Let a sequence $(a_n)_{n=0}^\infty$ be defined recursively $a_{n+1} = (1-a_n)^{\frac1p}$, where $p>1$, $0<a_0<(1-a_0)^{\frac1p}$. Let $a$ be the unique real root of $a=(1-a)^{\frac1p}$, $0<a<1$. It is clear $0<a_0<(1-a_0)^{\frac1p}\Leftrightarrow 0<a_0<a$. Prove

1) $a_{2k-2}<a_{2k}<a<a_{2k+1}<a_{2k-1}$ and $a_{2k+1}-a<a-a_{2k}$.

2) $\lim\limits_{n\to\infty}a_n=a$.

Define $f(x):=(1-x)^{\frac1p}$. Consider $f^2$. When $p=2$, $a_{n+2}=f^2(a_n)=\big(1-(1-a_n)^{\frac12}\big)^{\frac12}$. $a_{n+2}>a_n\Leftrightarrow (1-a_n)(1+a_n)^2>1\Leftrightarrow a_n<(1-a_n)^{\frac12}$, and the conclusion is proved. But I am having difficulty generalizing this method to arbitrary $p>1$.

I also suspect that there is a general method to solve this kind of problem.

  • $\begingroup$ just be a little more careful with the tags..it is not a dynamical systems question i believe. $\endgroup$ – Marios Gretsas Sep 11 '17 at 11:00
  • $\begingroup$ @MariosGretsas: I think it can be considered that, as it can be considered as the fixed point solution of $x=(1-x)^{\frac1p}$. What would you suggest as a better tag? $\endgroup$ – Hans Sep 11 '17 at 15:11
  • $\begingroup$ Personally for me just ''real analysis'' or calculus if fine $\endgroup$ – Marios Gretsas Sep 11 '17 at 15:13
  • $\begingroup$ you need to prove $(1-(1-a_0)^{\frac 1p})^{\frac 1p} \geq a_0$ which does not look like a cakewalk. $\endgroup$ – Gabriel Romon Sep 11 '17 at 17:31
  • $\begingroup$ @GabrielRomon: That is exactly what I have said in my note below the question. $\endgroup$ – Hans Sep 11 '17 at 17:44

We adopt the approach described in the note below the original question, i.e., by examining the composite $f^2$ of the original mapping $f$. We prove a more general statement than the original question. To this end, we prepare the following lemmas.

Lemma 1: Suppose $g(x)$ defined on $[0,\infty)$ strictly increases with $x$ and $g(0)>0$. $a:=\inf\{x|g(x)< x\}>g(0)$ and $g(a)=a$ thus $a=\inf\{x|g(x)\le x\}$.

Proof: By contradiction. QED

Lemma 2: Assume the same condition as in Lemma 1. $g^n(0)$ increases with $n$ and $g^n(0)<a$ $\forall n\in \mathbf N$ and thus $\lim\limits_{n\to\infty}g^n(0)$ exists and is bounded from above by $a$.

Proof: Choose an arbitrary $x_1\in[0,a)$. By the definition of $a$, $g(x_1)>x_1$. Since $g(x)$ increases with $x$, $g(x_1)<g(x)<g^2(x_1),\ \forall x\in(x_1,g(x_1))$. So $x<g(x_1)<g(x),\ \forall x\in(x_1,g(x_1)]$ (Note the statement is true for $x=g(x_1)$). By the definition of $a$ again, $g(x_1)<a$. Then $[x_1,g(x_1)]\subset(0,a)$. In conclusion, we have shown $x_1\in (0,a)\implies x_1<g(x_1)<a$, or $g^n(x_1)$ increases but is bounded from above by $a$. $g^n(x_1)$ has a limit and is bounded from above by $a$.


Lemma 3: Assume the same condition as in Lemma 1. In addition, $g$ is continuous. $\lim\limits_{n\to\infty}g^n(0)=a$.

Proof: Suppose the limit is $a'<a$. By the definition of $a$, $a'<g(a')$. But $g$ is continuous, $a'=\lim\limits_{n\to\infty} g(g^n(x_1))=g(\lim\limits_{n\to\infty}g^n(x_1))=g(a')>a',$ a contradiction.


$f^2(x)$ increases with $x$ for monotonic $f$. Set $g=f^2$. Adding $g(x_1)>x_1$ for some $x_1$ would allow the application of Lemmas 1 and 2. Assuming continuity of $f$, and say $f(0)\ge 0$ and $\frac d{dx}g(0)=f'(f(0))f'(0)>1$, would guarantee just the aforementioned condition, as well as the application of Lemma 3.

The particular form of $f(x)=(1-x)^{\frac1p},\ \forall x\in[0,1],\ p>1$ in the original question is one such example. For this particular example, we show further that this $a$ is exactly the unique solution of $x=f(x)$. $g(x)=(1-(1-x)^{\frac1p})^{\frac1p}$. We now show there is a unique root for $g(x)-x=0,\ x\in(0,1)$ and that root is the unique root of $f(x)=x$. $$g''(x)= -\frac{p^2-1}{p^4} \big(1-(1-x)^{\frac1p}\big)^{\frac1p-2}(1- x)^{\frac1p-2}\Big((1-x)^{\frac1p}-\frac p{p + 1}\Big).$$ $(g(x)-x)''$ is strictly increasing with a unique root $x_0=1-\Big(1-\frac 1{p + 1}\Big)^p\in(0,1)$. This implies $g(x)<0\Longleftrightarrow x\in(0,a)$ and $g(x)>0\Longleftrightarrow x\in(a,1)$. With $g(0)=g(1)-1=0$, $g(x)-x=0$ therefore has only $1$ root in the open interval $(0,1)$.

$f(x):=(1-x)^{\frac1p},\ \forall x\in[0,1],\ p>1$. To prove point 1) $a-a_0>a_1-a$, we show $0<-f'(x)<1,\ x[0,x_0]$ where $x_0=f(x_0)$.

Because $-f'(x) = \frac1p(1-x)^{\frac1p-1}$ increases from $-f'(0)=\frac1p<1$ to $-f'(1)=\infty$ as $x$ increases from $0$ to $1$, $-f'(x_2)=1\text{ for some }x_2\in(0,1)\Longrightarrow -f'(x)<1\ \forall x[0,x_2)$. To show $-f'(x_0)<1$ it is sufficient to show $x_0<x_2$. To show the latter, it is sufficient to show $f(x_2)<x_2$.

Solving $-f'(x)=1$ we obtain $x_2=1-p^{-\frac p{p-1}}$. We want to show $$p^{-\frac1{p-1}}=f(x_2)<x_2=1-p^{-\frac p{p-1}} \tag1$$ or $\displaystyle\frac{\ln(p+1)}p<\frac{\ln p}{p-1}$. It is sufficient to show $h(p)=\frac{\ln(p)}{p-1}$ is a decreasing function in $p\in(1,\infty)$. Since $$h'(p) = \frac1{(p-1)^2}\Big(1-\frac1p-\ln p\Big)=\frac1{(p-1)^2}\int_1^p \Big(\frac 1 {u^2}-\frac1u\Big)\,du<0,$$ we are done.

Alternatively, we prove inequality (1) as follows. Inequality (1) is equivalent to $$p^{\frac p{p-1}}=\big(1+(p-1)\big)^{\frac p{p-1}}>(p-1)\frac p{p-1}=p+1.$$ But the inequality in the middle is a special case of the Bernoulli's inequality.

  • $\begingroup$ How does one conclude that $x_1<g(x_1)\forall x_1\in(0,a)$? $\endgroup$ – Simply Beautiful Art Sep 11 '17 at 21:06
  • $\begingroup$ @SimplyBeautifulArt: Are you referring to the second sentence of the second paragraph? If so, see the first half clause of the second sentence, i.e., "by the definition of $a$". $\endgroup$ – Hans Sep 11 '17 at 21:09
  • $\begingroup$ Oh... interesting approach. $\endgroup$ – Simply Beautiful Art Sep 11 '17 at 21:11
  • $\begingroup$ Actually, your second paragraph has an error. "Since $g$ increases, $g(x_1)<g(x)<g^2(x_1),~\forall x\in(x_1,g(x_1))$". You implicitly assume that $x_1<g(x_1)$ before proving it, since$$g(x_1)<g(x)<g^2(x_1)$$requires$$g(x_1)<g^2(x_1)$$to be true. $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 14:59
  • $\begingroup$ @SimplyBeautifulArt: We had gone through why $x_1<g(x_1)$ in the subsequent discussion following your very first comment. Do you agree? If so, since $g$ increases, we have $g(u)<g(v),\ \forall u<v$. Let $u=x_1$ and $v=g(x_1)$. You have the desired result. $\endgroup$ – Hans Sep 12 '17 at 17:13

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