3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $a$.

QED

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.

QED

$f^2(x)$ 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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.