Explicit examples in ergodic optimization

Abstract

Denote by T the transformation $T(x)= 2 \,x$ (mod 1). Given a potential $A:S^1 \rightarrow {\mathbb{R}}$ we are interested in exhibiting in several examples the explicit expression for the calibrated subaction $V: S^1 \rightarrow {\mathbb{R}}$ for A. The action of the 1/2 iterative procedure ${\mathcal{G}}$, acting on continuous functions $f: S^1 \rightarrow {\mathbb{R}}$, was analyzed in a companion paper. Given an initial condition $f_0$, the sequence, ${\mathcal{G}}^n(f_0)$ will converge to a subaction. The sharp numerical evidence obtained from this iteration allow us to guess explicit expressions for the subaction in several worked examples: among them for $A(x) = \sin ^2 ( 2 \pi x)$ and $A(x) = \sin ( 2 \pi x)$. Here, among other things, we present piecewise analytical expressions for several calibrated subactions. The iterative procedure can also be applied to the estimation of the joint spectral radius of matrices. We also analyze the iteration of ${\mathcal{G}}$ when the subaction is not unique. Moreover, we briefly present the version of the 1/2 iterative procedure for the estimation of the main eigenfunction of the Ruelle operator.

Appendix 1

1.1 The subaction equation in the case $A(x)=\sin ^2(2 \pi x)$

In this section we consider the case $A(x)=\sin ^2(2 \pi x)$ which was initially discussed on Sect. 5. We want to give more details on the proofs. We want to show first that $V(x)=\sup \,\{\,V_1(x),V_2(x)\,\}$ is a calibrated subaction for A, when $V_1$ and $V_2$ are described by (24). Remember that for all x we have $V_1(x)= V_2(1-x).$ Later we will present the power expansion for $V_2$ which will show can be described by (25).

Lemma 12

If $V_2(x)=\lim _{n \rightarrow +\infty }V_2^{n*}(x)$, then $V_2(x)=\sum _{i=0}^{N}\left( F\circ \eta ^i(x)-2{\hat{m}}(A)\right) + \epsilon _N(x)$, where $|\epsilon _N(x) | \le 2\pi \sum _{i=N}^{+\infty }\frac{1}{4^i}= \frac{2 \pi }{3 \cdot 4^{N-1}} \le \frac{2}{3 \cdot 4^{N-2}}.$

Proof

We just have to use the property that $\sin ^2$ has Lipchitz constant equal 2. $\square$

We want to show that $V_2$ indeed satisfies (22).

Lemma 13

If $V_2(x)=\lim _{n \rightarrow +\infty }V_2^{n*}(x)$, then

$$\begin{aligned} V_2(x)=V_2(\eta (x))+A\left( \frac{x}{2}\right) + A\left( \frac{x}{4}+\frac{1}{2}\right) -2\,{\hat{m}}(A). \end{aligned}$$

Proof

Denote $H(x)=A\left( \frac{x}{2}\right) + A\left( \frac{x}{4}+\frac{1}{2}\right) -2\,{\hat{m}}(A)$. Then, $V_2(x)=\sum _{i=0}^{+\infty }H(\eta ^i(x))$ and $V_2(\eta (x))= \sum _{i=1}^{+\infty }H(\eta ^i(x)).$ Therefore, $V_2(\eta (x))=\sum _{i=0}^{+\infty }(H(\eta ^i(x)) -H(x).$ From this follows $V_2(\eta (x))=V_2(x)- H(x) ,$ and, finally $V_2(x)=V_2(\eta (x))+A(\frac{x}{2})+ A(\frac{x}{4}+\frac{1}{2})-2{\hat{m}}(A)$. $\square$

Lemma 14

If $V_2(x)=\lim _{n \rightarrow +\infty }V_2^{n*}(x)$ and ${\hat{m}}(A)=\frac{A(1/3)+A(2/3)}{2}$, then the function $V_1(x)=V_2((x+1)/2)+A((x+1)/2)-{\hat{m}}(A)$ satisfies $V_1(x/2)+A(x/2)=V_2(x)+{\hat{m}}(A).$

Proof

From the relation between $V_1$ and $V_2$ we have $V_2((x+1)/2)+A((x+1)/2)=V_1(x)+{\hat{m}}(A).$ Taking composition with $\tau _1(x)=x/2$ we get

$$\begin{aligned} V_1(x/2)+A(x/2)= & {} V_2(x/4 + 1/2) +A(x/4 +1/2) + A(x/2) -{\hat{m}}(A) \nonumber \\= & {} V_2(\eta (x))+A(x/2) +A(x/4 +1/2) -{\hat{m}}(A). \end{aligned}$$

(41)

From Lemma 13 we obtain $V_2(\eta (x))-V_2(x)=2{\hat{m}}(A)-(A(x/2) +A(x/4 +1/2)),$ therefore, adding and subtrating $V_2(x)$ in (41) we have

$$\begin{aligned} V_1(x/2)+A(x/2)= & {} V_2(\eta (x))-V_2(x)+V_2(x)+A(x/2) +A(x/4 +1/2) -{\hat{m}}(A)\\ {}= & {} 2{\hat{m}}(A)-(A(x/2)+A(x/4+1/2))+V_2(x)+A(x/2) +A(x/4 +1/2) -{\hat{m}}(A). \end{aligned}$$

Finally, $V_1(x/2)+A(x/2)=V_2(x)+{\hat{m}}(A).$ $\square$

Now we need some differentiability results for $V_1$ e $V_2$.

Proposition 15

$V_2(x)$ is differentiable in [0, 1] and $V_2'(x)=$ $\sum _{i=0}^{+\infty } 2 \pi (\eta ^i)'(x)\left( \sin \left( \pi \eta ^i(x) \right) \cos (\pi \eta ^i (x)) +\frac{1}{2}\sin \left( \frac{\pi \eta ^i(x)}{2} \right) \cos \left( \frac{\pi \eta ^i (x)}{2} \right) \right) .$

We leave the proof for the reader.

From the last proposition we get

$$\begin{aligned} V_2^{\prime }(x)=\sum _{i=0}^{+\infty } 2 \pi \frac{1}{4^i}\left( \sin \left( \pi \eta ^i(x) \right) \cos (\pi \eta ^i (x)) +\frac{1}{2}\sin \left( \frac{\pi \eta ^i(x)}{2} \right) \cos \left( \frac{\pi \eta ^i (x)}{2} \right) \right) . \end{aligned}$$

Lemma 16

$V_2'(x)=\varphi _N(x)+ \xi _N(x)$, where $|\xi _N(x)|\le 3\pi \sum _{i=N}^{+\infty } |\frac{1}{4^i}|= \frac{\pi }{4^{N-1}}$, $\varphi _N(x)=\sum _{i=0}^{N} 2 \pi \frac{1}{4^i}\left( \sin \left( \pi \eta ^i(x) \right) \cos (\pi \eta ^i (x)) +\frac{1}{2}\sin \left( \frac{\pi \eta ^i(x)}{2} \right) \cos \left( \frac{\pi \eta ^i (x)}{2} \right) \right) .$

We leave the proof for the reader.

$I_E$ denotes the indicator function of the interval E.

Theorem 17

Taking $V_2(x)=\lim _{n \rightarrow +\infty }V_2^{n*}(x)$ and $V_1(x)=V_2((x+1)/2)+A((x+1)/2)-{\hat{m}}(A)$, we get that $V(x)=V_1(x)I_{[0,1/2)}(x)+ V_2(x) I_{[1/2,1]}(x).$ is a calibrated subaction for A, when ${\hat{m}}(A)=\frac{A(1/3)+A(2/3)}{2}=m(A)$.

Proof

We have to show that $\max _{T(y)=x}[A(y)+V(y)]=\max \{V_1(x/2)+A(x/2),V_2((x+1)/2) + A((x+1)/2) \}.$ As $V_1(u/2)+A(u/2)=V_2(u)+{\hat{m}}(A),$ and, $V_1(x)=V_2(1-x)$, then, we have to show that

$$\begin{aligned} \max _{T(y)=x}[A(y)+V(y)]= \max \{V_2(x)+{\hat{m}}(A), V_2(1-x)+{\hat{m}}(A) \} \end{aligned}$$

(42)

We will show first that if $u \in [0,1/2]$, then

$$\begin{aligned} V_2(u)+{\hat{m}}(A) \le V_2(1-u)+{\hat{m}}(A)=V_1(u)+{\hat{m}}(A) . \end{aligned}$$

Denote $\gamma (u)=V_2(u)-V_2(1-u)$. By Lemma 16 we get

$$\begin{aligned} \gamma '(u)= & {} V_2'(u)+V_2'(1-u)= \varphi _N(1-u)+\varphi _N(u) + (\xi _N(1-u)+\xi (u))\\\ge & {} \varphi _N(1-u)+\varphi _N(u) - 2\frac{\pi }{4^{N-1}}. \end{aligned}$$

Taking $N=4$ it is easy to se that if $u \in [0.1,0.9]$ then $\gamma '(u)> 0$. The function $\gamma$ is monotone increasing from 0.1 to 0.9 and $\gamma (1/2)=0$. Then $\gamma$ is negative on the interval [0.1, 0.5]. A similar argument can also handle the case $x\in [0,0.1]$. We use Lemma 12, the fact that $\gamma (u)=V_2(u)-V_2(1-u)$ and the control of the error $|\epsilon _N(x) |$. Then, finally we get that $\gamma$ is also negative in [0, 0.1] and is positive for $x\in [0.9,1]$. From the above we get $\max _{T(y)=u}[A(y)+V(y)]= V_2(1-u)+{\hat{m}}(A),\quad u \in [0,1/2]$ and $\max _{T(y)=u}[A(y)+V(y)]= V_2(u)+{\hat{m}}(A),\quad u \in [0,1/2].$ Therefore, for all $x \in [0,1]$ we get $\max _{T(y)=x}[A(y)+V(y)]= V(x)+ {\hat{m}}(A)$ Then, V is a calibrated subaction. $\square$

Now we will express $V_2$ in power series. Our final result will be given by expression (46). Using the property $\sin ^2(x)=\frac{1-\cos (2\pi x)}{2}$, we get

$$\begin{aligned} {V_2(x+2/3)=\frac{1}{2} \sum\limits _{i=0}^{+\infty }\left( \sin \left( \frac{4\pi }{3}\right) \sin \left( 2\pi \left( -\frac{1}{2}\right) ^ix\right) - \cos \left( \frac{4\pi }{3}\right) (\,\cos \left( 2\pi \left( -\frac{1}{2}\right) ^i\,x\,-1\,)\right) \right) }. \end{aligned}$$

Now, define

$$\begin{aligned} {M(x)=\frac{\sin (4\pi /3)}{2}\sum\limits_{i=0}^{+\infty }(\,\sin (2\pi (-1/2)^ix)\,-\, \sin (0)\,)} \end{aligned}$$

and

$$\begin{aligned} {}^{Q(x)= \frac{-\cos (4\pi /3)}{2}\sum _{i=0}^{+\infty }(\cos (2\pi (-1/2)^i\,x\,-\, \cos (0)).} \end{aligned}$$

We will express later $V_2$ as $V_2(x)=Q(x-2/3)+M(x-2/3).$

Lemma 18

M and Q are uniformly convergent in each interval $[-a,a]$.

Proof

As the function $\sin$ is Lipschitz, then, there is a constant C, such that,

$$\begin{aligned} |\sin (x)-\sin (y)|\le C|x-y| \le 2aC, \end{aligned}$$

and $\sum _{i=0}^{+\infty }\left| \sin \left( 2\pi \left( -\frac{1}{2} \right) ^i x\right) \right| \le \sum _{i=0}^{+\infty }2\,a\,C\,\left| 2\pi \left( -\frac{1}{2} \right) ^i\right| \le +\infty .$ For Q we use an analogous argument. $\square$

As $\cos (x)=\sum _{k=0}^{+\infty }\frac{(-1)^{k}x^{2k}}{(2k)!}$ one can write Q as

$$\begin{aligned} {Q(x)=\frac{-\cos (4\pi /3)}{2} \sum _{k=1}^{+\infty } \sum _{i=0}^{+\infty }\left( \frac{(-1)^k(2\pi x)^{2k}}{2^{2ik}(2k)!}\right) .} \end{aligned}$$

(43)

Finally, we get $Q(x)=\frac{-\cos (4\pi /3)}{2}\sum _{k=1}^{+\infty }\frac{(-1)^k(2\pi x)^{2k}}{(2k)!}\frac{2^{2k}}{2^{2k}-1}.$ Proceeding in analogous way we get $M(x)=\frac{\sin (4\pi /3)}{2}\sum _{k=0}^{+\infty }\frac{(-1)^k(2\pi x)^{2k+1}}{(2k+1)!}\frac{2^{2k+1}}{2^{2k+1}+1}.$

Proposition 19

For a fixed $0<\varepsilon <1$, if $x \in [-1+\varepsilon ,1-\varepsilon ]$, we can exchange the order in the sum of (43) and we get

$$\begin{aligned} {Q(x)=\frac{-\cos (4\pi /3)}{2}\sum _{k=1}^{+\infty }\frac{(-1)^k(2\pi x)^{2k}}{(2k)!}\frac{2^{2k}}{2^{2k}-1}.} \end{aligned}$$

Proof

Note that if $|x|<1$ there exists a constant K (the coefficients on the power series of $\cos$ are decreasing) such that

$$\begin{aligned}\left| \sum _{k=1}^{+\infty }\frac{(-1)^k (2\pi x)^{2k}}{2^{2ik}(2k)!}\right| &\le \sum _{k=1}^{+\infty }\left| \frac{(2\pi x)^{2k}}{2^{2ik}(2k)!}\right| \le \frac{1}{2^i}\sum _{k=1}^{+ \infty }\left( K\, x^{2k} \right) \\&=\frac{K}{2^i} \left( \frac{x^2}{1-x^2}\right) \le \frac{K}{2^i}\left( \frac{|1-\varepsilon |^2}{1-|1-\varepsilon |^2}\right) . \end{aligned}$$

We can exchange the order on the double sum: $\forall x \in [-1+\varepsilon ,1-\varepsilon ]$,

$$\begin{aligned} {\sum _{i=0}^{+ \infty }\sum _{k=1}^{+\infty }\left| \frac{(-1)^k (2\pi x)^{2k}}{2^{2ik}(2k)!}\right| \le \sum _{i=0}^{+ \infty }\frac{K}{2^i} \left( \frac{x^2}{1-x^2} \right) \le 2K\,\left( \frac{|1-\varepsilon |^2}{1-|1-\varepsilon |^2}\right) < +\infty .} \end{aligned}$$

Note that $(x-2/3)\in [-2/3,1/3]$. Then,

$$\begin{aligned} {Q(x-2/3)=\frac{-\cos (4\pi /3)}{2}\sum _{k=1}^{+\infty }\frac{(-1)^k(2\pi (x-2/3))^{2k}}{(2k)!}\frac{2^{2k}}{2^{2k}-1}.} \end{aligned}$$

(44)

In the same way we get

$$\begin{aligned} {M(x-2/3)=\frac{\sin (4\pi /3)}{2}\sum _{k=0}^{+\infty }\frac{(-1)^k(2\pi (x-2/3))^{2k+1}}{(2k+1)!}\frac{2^{2k+1}}{2^{2k+1}+1}.} \end{aligned}$$

(45)

$\square$

As $V_2(x+2/3)= M(x) + Q(x)$, then, $V_2(x)=Q(x-2/3)+M(x-2/3).$ Finally, from (44) and (45) the power series expression of $V_2$ around 2/3 is given by

$$\begin{aligned}V_2(x)&= \frac{\sin (4\pi /3)}{2}\sum _{k=0}^{+\infty }\frac{(-1)^k(2\pi \left( x- \frac{2}{3} \right) )^{2k+1}}{(2k+1)!} \frac{2^{2k+1}}{2^{2k+1}+1} \nonumber \\&\quad -\frac{\cos (4\pi /3)}{2}\sum _{k=1}^{+\infty }\frac{(-1)^k(2\pi \left( x- \frac{2}{3} \right) )^{2k}}{(2k)!}\frac{2^{2k}}{2^{2k}-1} \end{aligned}$$

(46)

We can express the power series of $V_1$ around 1/3 from $V_1(x)=V_2(1-x)$.

1.2 The involution kernel for a map with a indifferent fixed point

In this section, we show some results claimed on Sect. 3.

Consider $f:[0,1]\rightarrow [0,1]$, where

$$\begin{aligned} \left\{ \begin{array}{ll} f(y)= \frac{y}{1-y},&{} {\text{if}},\; 0\le y \le \frac{1}{2},\\ f(y)= 2- \frac{1}{y},&{} {\text{if}},\; \frac{1}{2}< y\le 1,\\ \end{array} \right. \end{aligned}$$

and the potential $A(y) = \log f '(y)$, which is given by the expression

$$\begin{aligned} \left\{ \begin{array}{ll} f'(y)= \frac{1}{(1-y)^2}, &{} {\text{if}},\; 0\le y \le \frac{1}{2},\\ f'(y)= \frac{1}{y^2},&{} {\text{if}},\; \frac{1}{2}< y\le 1.\\ \end{array} \right. \end{aligned}$$

We want to derive the involution kernel for A. We claim the involution kernel for such A is $W(y,x) = 2 \log (x + y - 2 xy).$ We will show that

$$\begin{aligned} A(F^{-1}(y,x))+ W( F^{-1}(y,x) )- W(y,x) =A(y). \end{aligned}$$

We denote $R_0\subset [0,1]^2$ the cylinder $0< y< 1/2$, and $R_1\subset [0,1]^2$ the cylinder $1/2< y< 1$. Restricted to $R_0$, the inverse $F^{-1}(y,x)$ is given by $F^{-1}(x,y) =(\frac{y}{ 1-y}, \frac{x}{1+ x}).$ From this we get, for $(y,x) \in R_0$, $A(F^{-1}(y,x))= \log (1+x)^{-2}.$ Moreover, in this case, for (y, x) in the cylinder $R_0$,

$$\begin{aligned} { W( F^{-1}(y,x) )= 2 \,\log \left( \frac{y}{1-y} + \frac{x}{1+x} - 2\,\frac{\,y\, \,\,x}{(1-y)\,\, (1+x)}\,\right) = \, 2 \log \left( \frac{x + y - 2 x\, y }{(1-y)\,\, (1+x)} \right) .} \end{aligned}$$

Therefore, for $0<y<1/2$, we have

$$\begin{aligned}&{A(F^{-1}(y,x))+ W( F^{-1}(y,x) )- W(y,x) } \\&={\log (\, (1+x)^{-2}\, \frac{( x+y - 2 x\, y)^{-2} }{(1-y)^{-2} \, (1 + x)^{-2}} \, \frac{1}{ ( x+y - 2 x\, y)^{-2} } \,)= 2\, \log (1-y)= A(y).} \end{aligned}$$

Now we have to consider the cylinder $R_1$, where $1/2< y<1$. In this case, $F^{-1}(y,x)= ( 2- \frac{1}{y} , \frac{1}{2-x}).$ Therefore, $F^{-1}(y,x)= \, 2 \, \log (2-x),$ and,

$$\begin{aligned} W( F^{-1}(y,x) )&=2\, \log (\, \frac{2 y -1}{y} \,-\, \frac{1}{2-x} \, + \, 2 \frac{(2 \, y -1)}{y \, (2-x)}\,) \\&= 2\, \log ( \, \frac{(2 y-1)\, (2-x) + y - 2\, (2y-1)}{y \, (2-x)} \,)\,=\, 2\, \log ( \,\frac{x+y - 2\, x\, y}{y \, (2-x)} \,) \end{aligned}$$

Finally, for $1/2<y<1$, we have

$$\begin{aligned}&{A(F^{-1}(y,x))+ W( F^{-1}(y,x) )- W(y,x)} \\&={ \log (\, (2-x)^{-2}\, \frac{( x+y - 2 x\, y)^{-2} }{y^{-2} \, (2- x)^{-2}} \, \frac{1}{ ( x+y - 2 x\, y)^{-2} } \,)= 2\, \log y= A(y).} \end{aligned}$$

This shows that $W(y,x) = 2 \log (x + y - 2 xy)$ is the involution kernel for $\log f'(y)$.

We thank the referee for his careful reading which helped us to improve the reading of the text