The ESS for evolutionary matrix games under time constraints and its relationship with the asymptotically stable rest point of the replicator dynamics

Abstract

Recently we interpreted the notion of ESS for matrix games under time constraints and investigated the corresponding state in the polymorphic situation. Now we give two further static (monomorphic) characterizations which are the appropriate analogues of those known for classical evolutionary matrix games. Namely, it is verified that an ESS can be described as a neighbourhood invader strategy independently of the dimension of the strategy space in our non-linear situation too, that is, a strategy is an ESS if and only if it is able to invade and completely replace any monomorphic population which totally consists of individuals following a strategy close to the ESS. With the neighbourhood invader property at hand, we establish a dynamic characterization under the replicator dynamics in two dimensions which corresponds to the strong stability concept for classical evolutionary matrix games. Besides, in some special cases, we also prove the stability of the corresponding rest point in higher dimensions.

Appendix

In this part we cite or claim some technical statements for the convenience of the reader and prove the new assertions appeared in the previous sections.

1.1 Auxiliary statements

Proof of Lemma 2.3

(i)$\Leftrightarrow$(ii) Consider a NE $\mathbf p.$ By (2.5), we have

$$\begin{aligned}{}[1+\mathbf e_i T\rho (\mathbf p)\mathbf p] \frac{\mathbf pA\mathbf p}{1+\mathbf p T\rho (\mathbf p)\mathbf p} \ge \mathbf e_{i}A\mathbf p, \quad i\in \mathrm {supp}(\mathbf p). \end{aligned}$$

Multiplying by $p_{i}$ we get

$$\begin{aligned} {[}p_i+p_i \mathbf e_i T\rho (\mathbf p)\mathbf p] \frac{\mathbf pA\mathbf p}{1+\mathbf p T\rho (\mathbf p)\mathbf p} \ge p_i \mathbf e_{i}A\mathbf p\quad i\in \mathrm {supp}(\mathbf p). \end{aligned}$$

Since $\mathbf p=\sum _{i\in \mathrm {supp}(\mathbf p)}p_i\mathbf e_i$ it follows that if we take the sum of the previous inequalities for $i\in \mathrm {supp}(\mathbf p)$ we obtain that

$$\begin{aligned} \sum _{i\in \mathrm {supp}(\mathbf p)} [p_i+p_i \mathbf e_i T\rho (\mathbf p)\mathbf p] \frac{\mathbf pA\mathbf p}{1+\mathbf p T\rho (\mathbf p)\mathbf p}&= [\mathbf p+ \mathbf p T\rho (\mathbf p)\mathbf p] \frac{\mathbf pA\mathbf p}{1+\mathbf p T\rho (\mathbf p)\mathbf p}\\&=\mathbf pA\mathbf p=\sum _{i\in \mathrm {supp}(\mathbf p)}p_i \mathbf e_{i}A\mathbf p, \end{aligned}$$

that is, the sum of the left-hand sides of the inequalities is equal to the sum of the right-hand sides of the inequalities. This is possibile only if there is an equality in (2.5) for every $i\in \mathrm {supp}(\mathbf p)$.

Now assume that (2.6) holds and $\mathbf q$ is an arbitrary strategy. It is clear that (2.6) is equivalent to the inequalities

$$\begin{aligned}{}[1+\mathbf e_{i}T\rho (\mathbf p) \mathbf p] \frac{\mathbf pA\mathbf p}{1+\,\mathbf pT\rho (\mathbf p) \mathbf p}\ge \mathbf e_{i}A\mathbf p \quad \text {with equality if } i\in \mathrm {supp}(\mathbf p). \end{aligned}$$

(A.13)

Multiply by $q_i$ the ith inequality in (A.13). Taking the sum from $i=1$ to $i=N$ we get that

$$\begin{aligned} {[}1+\mathbf q T\rho (\mathbf p)\mathbf p] \frac{\mathbf pA\mathbf p}{1+\mathbf p T\rho (\mathbf p)\mathbf p} \ge \mathbf qA\mathbf p\quad \text {with equality if } \mathrm {supp}(\mathbf q)\subset \mathrm {supp}(\mathbf p) \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{\mathbf pA\mathbf p}{1+\mathbf p T\rho (\mathbf p)\mathbf p} \ge \frac{\mathbf qA\mathbf p}{[1+\mathbf q T\rho (\mathbf p)\mathbf p]}\quad \text {with equality if } \mathrm {supp}(\mathbf q)\subset \mathrm {supp}(\mathbf p). \end{aligned}$$

This means that (2.5) holds for every $\mathbf q\in S_N$. Therefore $\mathbf p$ is a NE.

(i)$\Leftrightarrow$(iii) We have just seen that if $\mathbf p$ is a NE then $\omega _{\mathbf p}(\mathbf p,\mathbf e_i,0)=\omega _{\mathbf e_i}(\mathbf p, \mathbf e_i,0)$ for every $i\in \mathrm {supp}(\mathbf p)$ which implies that $\omega _{\mathbf p}(\mathbf p,\mathbf q,0)=\omega _{\mathbf q}(\mathbf p,\mathbf q,0)$ whenever $\mathrm {supp}(\mathbf q)\subset \mathrm {supp}(\mathbf p)$.

If $\omega _{\mathbf p}(\mathbf p,\mathbf q,0)\ge \omega _{\mathbf q}(\mathbf p,\mathbf q,0)$ for every $\mathbf q\in S_N$ with equality if $\mathrm {supp}(\mathbf q)\subset \mathrm {supp}(\mathbf p)$ then (2.5) holds, so $\mathbf p$ is a NE. $\square$

Lemma A.1

(Garay et al. 2017, Lemma 2) The following system of nonlinear equations in n variables,

$$\begin{aligned} u_i=\frac{1}{1+\sum _{j=1}^n c_{ij}u_j},\quad 1\le i\le n, \end{aligned}$$

(A.14)

where the coefficients $c_{ij}$ are positive numbers, has a unique solution in the unite hypercube $[0,1]^{n}$.

Remark A.2

In Garay et al. (2017), Lemma 2 $c_{ij}$ is assumed to be positive for every i, j but considering the proof a bit further one can see the validity of the lemma with non-negative $c_{ij}$-s too.

As claimed by the next lemma, the solution of the previous equation system varies continuously with the coefficients.

Lemma A.3

[Garay et al. 2018, Lemma 6.2 (i)] The solution $\mathbf {u}=(u_1,u_2,\ldots ,u_n)\in [0,1]^{n}$ of (A.14) is a continuous function in

$$\begin{aligned} \mathbf {c}:=(c_{11},\ldots , c_{1n},c_{21},\ldots ,c_{2n},\ldots , c_{n1},\ldots ,c_{nn})\in \mathbb {R}_{\ge 0}^{n^2}. \end{aligned}$$

Lemma A.4

(Garay et al. 2018, Corollary 6.3) Consider a population of phenotypes $\mathbf p^*,\mathbf p_1,\ldots ,\mathbf p_n\in S_N$ with frequencies $1-x,x_1,\ldots ,x_n$ where $x=x_1+x_2+\cdots +x_n$.

• (i) The active parts $\varrho ^*$ and $\varrho _i$ ($i=1,\ldots ,n$) of the different phenotypes [see the definition at (2.8)] continuously depend on $\mathbf {x}=(1-x,x_1,\ldots ,x_n)$.

• (ii) If $\mathbf {y}$ is another frequency distribution such that $\bar{\mathbf h}(\mathbf {y}) =\bar{\mathbf h}(\mathbf {x})=:\bar{\mathbf h}$, then both $\bar{\varrho }(\mathbf {x})=\bar{\varrho }(\mathbf {y})$, $\varrho ^*(\mathbf {x})=\varrho ^*(\mathbf {y})$ and $\varrho _i(\mathbf {x})=\varrho _i(\mathbf {y})$ ($1\le i\le n$).

• (iii) If $\bar{\mathbf h}$ can be uniquely represented as a convex combination of $\mathbf p^*,\mathbf p_1,\ldots ,\mathbf p_n$⁷ then $x_i$ must be equal to $y_i$ for every i.

Obviously, if there exists only two phenotypes with positive frequency then the average strategy of the active subpopulation is a convex combination of the two phenotypes. Conversely, if a strategy is a convex combination of the two phenotypes, does there exist a composition which mix the strategy? The next lemma gives the precise answer.

Lemma A.5

(Garay et al. 2018, Lemma 6.4) Let $\mathbf p,\mathbf q\in S_2$. Denote by $\varrho _{\mathbf p}(\varepsilon ),\varrho _{\mathbf q}(\varepsilon )$ the unique solution in $[0,1]\times [0,1]$ of the system

$$\begin{aligned} \varrho _{\mathbf p}= & {} \frac{1}{1+\mathbf p T [(1-\varepsilon )\varrho _{\mathbf p} \mathbf p+\varepsilon \varrho _{\mathbf q} \mathbf q]}, \\ \varrho _{\mathbf q}= & {} \frac{1}{1+\mathbf q T [(1-\varepsilon )\varrho _{\mathbf p} \mathbf p+\varepsilon \varrho _{\mathbf q} \mathbf q]}. \end{aligned}$$

Furthermore, $\bar{\varrho }(\varepsilon ):=(1-\varepsilon )\varrho _{\mathbf p}(\varepsilon )+ \varepsilon \varrho _{\mathbf q}(\varepsilon )$ and

$$\begin{aligned} \mathbf r(\varepsilon ):=\frac{1}{\bar{\varrho }(\varepsilon )} [(1-\varepsilon )\varrho _{\mathbf p}(\varepsilon )\mathbf p+ \varepsilon \varrho _{\mathbf q}(\varepsilon )\mathbf q]. \end{aligned}$$

Then $\mathbf r(0)=\mathbf p$, $\mathbf r(1)=\mathbf q$ and $\mathbf r(\varepsilon )$ uniquely runs through the line segment between $\mathbf p$ and $\mathbf q$ as $\varepsilon$ runs from 0 to 1 in such a way that $0\le \varepsilon _1<\varepsilon _2\le 1$ implies that $||\mathbf r(\varepsilon _1)-\mathbf p||< ||\mathbf r(\varepsilon _2)-\mathbf p||$.

1.2 Proof of the main theorems

Now, we are ready to prove the two important static characterizations of the notion of UESS.

Proof of Theorem 3.1

Assume that $\mathbf p$ is a UESS and let $\varepsilon _0$ be the uniform threshold number in Definition 2.1. Denote by F the union of those faces⁸ of $S_N$ which do not contain $\mathbf p$.

Necessity. We claim that

$$\begin{aligned} \min _{\hat{\mathbf q}\in F}\bigg |\bigg |\frac{1}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon _0)} \left[ (1-\varepsilon _0)\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon _0)\mathbf p+ \varepsilon _0\rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\varepsilon _0) \mathbf p\right] -\mathbf p\bigg |\bigg | \end{aligned}$$

is a suitable choice for $\delta$ (do not forget a positive function continuous on a compact set has a positive minimum). Indeed, let $\mathbf q$ be a strategy for which $0<||\mathbf p-\mathbf q||<\delta$ and take the type $\mathbf q$ with frequency $\eta (\in (0,1])$. Denote by $\hat{\mathbf q}$ the point in F which can be represented in a form $\mathbf p+\tau (\mathbf q-\mathbf p)$ with some $\tau >0$ (that is $\hat{\mathbf q}$ is the common point of the boundary of F and the half line from $\mathbf p$ through $\mathbf q$).

By Lemma A.5, for any $\eta \in (0,1]$ there is precisely one $\varepsilon =\varepsilon (\eta )\in (0,1]$ such that

$$\begin{aligned} \mathbf r(\hat{\mathbf q},\varepsilon )&:= \frac{(1-\varepsilon ) \rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )} \mathbf p+\frac{\varepsilon \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\varepsilon )}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )}\hat{\mathbf q}\\&=\frac{(1-\eta ) \rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )}{\bar{\rho }(\mathbf p,\mathbf q,\eta )} \mathbf p+\frac{\eta \rho _{\mathbf q}(\mathbf p,\mathbf q,\eta )}{\bar{\rho }(\mathbf p,\mathbf q,\eta )}\mathbf q=: \mathbf r(\mathbf q,\eta ). \end{aligned}$$

From this fact, a similar argument (the uniqueness provided by Lemma A.1) as in the proof of Lemma A.4(ii) shows that $\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon (\eta ))= \bar{\rho }(\mathbf p,\mathbf q,\eta )$ and $\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon (\eta )) =\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )$,⁹ respectively. Hence, one can readily infer that

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )&= \rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )\mathbf p A \bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )\mathbf r(\hat{\mathbf q},\varepsilon )\\&= \rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )\mathbf p A \bar{\rho }(\mathbf p,\mathbf q,\eta )\mathbf r(\mathbf q,\eta )=\omega _{\mathbf p}(\mathbf p,\mathbf q,\eta ), \end{aligned}$$

and, similarly,

$$\begin{aligned} \bar{\omega }(\mathbf p,\hat{\mathbf q},\varepsilon )&=\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )\mathbf r(\hat{\mathbf q},\varepsilon )A \bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )\mathbf r(\hat{\mathbf q},\varepsilon ) \\&=\bar{\rho }(\mathbf p,\mathbf {q},\eta )\mathbf r(\mathbf {q},\eta )A \bar{\rho }(\mathbf p,\mathbf q,\eta )\mathbf r(\mathbf q,\eta )=\bar{\omega }(\mathbf p,\mathbf q,\eta ). \end{aligned}$$

Considering the choice of $\delta$ and Lemma A.5 we infer from the fact that $||\mathbf p-\mathbf r\big (\hat{\mathbf q},\varepsilon (\eta )\big )|| =||\mathbf p-\mathbf r(\mathbf q,\eta )||\le ||\mathbf p-\mathbf q||<\delta$ that $0<\varepsilon (\eta )\le \varepsilon _0$. We thus obtain

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\mathbf q,\eta )=\omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon (\eta )) >\bar{\omega }(\mathbf p,\hat{\mathbf q},\varepsilon (\eta ))= \bar{\omega }(\mathbf p,\mathbf q,\eta ) \end{aligned}$$

for every $\mathbf q$ with $||\mathbf q-\mathbf p||<\delta$ and for every $\eta \in (0,1]$.

Sufficiency. Let us turn to the other direction. Let $\hat{\mathbf q}\in F$ and let $\mathbf q\not =\hat{\mathbf q}$ be a point on the segment between $\mathbf p$ and $\hat{\mathbf q}.$ We have seen in the proof of the previous direction that for any $\eta \in (0,1]$ there is a unique $\varepsilon =\varepsilon (\eta )$ such that $\mathbf r(\hat{\mathbf q},\varepsilon )=\mathbf r(\mathbf q,\eta )$. We show that $\varepsilon <\eta$.

Writing out the definition of $\mathbf r(\hat{\mathbf q},\varepsilon )$ and $\mathbf r(\mathbf q,\eta )$ their equality means

$$\begin{aligned}&\frac{1}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )} [(1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )\mathbf p + \varepsilon \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\varepsilon ) \hat{\mathbf q}]\nonumber \\&\quad = \frac{1}{\bar{\rho }(\mathbf p,\mathbf q,\eta )} [(1-\eta )\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )\mathbf p +\eta \rho _{\mathbf q}(\mathbf p,\mathbf q,\eta ) \mathbf q]. \end{aligned}$$

(A.15)

By Lemma A.5 there is a $\theta \in (0,1)$ such that $\mathbf r(\hat{\mathbf q},\theta )=\mathbf q$. Replace $\mathbf q$ with this representation in (A.15). We get

$$\begin{aligned}&\frac{1}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )} [(1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )\mathbf p + \varepsilon \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\varepsilon ) \hat{\mathbf q}]\\&\quad = \frac{1}{\bar{\rho }(\mathbf p,\mathbf q,\eta )} \left[ (1-\eta )\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )\mathbf p\right. \\&\left. \qquad +\,\eta \rho _{\mathbf q}(\mathbf p,\mathbf q,\eta ) \underbrace{\left( \frac{1}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\theta )} [(1-\theta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\theta )\mathbf p + \theta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\theta ) \hat{\mathbf q}]\right) } _{\mathbf r(\hat{\mathbf q},\theta )=\mathbf q}\right] . \end{aligned}$$

Lemma A.4(iii) yields the coefficients of $\mathbf p$ in the two sides to be equal:

$$\begin{aligned}&\frac{(1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon )} = \frac{1}{\bar{\rho }(\mathbf p,\mathbf q,\eta )} \bigg [ (1-\eta )\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )\left. \right. \\&\quad \left. +\, \eta \rho _{\mathbf q}(\mathbf p,\mathbf q,\eta ) (1-\theta )\frac{\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\theta )}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\theta )}\right] . \end{aligned}$$

Since $\bar{\rho }(\mathbf p,\hat{ \mathbf q},\varepsilon ) =\bar{\rho }(\mathbf p,\mathbf q,\eta )$ (see the proof of the previous direction) it can be simplified by $\bar{\rho }(\mathbf p,\hat{\mathbf q},\varepsilon ) =\bar{\rho }(\mathbf p,\mathbf q,\eta )$ which results in

$$\begin{aligned} (1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon ) = (1-\eta )\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )+ \underbrace{\eta \rho _{\mathbf q}(\mathbf p,\mathbf q,\eta ) (1-\theta )\frac{\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\theta )}{\bar{\rho }(\mathbf p,\hat{\mathbf q},\theta )}}_{>0}. \end{aligned}$$

Since the second term of the right-hand side is positive it is inferred that

$$\begin{aligned} (1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon ) > (1-\eta )\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta ). \end{aligned}$$

Recalling that $\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon ) =\rho _{\mathbf p}(\mathbf p,\mathbf q,\eta )$ (see the proof of the previous direction) one can conclude that $\varepsilon <\eta$. This immediately implies if for some $\varepsilon _0(\hat{\mathbf q})>0$ we have that

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )> \bar{\omega } (\mathbf p,\hat{\mathbf q},\varepsilon ) \end{aligned}$$

whenever $0<\varepsilon <\varepsilon _0(\hat{\mathbf q})$ then, also,

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\mathbf q,\eta )> \bar{\omega } (\mathbf p,\mathbf q,\eta ) \end{aligned}$$

whenever $0<\eta <\varepsilon _0(\hat{\mathbf q})$, so such an $\varepsilon _0(\hat{\mathbf q})$ is uniform on the segment between $\hat{\mathbf q}$ and $\mathbf p$.

It is clear from the argument made hitherto that the half of the unique $\varepsilon >0$ for which $||\mathbf r(\hat{\mathbf q},\varepsilon )-\mathbf p||=\delta$ is an appropriate choice for such an $\varepsilon _0(\hat{\mathbf q})$ (if $||\hat{\mathbf q}-\mathbf p||<\delta$ let $\varepsilon _0(\hat{\mathbf q})$ set to be, say, 1 / 2). With this choice of $\varepsilon _0(\hat{\mathbf q})$ we have

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\varepsilon )> \bar{\omega } (\mathbf p,\hat{\mathbf q},\varepsilon ), \end{aligned}$$

on the one hand, and

$$\begin{aligned} ||\mathbf r(\hat{\mathbf q},\varepsilon )-\mathbf p||<\delta , \end{aligned}$$

on the other, for any $0<\varepsilon \le \text {(sic!)}\, \varepsilon _0(\hat{\mathbf q})$. Since, by Lemma A.3, $\rho _{\mathbf q},\rho _{\mathbf p},\bar{\rho }$ are continuous in $(\mathbf q,\varepsilon )\in S_N\times [0,1]$ it follows that $\omega _{\mathbf q},\omega _{\mathbf p},\bar{\omega }$ and $\mathbf r(\mathbf q,\varepsilon )$ are also continuous. Consequently, for every $\hat{\mathbf q}\in F$, there exists a $\lambda =\lambda (\hat{\mathbf q})$ such that

$$\begin{aligned} ||\mathbf r\big (\hat{\mathbf s},\varepsilon _0(\hat{\mathbf q})\big )-\mathbf p||<\delta , \end{aligned}$$

whenever $\hat{\mathbf s}\in S_N$ and $||\hat{\mathbf s}-\hat{\mathbf q}|| <\lambda (\hat{\mathbf q})$. Taking into account the definition of $\delta$ and Lemma A.5, we find it is also valid that

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\hat{\mathbf s},\varepsilon )> \bar{\omega }(\mathbf p,\hat{\mathbf s},\varepsilon ), \end{aligned}$$

whenever $0<\varepsilon \le \text {(sic!)}\, \varepsilon _0(\hat{\mathbf q})$. By the compactness of F, there are (finitely many) $\hat{\mathbf q}_1$, $\ldots$, $\hat{\mathbf q}_k\in F$ such that the union of the open ball with centers at $\hat{\mathbf q}_1$, $\ldots$, $\hat{\mathbf q}_k$ and with radius $\lambda (\hat{\mathbf q}_1)$, $\ldots$, $\lambda (\hat{\mathbf q}_k)$ in turn covers F. Then

$$\begin{aligned} \min _{1\le i\le k}\varepsilon _0(\hat{\mathbf q}_i) \end{aligned}$$

is a suitable choice for $\varepsilon _0$ in the definition of the UESS. $\square$

Proof of Theorem 3.2

The necessity is clear. The problematic direction is the sufficiency. Assume that (3.1) holds with $\varepsilon =1$ for any $\mathbf u$ with $0<||\mathbf u-\mathbf p||<\delta$ and let $\mathbf q$ be an arbitrary element of $S_n$ with $0<||\mathbf q-\mathbf p||<\delta$. We validate that, for any $\varepsilon \in (0,1]$, (3.1) holds. For $\varepsilon =1$, this is just the assumption. For $0<\varepsilon <1$, let

$$\begin{aligned} \mathbf r=\mathbf r(\varepsilon ):=\frac{1}{{\bar{\rho }}(\mathbf p, \mathbf q,\varepsilon )} [(1-\varepsilon )\rho _\mathbf p(\mathbf p, \mathbf q,\varepsilon )\mathbf p+ \varepsilon \rho _\mathbf q(\mathbf p, \mathbf q,\varepsilon ) \mathbf q]. \end{aligned}$$

By Lemma A.5, $\mathbf r(\varepsilon )$ is on the segment matching $\mathbf p$ and $\mathbf q$ such that

$$\begin{aligned} ||\mathbf r(\varepsilon )-\mathbf p||<||\underbrace{\mathbf r(1)}_{=\mathbf q}-\mathbf p||<\delta . \end{aligned}$$

On the other hand, by Proposition 2.6, the mean fitness of the population consisting of $\mathbf p$ and $\mathbf q$ individuals with proportions $(1-\varepsilon )$ and $\varepsilon$, respectively, corresponds to the fitness of a population consisting of only $\mathbf r(\varepsilon )$ individuals which can be viewed as a population consisting of $\mathbf p$ and $\mathbf r(\varepsilon )$ individuals with proportions 0 and 1, respectively. Considering Proposition 2.6, the last interpretation also shows that the fitness of a $\mathbf p$ individual in a population consisting of only $\mathbf r(\varepsilon )$ individuals is equal to that in population of the $\mathbf p,\mathbf q$ individuals. Formally, this means that ${\bar{\omega }}(\mathbf p,\mathbf q,\varepsilon )=\omega _{\mathbf r(\varepsilon )}(\mathbf p,\mathbf r(\varepsilon ),1).$ and $\omega _{\mathbf p}(\mathbf p,\mathbf r(\varepsilon ),1)= \omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon ).$ By assumption, $\omega _{\mathbf r(\varepsilon )}(\mathbf p,\mathbf r(\varepsilon ),1)< \omega _{\mathbf p}(\mathbf p,\mathbf r(\varepsilon ),1)$ because $||\mathbf r(\varepsilon )-\mathbf p||< \delta$. We immediately conclude that ${\bar{\omega }}(\mathbf p,\mathbf q,\varepsilon )< \omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )$ which can be possible if and only if $\omega _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon )< \omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )$. $\square$

Proof of Theorem 3.3

Assume that $\mathbf p^*\in S_2$ is a UESS. We prove that the set of states $\mathbf x=(x_1,\ldots ,x_n)$ with $\sum [x_i \varrho _i(\mathbf x)/{\bar{\varrho }}(\mathbf x)] \mathbf p_i=\mathbf p^*$ is locally asymptotically stable. Consider the phenotypes $\mathbf p_1,\ldots , \mathbf p_k,\mathbf p_{k+1},\ldots ,\mathbf p_n$ in the ascending order of the second coordinates, that is, $p_1^2<p_2^2<\cdots<p_k^2\le p^*_2<p_{k+1}^2<\cdots <p_n^2$ where $p_i^1\mathbf e_1+p_i^2\mathbf e_2=\mathbf p_i$ ($i=1,2,\ldots ,n$) and $p^*_1 \mathbf e_1+p^*_2\mathbf e_2=\mathbf p^*$.

Assume that $\mathbf x\in S_n$ is a state such that

$$\begin{aligned} 0<||\bar{\mathbf h}(\mathbf x)- \mathbf p^*||<\min (\delta ,\min \{||\mathbf p_i-\mathbf p^*||: i=1,\ldots ,n\,\text {and}\,\mathbf p_i\not =\mathbf p^*\})=:\delta ' \end{aligned}$$

where

$$\begin{aligned} \bar{\mathbf h}(\mathbf x)=\sum _{i=1}^nx_i\frac{\varrho _i(\mathbf x)}{\bar{\varrho }(\mathbf x)}\mathbf p_i \end{aligned}$$

and $\delta$ comes from Theorem 3.2. This implies that

$$\begin{aligned} \varrho ^*(\mathbf x)\mathbf p^*A\bar{\varrho }(\mathbf x) \bar{\mathbf h}(\mathbf x)>\bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)A\bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)={\bar{W}}(\mathbf x) \end{aligned}$$

(A.16)

where

$$\begin{aligned} \varrho ^*(\mathbf x):=\frac{1}{1+\mathbf p^*T\bar{\varrho }(\mathbf x) \bar{\mathbf h}(\mathbf x)}= \rho _{\mathbf p^*}(\mathbf p^*,\bar{\mathbf h}(\mathbf x),1). \end{aligned}$$

It can be assumed that the second coordinate ${\bar{h}}_2(\mathbf x)$ of $\bar{\mathbf h}(\mathbf x)$ is strictly greater than $p^*_2$ (the case when ${\bar{h}}_2(\mathbf x)$ is strictly smaller can be treated in a similar way and when ${\bar{h}}_2(\mathbf x)=p^*_2$ then $\mathbf x$ is an equilibrium point of the replicator dynamics).

Observation 1

We prove that $W_i>{\bar{W}}$ for $1\le i\le k$ and $W_i<{\bar{W}}$ for $k+1\le i\le n$.

Recall the assumption that $p_2^*<{\bar{h}}_2$. Consequently, if $1\le i\le k$ then there is an $0\le \alpha _i= \alpha _i(\mathbf x)\le 1$ such that $\mathbf p^*=\alpha _i\mathbf p_i+(1-\alpha _i)\bar{\mathbf h}$ (it is easy to check that $\alpha _i=(p_1^*-{\bar{h}}_1)/(p_i^1-{\bar{h}}_1)$). Therefore we have

$$\begin{aligned} \alpha _i\frac{\varrho ^*}{\varrho _{i}}\varrho _{i}\mathbf p_i A\bar{\varrho }\bar{\mathbf h}+ (1-\alpha _i)\frac{\varrho ^*}{{\bar{\varrho }}}\bar{\varrho }\bar{\mathbf h} A\bar{\varrho }\bar{\mathbf h}= \varrho ^*\mathbf p^*A\bar{\varrho }\bar{\mathbf h} >\bar{\varrho }\bar{\mathbf h}A\bar{\varrho }\bar{\mathbf h}. \end{aligned}$$

(Note that $\alpha _i\varrho ^*/\varrho _i+(1-\alpha _i)\varrho ^*/{\bar{\varrho }}=1.$) From this we immediately infer that

$$\begin{aligned} W_i=\varrho _{i}\mathbf p_i A\bar{\varrho }\bar{\mathbf h}> \bar{\varrho }\bar{\mathbf h}A\bar{\varrho }\bar{\mathbf h}={\bar{W}}. \end{aligned}$$

(A.17)

Similarly, if $k+1\le i\le n$, then $\bar{ \mathbf h}$ lies on the line segment connecting $\mathbf p^*$ to $\mathbf p_i$. So there is a $0<\beta _i= \beta _i(\mathbf x)\le 1$ such that $\bar{\mathbf h}=\beta _i \mathbf p^*+(1-\beta _i)\mathbf p_i$ (it is easy to check that $\beta _i=({\bar{h}}_1-p_i^1)/(p_1^*-p_i^1)$). Therefore we have that

$$\begin{aligned} \varrho ^*\mathbf p^*A\bar{\varrho }\bar{\mathbf h} > \bar{\varrho }\bar{\mathbf h}A\bar{\varrho }\bar{\mathbf h}= \beta _i\frac{{\bar{\varrho }}}{\varrho ^*}\varrho ^*\mathbf p^* A\bar{\varrho }\bar{\mathbf h}+ (1-\beta _i)\frac{{\bar{\varrho }}}{\varrho _{i}} \varrho _{i}\mathbf p_i A\bar{\varrho }\bar{\mathbf h} \end{aligned}$$

which immediately implies that

$$\begin{aligned} {\bar{W}}= \bar{\varrho }\bar{\mathbf h}A\bar{\varrho }\bar{\mathbf h}> \varrho _{i}\mathbf p_i A\bar{\varrho }\bar{\mathbf h}=W_i. \end{aligned}$$

(A.18)

In summary, if $0<||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||<\delta '$ and $p_2^*<\bar{h}_2(x)$ then

• for i with $p_i^2\le p_2^*$ and $x_i>0$, we have that $\dot{x}_i=x_i[W_i(\mathbf x)-{\bar{W}}(\mathbf x)]>0$, and

• for i with $p_i^2>p_2^*$ and $x_i>0$, we have that $\dot{x}_i=x_i[W_i(\mathbf x)-{\bar{W}}(\mathbf x)]<0$, respectively.

Observation 2

Let $\mathbf x,\mathbf y\in S_n$ be states such that

• (i) $x_1\le y_1,\ldots ,x_k\le y_k$ but $\sum _{i=1}^kx_i<\sum _{i=1}^ky_i$;

• (ii) $y_{k+1}\le x_{k+1},\ldots ,y_n\le x_n$ but $\sum _{i=k+1}^ny_i<\sum _{i=k+1}^nx_i$;

• (iii) $p_2^*<\bar{h}_2(\mathbf x)$, $p_2^*<\bar{h}_2(\mathbf y)$; and

• (iv) $0<||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||<\delta '$, $0<||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||<\delta '$, respectively.

We prove that $||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||>||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||$. Suppose the contrary, that is, there are states $\mathbf x,\mathbf y$ satisfying the previous conditions (i)–(iv), but with $||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||\le ||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||$. Consider a state $\mathbf z^0\in S_n$ with $z_i^0=0$ if $1\le i\le k$ and with $z_i^0\ge x_i$ if $k+1\le i\le n$. Then $p_{k+1}^2\le {\bar{h}}_2(\mathbf z^0)$. Hence

$$\begin{aligned} ||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||<\delta '\le ||\bar{\mathbf h}(\mathbf z^0)-\mathbf p^*||. \end{aligned}$$

Now take $S_n\ni (z_1,\ldots ,z_n)=\mathbf z=\mathbf z^0$ and start to increase the first coordinate of $\mathbf z$ in the following way:

• $z_1$ cannot be greater, then $x_1$;

• as $z_1$ is increasing $z_{k+1},\ldots ,z_n$ is decreasing but $z_i$ cannot be less, than $x_i$ if $k+1\le i\le n$, say, first we decrease $z_n$ until $z_n=x_n$ then we decrease $z_{n-1}$ until $z_{n-1}=x_{n-1}$ and so on;

• if for some $0<z_1\le x_1$ we have that $||\mathbf p^*-\bar{\mathbf h}(\mathbf z)||=||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||$, then we stop;

• if for every $0<z_1\le x_1$ we have that $||\mathbf p^*-\bar{\mathbf h}(\mathbf z)||>||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||$, then we set $z_1$ to be $x_1$ and start to increase $z_2$ repeating the process replacing the index 1 with index 2;

• if we do not find a $\mathbf z$ with $||\mathbf p^*-\bar{\mathbf h}(\mathbf z)||=|| \mathbf p^*-\bar{\mathbf h}(\mathbf y)||$ by moving $z_2$, then we set $z_2$ to be $x_2$ and start to increase $z_3$ and so on.

As $\bar{\mathbf h}$ is continuous in $\mathbf z$ and

$$\begin{aligned} ||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||\le ||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||<||\mathbf p^*-\bar{\mathbf h}(\mathbf z^0)|| \end{aligned}$$

we must find a $\mathbf z\in S_n$ such that (i) $z_i\le x_i\le y_i$ if $1\le i \le k$ but $\sum _{i=1}^kz_i<\sum _{i=1}^ky_i$, (ii) $y_i\le x_i\le z_i$ if $k+1\le i\le n$ but $\sum _{i=k+1}^ny_i<\sum _{i=k+1}^nz_i$, (iii) $||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||=||\mathbf p^*-\bar{\mathbf h}(\mathbf z)||$, (iv) $p_2^*<{\bar{h}}_2(\mathbf z)$. Since $p_2^*<{\bar{h}}_2(\mathbf y)$ also holds $\bar{\mathbf h}(\mathbf z)$ must be equal to $\bar{\mathbf h}(\mathbf y)$. As ${\bar{\varrho }}$ is the solution in [0, 1] of the equation ${\bar{\varrho }}=1/(1+\bar{\mathbf h}T{\bar{\varrho }} \bar{\mathbf h})$ we also have

$$\begin{aligned} {\bar{\varrho }}(\mathbf z)=\frac{-1+\sqrt{1+4\bar{\mathbf h}(\mathbf z)T \bar{\mathbf h}(\mathbf z)}}{2\bar{\mathbf h}(\mathbf z)T\bar{\mathbf h}(\mathbf z)}= \frac{-1+\sqrt{1+4\bar{\mathbf h}(\mathbf y)T\bar{\mathbf h}(\mathbf y)}}{2\bar{\mathbf h}(\mathbf z)T\bar{\mathbf h}(\mathbf z)} ={\bar{\varrho }}(\mathbf y). \end{aligned}$$

Hence

$$\begin{aligned} \varrho _i(\mathbf z)=\frac{1}{1+\mathbf p_i T{\bar{\varrho }}(\mathbf z) \bar{\mathbf h}(\mathbf z)}= \frac{1}{1+\mathbf p_i T{\bar{\varrho }}(\mathbf y) \bar{\mathbf h}(\mathbf y)}= \varrho _i(\mathbf y). \end{aligned}$$

Consequently, we get that

$$\begin{aligned} 0&={\bar{h}}_2(\mathbf z)-{\bar{h}}_2(\mathbf y)=\sum _{i=1}^n z_i\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}p_i^2-\sum _{i=1}^n y_i\frac{\varrho _i(\mathbf y)}{{\bar{\varrho }}(\mathbf y)}p_i^2=\sum _{i=1}^n (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}p_i^2\nonumber \\&\ge p_{k+1}^2\underbrace{\sum _{i=k+1}^n (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}}_{> 0}+p_{k}^2 \underbrace{\sum _{i=1}^k (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}}_{< 0}\nonumber \\&=[p_{k}^2+(p_{k+1}^2-p_{k}^2)]\sum _{i=k+1}^n (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}+p_{k}^2 \sum _{i=1}^k (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}\nonumber \\&=(p_{k+1}^2-p_{k}^2) \sum _{i=k+1}^n (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}+p_{k}^2 \sum _{i=1}^n (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}. \end{aligned}$$

(A.19)

Since $\sum _{i=1}^nz_i\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}= \sum _{i=1}^ny_i\frac{\varrho _i(\mathbf y)}{{\bar{\varrho }}(\mathbf y)}=1$, ${\bar{\varrho }}(\mathbf y)={\bar{\varrho }}(\mathbf z)$ and $\varrho _i(\mathbf y)=\varrho _i(\mathbf z)$ ($i=1,\ldots ,n$) it follows that $\sum _{i=1}^n(z_i-y_i) \frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}=0$ and (A.19) can be continued as

$$\begin{aligned} =(p_{k+1}^2-p_{k}^2) \sum _{i=k+1}^n (z_i-y_i)\frac{\varrho _i(\mathbf z)}{{\bar{\varrho }}(\mathbf z)}>0 \end{aligned}$$

which is a contradiction. This validates that $||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||>||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||$ providing that $x_1\le y_1,\ldots ,x_k\le y_k$, $\sum _{i=1}^kx_i<\sum _{i=1}^ky_i$, $y_{k+1}\le x_{k+1}, \ldots ,y_n\le x_n$, $\sum _{i=k+1}^ny_i<\sum _{i=k+1}^nx_i$, $p_2^*<{\bar{h}}_2(\mathbf x)$, $p_2^*<{\bar{h}}_2(\mathbf y)$, $0<||\mathbf p^*-\bar{\mathbf h}(\mathbf x)||<\delta '$ and $0<||\mathbf p^*-\bar{\mathbf h}(\mathbf y)||<\delta '$.

Conclusion

In summary, if $0<||\bar{\mathbf h}(\mathbf x)-\mathbf p^*||<\delta '$ and $p_2^*<\bar{h}_2(\mathbf x)$ it follows that for a positive $x_i$ the expression $x_i[W_i(\mathbf x)-{\bar{W}}(\mathbf x)]$ is strictly positive or strictly negative according as $1\le i\le k$ or $k+1 \le i \le n$ (Observation 1). This means that if $x_i>0$ then $x_i$ strictly increases for $1\le i \le k$ and $x_i$ strictly decreases for $k+1\le i\le n$, respectively, which, by Observation 2, implies that ${\bar{h}}_2(\mathbf x)$ has to strictly decrease until reaching $p_2^*$. If ${\bar{h}}_2(\mathbf x)<p_2^*$ then a similar argument shows that ${\bar{h}}_2(\mathbf x)$ tends to $p_2^*$ strictly increasingly. Consequently, $\bar{\mathbf h}(\mathbf x)\rightarrow \mathbf p^*$ in a monotone way. $\square$

Proof of Theorem 3.4

We trace back the proof to Garay et al. (2018, Theorem 4.2) which says that $\mathbf p^*$ is UESS providing that the state $\mathbf y \in S_2$ with

$$\begin{aligned} y_1\frac{\varrho _{\mathbf e_1}(\mathbf y,\mathbf e_1,\mathbf e_2)}{\bar{\varrho }(\mathbf y,\mathbf e_1,\mathbf e_2)}\mathbf e_1+y_2\frac{\varrho _{\mathbf e_2}(\mathbf y,\mathbf e_1,\mathbf e_2)}{\bar{\varrho }(\mathbf y,\mathbf e_1,\mathbf e_2)}\mathbf e_2=\mathbf p^* \end{aligned}$$

is a locally asymptotically stable rest point of the replicator dynamics with respect to $\mathbf e_1,\mathbf e_2\in S_2$. (Here $\varrho _{\mathbf e_1}=\varrho _{\mathbf e_1}(\mathbf y,\mathbf e_1,\mathbf e_2)$ and $\varrho _{\mathbf e_2}=\varrho _{\mathbf e_2}(\mathbf y,\mathbf e_1,\mathbf e_2)$ are the solution of the equation system

$$\begin{aligned} \varrho _{\mathbf e_i}=\frac{1}{1+\mathbf e_i T(y_1\varrho _{\mathbf e_1}\mathbf e_1+y_2\varrho _{\mathbf e_2}\mathbf e_2)}, \quad i=1,2 \end{aligned}$$

and ${\bar{\varrho }}=y_1\varrho _{\mathbf e_1}+y_2\varrho _{\mathbf e_2}.$)

It is not too difficult to see that this statement remains true if we replace $\mathbf e_1,\mathbf e_2$ with strategies $\mathbf q_1\not =\mathbf q_2$ with $q_1^2<p^*_2<q_2^2$ where $\mathbf q_i=(q_i^1,q_i^2)=q_i^1\mathbf e_1+q_i^2\mathbf e_2$ and $\mathbf p^*=(p_1^*,p_2^*)=p_1^*\mathbf e_1+p_2^*\mathbf e_2$.

It can be assumed that $p_1^2<p^*_2<p_2^2$ or change the order of $\mathbf p_1,\ldots ,\mathbf p_n$. For the replicator dynamics with respect to $\mathbf p_1,\ldots ,\mathbf p_n$, consider the initial value problem with $x_i(0)=0$ for $i=3,4,\ldots ,n$ and with $x_1(0),x_2(0)$ such that $||\bar{\mathbf h}\big (\mathbf x(0)\big )-\mathbf p^*||<\delta$ where $\delta$ comes from the remark about the local asymptotic stability of set G after Theorem 3.3. Because of the uniqueness of solutions of differential equations with continuously differentiable right-hand side [see e.g. p. 144 in Hirsch et al. (2004)] it follows that $x_1(t)=y_1(t)$, $x_2(t)=y_2(t)$ and $x_i(t)=0$ ($t\ge 0$, $i=2,3,\ldots ,n$) where $y_1(t),y_2(t)$ is the solution of the initial value problem $y_1(0)=x_1(0),$$y_2(0)=x_2(0)$ for the replicator dynamics with respect to $\mathbf p_1,\mathbf p_2$. To finish the proof one should only apply (Garay et al. 2018, Theorem 4.2) for the replicator dynamics with respect to $\mathbf p_1,\mathbf p_2$. $\square$

1.3 Proof of the statements of Sect. 4

Proof of Theorem 4.1

By Remark 2.4, we can assume that $\mathbf p=\mathbf e_1$. Let F denote the surface of $S_N$ determined by the vertices $\mathbf e_2,\ldots ,\mathbf e_N$ that is $F=\{\mathbf q\in S_N\,:\,q_1=0\}$. By Definition 2.2, we have that

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\mathbf q,0)-\omega _{\mathbf q}(\mathbf p,\mathbf q,0)>0 \end{aligned}$$

for every $\mathbf q \not =\mathbf p$, in particular, for every $\mathbf q\in F$. Since $F\times \{0\}$ is a compact set from the continuity of $\omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )-\omega _{\mathbf q} (\mathbf p,\mathbf q,\varepsilon )$ in $(\mathbf q,\varepsilon )$ we infer the existence of an $\varepsilon _0>0$ such that

$$\begin{aligned} \omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )-\omega _{\hat{\mathbf q}} (\mathbf p,\hat{\mathbf q},\eta )>0 \end{aligned}$$

(A.20)

for any $\hat{\mathbf q}\in F$ and $0\le \eta \le \varepsilon _0.$

We prove that Definition 2.1 holds with this $\varepsilon _0$. To see this let $\mathbf q$ be an arbitrary strategy distinct from $\mathbf p$ and $0\le \varepsilon \le \varepsilon _0$. Take the strategy $\hat{\mathbf q}\in F$ such that $\mathbf q$ lies on the segment between $\mathbf p$ and $\hat{\mathbf q}$. Then, using Lemma A.5 as in the proof of Theorem 3.1, we infer that there is an $\eta =\eta (\varepsilon )\le \varepsilon$ such that

$$\begin{aligned}&\frac{1}{{\bar{\rho }}(\mathbf p,\mathbf q,\varepsilon )} [(1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )\mathbf p+ \varepsilon \rho _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon ) \mathbf q]\\&\quad =\frac{1}{{\bar{\rho }}(\mathbf p,\hat{\mathbf q},\eta )} [(1-\eta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )\mathbf p+ \eta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta ) \hat{\mathbf q}] \end{aligned}$$

where ${\bar{\rho }}(\mathbf p,\mathbf q,\varepsilon )= (1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )+ \varepsilon \rho _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon )$. Note that, as in the necessity part of the proof of Theorem 3.1, this implies that ${\bar{\rho }}(\mathbf p,\mathbf q,\varepsilon )= {\bar{\rho }}(\mathbf p,\hat{\mathbf q},\eta )$, $\rho _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )= \rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )$ and $\omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )= \omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )$, respectively.

Therefore we have

$$\begin{aligned}&(1-\varepsilon )\omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )+ \varepsilon \omega _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon )\\&\quad =[(1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )\mathbf p+ \varepsilon \rho _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon )\mathbf q]A [ (1-\varepsilon )\rho _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon )\mathbf p+ \varepsilon \rho _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon )\mathbf q]\\&\quad = [(1-\eta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )\mathbf p+ \eta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta ) \hat{\mathbf q}]A [(1-\eta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )\mathbf p+ \eta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta ) \hat{\mathbf q}]\\&\quad =(1-\eta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )\mathbf pA [(1-\eta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )\mathbf p+ \eta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta ) \hat{\mathbf q}]\\&\qquad +\, \eta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta )\hat{\mathbf q}A [(1-\eta )\rho _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )\mathbf p+ \eta \rho _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta ) \hat{\mathbf q}]\\&\quad = (1-\eta )\omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )+ \eta \omega _{\hat{\mathbf q}}(\mathbf p,\hat{\mathbf q},\eta ). \end{aligned}$$

By (A.20), it can be continued as

$$\begin{aligned} <\omega _{\mathbf p}(\mathbf p,\hat{\mathbf q},\eta )= \omega _{\mathbf p}(\mathbf p,\mathbf q,\varepsilon ), \end{aligned}$$

so by comparing the rightmost side with the leftmost one we get that $\omega _{\mathbf q}(\mathbf p,\mathbf q,\varepsilon )<\omega _{\mathbf p}(\mathbf p,\mathbf q, \varepsilon )$ for every $\mathbf q\not =\mathbf p$ and $0\le \varepsilon \le \varepsilon _0$ which proves that $\mathbf p$ is a UESS. $\square$

Proof of Lemma 4.3

$\mathbf p^*$ being a UESS ensures the existence of an $\varepsilon _0>0$ such that

$$\begin{aligned}&\frac{\mathbf p^* A [(1-\varepsilon )\rho _{\mathbf p^*}(\mathbf p^*,\mathbf q, \varepsilon )\mathbf p^*+ \varepsilon \rho _\mathbf q(\mathbf p^*,\mathbf q,\varepsilon )\mathbf q]}{1+\mathbf p^* T [(1-\varepsilon )\rho _{\mathbf p^*}(\mathbf p^*,\mathbf q,\varepsilon )\mathbf p^*+ \varepsilon \rho _\mathbf q(\mathbf p^*,\mathbf q,\varepsilon )\mathbf q]}\nonumber \\&\quad >\frac{\mathbf q A [(1-\varepsilon ) \rho _{\mathbf p^*}(\mathbf p^*,\mathbf q,\varepsilon )\mathbf p^*+ \varepsilon \rho _\mathbf q(\mathbf p^*,\mathbf q,\varepsilon )\mathbf q]}{1+\mathbf q T [(1-\varepsilon )\rho _{\mathbf p^*}(\mathbf p^*,\mathbf q,\varepsilon )\mathbf p^*+ \varepsilon \rho _\mathbf q(\mathbf p^*,\mathbf q,\varepsilon )\mathbf q]} \end{aligned}$$

(A.21)

for every $\mathbf q\not = \mathbf p^*$ and every $0<\varepsilon \le \varepsilon _0$. Let $0<x\le \varepsilon _0$ and $\mathbf x:=(x^*,x_1,\ldots ,x_n)\in [0,1]^{n+1}$ such that $x_1+\cdots +x_n=x$ and $x^*=1-x$. Consider $\breve{\varrho }(\mathbf x)$, $\breve{\mathbf h}(\mathbf x)$ and $\breve{W}(\mathbf x)$, respectively, as in Propositions 2.6. It is clear that $\breve{\mathbf h}(\mathbf x)$ is a convex combination of $\mathbf p_1,\ldots ,\mathbf p_n$, so $\breve{\mathbf h}(\mathbf x)\not =\mathbf p^*$. By Proposition 2.6,

$$\begin{aligned} (1-x)\varrho ^*(\mathbf x)\mathbf p^*+_{k=1}^nx_k \varrho _k(\mathbf x)\mathbf p_k= (1-x)\varrho ^*(\mathbf x)\mathbf p^*+x\breve{\varrho }(\mathbf x) \breve{\mathbf h}(\mathbf x). \end{aligned}$$

(A.22)

If we set $\varepsilon =x$ in (A.21), (A.22) shows that

$$\begin{aligned} W^*(\mathbf x)&= \frac{\mathbf p^* A [(1-x)\varrho ^*(\mathbf x)\mathbf p^*+ x\breve{\varrho }(\mathbf x)\breve{\mathbf h}(\mathbf x)]}{1+\mathbf p^* T[(1-x)\varrho ^*(\mathbf x)\mathbf p^*+ x\breve{\varrho }(\mathbf x)\breve{\mathbf h}(\mathbf x)]}\\&> \frac{\breve{\mathbf h}(\mathbf x) A [(1-x)\varrho ^*(\mathbf x)\mathbf p^*+ x\breve{\varrho }(\mathbf x)\breve{\mathbf h}(\mathbf x)]}{1+\breve{\mathbf h}(\mathbf x) T[(1-x)\varrho ^*(\mathbf x)\mathbf p^*+ x\breve{\varrho }(\mathbf x)\breve{\mathbf h}(\mathbf x)]}=\breve{W}(\mathbf x). \end{aligned}$$

Hence $W^*(\mathbf x)>(1-x)W^*(\mathbf x)+x\breve{W}(\mathbf x)={\bar{W}}(\mathbf x)$ whenever $0<x<\varepsilon _0$, that is, $\varepsilon _0$ is a suitable choice for $\delta$. $\square$

Proof of Corollary 4.5

Let $\varepsilon _0$ be the same as in the previous proof. Assume that $0<||(1,0,\ldots ,0)-\mathbf x||<\varepsilon _0/n$. Then

$$\begin{aligned} x_1+\cdots +x_n\le n\sqrt{x_1^2+\cdots +x_n^2}&< n\sqrt{(1-x)^2+x_1^2+\cdots +x_n^2}\\&\quad = n||(1,0,\ldots ,0)-\mathbf x||\le \varepsilon _0. \end{aligned}$$

By the previous lemma, therefore, it is true that

$$\begin{aligned} (1,0,\ldots ,0)\cdot \big (W^*(\mathbf x),W_1(\mathbf x),\ldots ,W_n(\mathbf x)\big ) =W^*(\mathbf x)>{\bar{W}}(\mathbf x)=\mathbf x\mathbf W(\mathbf x) \end{aligned}$$

whenever $0<||(1,0,\ldots ,0)-\mathbf x||<\varepsilon _0/n$. $\square$

Proof of Lemma 4.6

• (a) See in Garay et al. (2018).

• (b) Note that, by Proposition 2.6 and 2.8, $\bar{\mathbf h}(\hat{\mathbf x})=\mathbf p^*$ and

  $$\begin{aligned} {\bar{W}}(\hat{\mathbf x})\big [=\breve{W}(\hat{\mathbf x})\big ]=\omega (\mathbf p^*)= \frac{\mathbf p^* A \mathbf p^* \rho (\mathbf p^*)}{1+\mathbf p^* T \mathbf p^* \rho (\mathbf p^*)}. \end{aligned}$$

  Since $\hat{\mathbf x}$ is an interior rest point the right hand side of the replicator dynamics can vanish if and only if $W_i(\hat{\mathbf x})={\bar{W}}(\mathbf x)$ for every i which, by Lemma 2.3, immediately implies that $\mathbf p^*$ is a NE.

• (c)–(d) Since $\hat{\mathbf x}$ is also a rest point it follows that if $\hat{x}_i>0$ then

  $$\begin{aligned} W_i(\hat{\mathbf x})={\bar{W}}(\hat{\mathbf x})= \frac{\mathbf p^* A \mathbf p^* \rho (\mathbf p^*)}{1+\mathbf p^* T \mathbf p^* \rho (\mathbf p^*)} \end{aligned}$$

  must hold. Therefore, if, contrary to our claim, $\mathbf p^*$ is not a NE then the equilibrium condition is hurt only for some i with $\hat{x}_i=0$. For such an index i, say $i_0$, there is an $\varepsilon >0$ such that

  $$\begin{aligned} \underbrace{\frac{\mathbf p_{i_0} A \mathbf p^* \rho (\mathbf p^*)}{1+\mathbf p_{i_0}T \mathbf p^* \rho (\mathbf p^*)}}_{W_{i_0}(\hat{\mathbf x})}- \underbrace{\frac{\mathbf p^* A \mathbf p^* \rho (\mathbf p^*)}{1+\mathbf p^* T \mathbf p^* \rho (\mathbf p^*)}}_{{\bar{W}}(\hat{\mathbf x})}= \varepsilon >0. \end{aligned}$$

  So, by continuity, $W_{i_0}(\mathbf x)-{\bar{W}}(\mathbf x)>\varepsilon /2$ in a bounded neighbourhood ${\mathbb {H}}$ of $\hat{\mathbf x}$. Henceforth, the way of the proofs of c) and d) branch off:

  • (c) Since $\hat{\mathbf x}$ is stable there is another neighbourhood H of $\hat{\mathbf x}$ such that for any solution starting from H remains in ${\mathbb {H}}$ forever. Take an arbitrary $\mathbf x\in H$ with $x_{i_0}\not =0$ and consider this $\mathbf x$ as an initial value. Then

    $$\begin{aligned} \log x_{i_0}(t)&=\log x_{i_0}(0)+ \int _0^t\frac{\dot{x}_{i_0}(s)}{x_{i_0}(s)}\,\mathrm {d}s\\&= \log x_{i_0}(0)+ \int _0^tW_{i_0}(\mathbf x(s))-{\bar{W}}(\mathbf x(s))\,\mathrm {d}s\\&>\log x_{i_0}(0)+\int _0^t\frac{\varepsilon }{2}\,\mathrm {d}s= \log x_{i_0}(0)+t\frac{\varepsilon }{2}\rightarrow \infty \quad \text {as}\quad t\rightarrow \infty \end{aligned}$$

    contradicting that $\mathbf x(t)$ remains in ${\mathbb {H}}$.

  • (d) Since $\hat{\mathbf x}$ is the only member of the $\omega$-limit of $\mathbf x(t)$ and $\mathbf x(t)$ is bounded it follows that $\mathbf x(t)\rightarrow \hat{\mathbf x}$, in particular, $x_{i_0}(t)\rightarrow {\hat{x}}_{i_0}=0$. Hence, there is a $t_0$ such that $\mathbf x(t)\in {\mathbb {H}}$ whenever $t\ge t_0$ from which we infer that $\log x_{i_0}(t)\rightarrow \infty$ as in the proof of c) which contradicts $x_{i_0}(t)$ tending to 0.

$\square$

Proof of Theorem 4.7

By Corollary 4.5 the vector $(1,0,\ldots ,0)\in S_{n+1}$ is a PSS of the replicator dynamics which is an asymptotically stable rest point of the replicator dynamics in accordance with the remark immediate after Definition 4.2. $\square$

In the next proof we again apply Theorem 3.2.

Proof of Theorem 4.9

We use the notation of Proposition 2.6. Since Proposition 2.8 says that $(1,0,\ldots ,0)\in S_{n+1}$ is a rest point (moreover every point $\mathbf x$ for which $\bar{\mathbf h}(\mathbf x)=\mathbf p^*$ is a rest point) we should only prove the stability of $(1,0,\ldots ,0)$. By Lemma A.4, $\bar{\mathbf h}(\mathbf x)$ is continuous in $\mathbf x=(x^*,x_1,\ldots ,x_n)$. On the other hand, $\mathbf p^*$ is a UESS. Therefore, $\omega _{\mathbf p^*}\big (\mathbf p^*,\bar{\mathbf h}(\mathbf x),1\big ) \ge \omega _{\bar{\mathbf h}(\mathbf x)}\big (\mathbf p^*,\bar{\mathbf h}(\mathbf x),1\big )$ in a neighbourhood of $(1,0,\ldots ,0)$ (see Theorem 3.2). (The strict inequality is not sure at all because $\bar{\mathbf h}(\mathbf x) =\mathbf p^*$ can occur!) By Proposition 2.6, $\omega _{\mathbf p^*}\big (\mathbf p^*,\bar{\mathbf h}(\mathbf x),1\big )=W^*(\mathbf x)$ and $\omega _{\bar{\mathbf h}(\mathbf x)}\big (\mathbf p^*,\bar{\mathbf h}(\mathbf x),1\big )={\bar{W}}(\mathbf x)$. Therefore, $W^*(\mathbf x)-{\bar{W}}(\mathbf x)\ge 0$ holds at least in a neighbourhood of $(1,0,\ldots ,0)$.

From here, follow the proof of Hofbauer and Sigmund (1998, Theorem 7.2.4). Take the function $P(\mathbf x)=-\log x^*$ which is positive semidefinite (that is $P(\mathbf x)\ge 0$) in a neighbourhood of $(1,0,\ldots ,0)$ in $S_{n+1}$. As regards the derivative of $P(\mathbf x)$ along the replicator dynamics we have

$$\begin{aligned} \dot{P}(\mathbf x)&=\left( \frac{\partial }{\partial x^*}P(\mathbf x) \right) x^*[W^*(\mathbf x)-{\bar{W}}(\mathbf x)]+\sum _{k=1}^{n} \underbrace{\left( \frac{\partial }{\partial x_k}P(\mathbf x) \right) }_{=0} x_k[W_k(\mathbf x)-{\bar{W}}(\mathbf x)]\\&=-[W^*(\mathbf x)-{\bar{W}}(\mathbf x)]\le 0, \end{aligned}$$

that is, $\dot{P}(\mathbf x)$ is negative semidefinite in a neighbourhood of $(1,0,\ldots ,0)$. By Lyapunov’s theorem on stability [see for example (Hirsch et al. 2004, p. 194) or (Hofbauer and Sigmund 1998, Theorem 2.6.1)], this ensures the local stability of $(1,0,\ldots ,0)$ in the simplex $S_{n+1}$. $\square$