\) and therefore \(g(p)>g(q)\). Then, the double inequality of Equation (57) becomes \[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})(f^{-1})^{\prime}(x_{0})+o(x-x_{0}). \tag{48}\]3 That is, \[f^{-1}(x)=f^{-1}(x_{0})+(x-x_{0})(f^{-1})^{\prime}(x_{0})+o(x-x_{0}). \tag{48}\]4 since \(g(q)-g(p)<0\).