In this section we first provide a class of functionals on the simplex that are Lyapunov functions for a d.s.q.o. Then we prove the convergence theorem for d.s.q.o.

A continuous functional given *φ* : *S*^{m}^{–1} → *R* is said to be a Lyapunov function for the operator *V* if the limit lim_{n →} _{∞}*φ*(*V*^{n}(*x*^{0})) exists along the trajectory {*x*^{0}, *V*(*x*^{0}), *V*^{2}(*x*^{0}), …, *V*^{n}(*x*^{0}), …}.

The Lyapunov function is considered to be very useful in the study of limit behavior of (discrete) dynamical systems.

*A continuous functional given by*

$$\phi (x)={\displaystyle \sum _{i=1}^{m}{x}_{i}^{2}}$$

*is a Lyapunov function for doubly stochastic operator V. Moreover*, *if x* ∈ *S*^{m}^{–1} *and Vx is not the permutation of x*, *then φ*(*Vx*) < *φ*(*x*).

*Proof*. Use the fact (see [11], prop. F.1, page 78.) that the function $\psi (x)={\displaystyle \sum _{i<j}{x}_{i}{x}_{j}}$, defined on the simplex, is Schur-concave, i.e. satisfies *ψ*(*x*) ≥ *ψ*(*y*) whenever *x* ≺ *y*. Moreover, if *x*, *y* ∈ *S*^{m – 1}, and *x* is not the permutation of *y*, then *ψ*(*x*) > *ψ*(*y*). Taking into consideration *φ*(*x*) = 1 – 2*ψ*(*x*), we obtain that x ≺ *y* implies *φ*(*x*) ≤ *φ*(*y*). Since *V* is doubly stochastic, then *Vx* ≺ *x* hence *φ*(*Vx*) ≺ *φ*(*x*) is bounded. Note that the sequence *a*_{n} = *φ*(*V*^{n}(*x*^{0})), where *V* is doubly stochastic and *x*^{0} ∈ *S*^{m}^{–1}, is monotone. Hence the lim_{n→∞} *φ*(*V*^{n}(*x*^{0})) exists.

Now we show the second part. If *x* ∈ *intS*^{m}^{–1}, then according to Lemma it follows that *V*(*x*) ∈ *intS*^{m}^{–1}. Since all components of *V*(*x*) and *x* are positive and *Vx* is not the permutation of *x*, then by applying the above mentioned fact, we get *φ*(*x*) < *φ*(*y*). □

Based on the above method one can provide larger class of Lyapunov functions.

*Any continuous symmetric convex (or concave) functional is a Lyapunov function for a doubly stochastic operator V*.

These theorems are very useful to study the dynamics of individual doubly stochastic operators. We will provide an application of these theorems in the next section.

*Assume that a d.s.q.o. V does not have any periodic points on intS*^{m}^{–1}. *Then the trajectory of any initial point on the interior of simplex is convergent under any d.s.q.o*.

*Proof*. Let *V* : *S*^{m}^{–1} → *S*^{m}^{–1} be d.s.q.o. Then we have

$$x\succ Vx\succ {V}^{2}x\succ \cdots $$

It means that

$$\begin{array}{c}{x}_{[1]}\ge {\left(Vx\right)}_{[1]}\ge {\left({V}^{2}x\right)}_{[1]}\ge \cdots ,\\ {x}_{[1]}+{x}_{[2]}\ge {\left(Vx\right)}_{[1]}+{\left(Vx\right)}_{[2]}\ge {\left({V}^{2}x\right)}_{[1]}+{\left({V}^{2}x\right)}_{[2]}\ge \cdots ,\\ \cdots ,{\displaystyle \sum _{i=1}^{k}{x}_{[i]}}\ge {\displaystyle \sum _{i=1}^{k}{\left(Vx\right)}_{[i]}}\ge {\displaystyle \sum _{i=1}^{k}{\left({V}^{2}x\right)}_{[i]}}\ge \cdots \end{array}$$

The sequences $\left\{{\displaystyle {\sum}_{i=1}^{k}{\left({V}^{n}\left(x\right)\right)}_{\left[i\right]},n=1,2,\dots \forall k=\overline{1,m}}\right\}$ are increasing and bounded, consequently convergent.

The last suggests that the following sequences are also convergent
$$\left\{{\left({V}^{n}\left(x\right)\right)}_{\left[k\right]},n=1,2,\dots \forall k=\overline{1,m}\right\}$$

Let us denote *y*_{k} = lim_{n → ∞} .(*V *^{n}x)_{[k]} and *y* = (*y*_{1}, *y*_{2}, . . ., *y*_{m}).

If *z* = (*z*_{1}, *z*_{2}, . . ., *z*_{m}) ∈ *ω*(*x*^{0}) then there exists {*x*.^{(nj )}}, such that (*V* ^{nj} *x*) → *z*.

Therefore we have (*V*^{nj} x)_{↓} → *z*_{↓}. On the other hand (*V*^{nj} x)↓ → *y*, since *y* is a limit of the sequence (*V*^{n}x), *n* =1,2,... That’s why *z*_{↓} = (*y*_{1}, *y*_{2}, …, *y*_{m}) *= y*. This means that the cardinality of *ω*(*x*^{0}) cannot be greater than *m*!.

Note that since the simplex is compact, then *ω*(*x*^{0}) ≠ ∅. Now, assume that the cardinality of it is greater than 1. Then there is a point *z* in *ω*(*x*^{0}) that is not fixed by *V*. Because the set *ω*(*x*^{0}) is *V* – invariant then any iterate of *z* under must belong to *ω*(*x*^{0}). But this set is finite and therefore there exist natural numbers *p*, *q* (and let us assume *p* > *q*) such that *V*^{p}(*z*) = *V*^{q}(*z*). But then the point *V*^{q}(*z*) is periodic with period *p* – *q*. However, by the assumption *V* does not have any periodic points. Hence *m* must be 1 and |*ω*(*x*^{0})| = 1. Therefore, *ω*(*x*^{0}) may consist of a single point only and this implies that the trajectory starting at *x*^{0} tends to that point. □

An operator *V* : *S*^{m-1} → *S*^{m–1} is a permutation if there exists a permutation matrix such that *Vx = Px* ∀*x* ∈ *S*^{m–}^{1}. It is evident that there are exactly *m*! permutation operators on *S*^{m−1}.

*Any non-permutation d.s.q.o. V* : *S*^{2} → *S*^{2} *on* 2*D simplex does not have periodic points on the interior of the simplex*.

*Proof*. Assume *V* has a periodic point *x*^{0} ∈ *intS*^{2} and *V*^{p}(*x*^{0}) = *x*^{0} for *p* > 1. Then
$$V\succ V\left({x}^{0}\right)\succ \dots \succ {V}^{p}\left({x}^{0}\right)={x}^{0}$$

Hence *V*(*x*^{0}) is some permutation of *x*^{0}.

Let *x*^{0} = (*x*, *y*, *z*), *x*, *y*, *z* > 0, *x* + *y* + *z* = 1 and
$$V\left({x}^{0}\right)=\left(\left({A}_{1}{x}^{0},{x}^{0}\right),\left({A}_{2}{x}^{0},{x}^{0}\right),\left({A}_{3}{x}^{0},{x}^{0}\right)\right)$$

where (·,·) denotes the usual inner product. Without loss of generality we may assume *x* ≥ *y* ≥ *z* and (*A*_{1}*x*^{0}, *x*^{0}) ≥ (*A*_{2}*x*^{0}, *x*^{0}) ≥ (*A*_{3}*x*^{0}, *x*^{0}). Then V(*x*^{0}) = *x*^{0} (not some permutation of *x*^{0}). Because *x* + *y* + *z* = 1 then one can assume *x* ≥ *y* ≥ *z* and one can also assume the second assumption by changing matrices *A*_{1}, *A*_{2}, *A*_{3} if necessary. By applying Theorems 2.6 and 2.4 of [8] we can represent each *A*_{k}, *k* = 1, 2, 3 as
$${A}_{k}=\frac{{S}_{k}+{S}_{k}^{T}}{2}$$

where *S*_{k} is a row-stochastic matrix and ${S}_{k}^{T}$ is its transpose. Not that
$$\begin{array}{ll}\left({A}_{k}{x}^{0},{x}^{0}\right)\hfill & =\left(\frac{{S}_{k}+{S}_{k}^{T}}{2}{x}^{0},{x}^{0}\right)\hfill \\ \hfill & =\frac{1}{2}\left({S}_{k}{x}^{0},{x}^{0}\right)+\frac{1}{2}\left({S}_{k}^{T}{x}^{0},{x}^{0}\right)\hfill \\ \hfill & =\frac{1}{2}\left({S}_{k}{x}^{0},{x}^{0}\right)+\frac{1}{2}\left({x}^{0},{S}_{k}{x}^{0}\right)\hfill \\ \hfill & =\left({S}_{k}{x}^{0},{x}^{0}\right)\hfill \end{array}$$

Hence *V*(*x*^{0}) = ((*S*_{1}*x*^{0}, *x*^{0}), (*S*_{2}*x*^{0}, *x*^{0}), (*S*_{3}*x*^{0}, *x*^{0})) = (*x*, *y*, *z*).

Let *S*_{k} = {*s*_{ij, k})_{i, j = 1, 3}. Since *V*(*x*^{0}) = *x*^{0} then (*S*_{1}*x*^{0}, *x*^{0}) = *x*. On the other hand, since *S*_{1} is row-stochastic we must have
$${s}_{i1,1}x+{s}_{i2,1}y+{s}_{i3,1}z\le x$$(7)

Therefore
$$\begin{array}{ll}x\hfill & =\left({s}_{1}{x}^{0},{x}^{0}\right)\hfill \\ \hfill & =\left({s}_{11,1}x+{s}_{12,1}y+{s}_{13,1}z\right)x+\left({s}_{21,1}x+{s}_{22,1}y+{s}_{23,1}z\right)y+\left({s}_{31,1}x+{s}_{32,1}y+{s}_{33,1}z\right)z\hfill \\ \hfill & \le xx+xy+xz\hfill \\ \hfill & =x\left(x+y+z\right)\hfill \\ \hfill & =x,\hfill \end{array}$$

and we must have the equality in (7). Recall that ${x}^{0}=\left(x,y,z\right)\ne \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ as $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ is a fixed point for *V*. Hence the equality in (7) is possible only in the following cases:
$${S}_{1}=\left(\begin{array}{lll}1\hfill & 0\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill \end{array}\right)$$(8)

or
$$x=y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{1}=\left(\begin{array}{lll}{s}_{11,1}\hfill & {s}_{12,1}\hfill & 0\hfill \\ {s}_{21,1}\hfill & {s}_{22,1}\hfill & 0\hfill \\ {s}_{31,1}\hfill & {s}_{32,1}\hfill & 0\hfill \end{array}\right)$$(9)

Similar arguments for *S*_{3} = (*s*_{ij},_{3})_{i, j} = _{1, 3} show that
$${s}_{i1,3}x+{s}_{i2,3}y+{s}_{i3,3}z\ge z$$(10)

with equality is only possible in the following cases:
$${S}_{3}=\left(\begin{array}{lll}0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right)$$(11)

or
$$y=z,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{3}=\left(\begin{array}{lll}0\hfill & {s}_{12,3}\hfill & {s}_{13,3}\hfill \\ 0\hfill & {s}_{22,3}\hfill & {s}_{23,3}\hfill \\ 0\hfill & {s}_{32,3}\hfill & {s}_{33,3}\hfill \end{array}\right)$$(12)

Not that (9) and (12) can not hold simultaneously, as it would imply *x* = *y* = *z* hence ${x}^{0}=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$.

If (8) and (12) hold, then *V* would fix the first and the third components of *x*^{0} = (*x*, *y*, *z*) which implies that it would also fix the second component of *x*^{0} = (*x*, *y*, *z*). By the same reasons (9) and (11) can not hold simultaneously. Finally (8) and (11) together would imply that ${S}_{2}=\left(\begin{array}{lll}0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \end{array}\right)$ and hence *V* = *Id* which is not impossible as *V* is not a permutation operator. □

Consider *V* : *S*^{2} → *S*^{2} given by
$$\begin{array}{ll}V\left(x\right)\hfill & ={y}^{2}+2xz\hfill \\ V\left(y\right)\hfill & ={z}^{2}+2xy\hfill \\ V\left(z\right)\hfill & ={x}^{2}+2yz\hfill \end{array}$$

This operator is d.s.q.o. by the classification theorem 2.6 of [8] or one can easily check that *Vx* ≺ *x*. One can see that *V*^{3}(1, 0, 0) = (1, 0, 0), that is, (1, 0, 0) is a 3-periodic point.

Combining the previous two theorems we obtain a very interesting corollary.

*Any non-permutation d.s.q.o. has convergent trajectory on the interior of the simplex*.

Theorem 3.5 fails in higher dimension. Consider the following example in 3D simplex.

*Consider the operator V* : *S*^{3} → *S*^{3} *given by*
$$Vx=Dx=\left(\frac{{x}_{2}+{x}_{4}}{2},\frac{{x}_{1}+{x}_{3}}{2},\frac{{x}_{2}+{x}_{4}}{2},\frac{{x}_{1}+{x}_{3}}{2}\right)$$

*where*
$$D=\left(\begin{array}{llll}0\hfill & \frac{1}{2}\hfill & 0\hfill & \frac{1}{2}\hfill \\ \frac{1}{2}\hfill & 1\hfill & \frac{1}{2}\hfill & 0\hfill \\ 0\hfill & \frac{1}{2}\hfill & 0\hfill & \frac{1}{2}\hfill \\ \frac{1}{2}\hfill & 0\hfill & \frac{1}{2}\hfill & 0\hfill \end{array}\right)$$

*For this operator*, *one has*
$${V}^{2k+1}={V}^{1}=\left(\frac{{x}_{2}+{x}_{4}}{2},\frac{{x}_{1}+{x}_{3}}{2},\frac{{x}_{2}+{x}_{4}}{2},\frac{{x}_{1}+{x}_{3}}{2}\right)$$

*and*
$${V}^{2k}={V}^{2}=\left(\frac{{x}_{1}+{x}_{3}}{2},\frac{{x}_{2}+{x}_{4}}{2},\frac{{x}_{1}+{x}_{3}}{2},\frac{{x}_{2}+{x}_{4}}{2}\right)$$

*Therefore*, *any point is either periodic or becomes periodic after one iteration. Notice that this linear operator is given by a doubly stochastic matrix. Hence Vx* ≺ *x and so V is doubly stochastic. One can make this operator a d.s.q.o. by multiplying each component to x*_{1} + *x*_{2} + *x*_{3} + *x*_{4} = 1.

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