# Convergence of recursive sequence

I have tried to do this exercise. Do you think my solution is ok? Is it possible to get more information about the convergence? Is there a better way to do it?

Let $f:[0,1]\mapsto[0,1]$ be a continous function such that $f(x)<x \ \forall x \in (0,1]$ and the right limit of the derivative in 0 is 1/2. Fixed a point $a_0 \in [0,1]$ consider the succession $a_{n+1}:=f(a_n)$. Study the series: $\sum_{n=0}^\infty a_n$.

Thank you everyone for the kind help!

GivAlz

PS I'm doing some exercises for a very important test which is about a lot of different stuff I might not even have seen in a course yet. So every kind comment, suggestion about somehow similar exercises or theorems/results/whatever that you think might help me out is more than welcome.

As requested by Samprity, this is my answer. This is a little bit long as I try to spell out as many details as possible.

Claim 1: $\quad f(0) = 0$.

• Assume the contrary. If $f(0) > 0$, the the fact $f$ is continuous implies for some $\epsilon > 0$, $f(x) > \frac{f(0)}{2}$ for $x \in [0,\epsilon)$. If one pick a $y$ such that $0 < y < \min(\epsilon,\frac{f(0)}{2})$, we will find $f(y) > \frac{f(0)}{2} > y$. This contradicts with the given condition $f(x) < x$ for $x > 0$.

Claim 2: $\quad a_n \to 0$ as $n \to \infty.$

• By Claim 1, $f(0) = 0$. Combine these with $f(x) < x$ over $(0,1]$, we find $f(x) \le x$ over $[0,1]$. This in turn implies the sequence $a_n$ determined by the recurrence relation $a_{n+1} = f(a_n)$ is monotonic decreasing.

• Since $a_n$ is also bounded below by $0$, the sequence $a_n$ converges. Let us call the limit $a$ and it is clear $a \ge 0$.

• Since $f(x)$ is a continuous function, we have $$f(a) = f(\lim_{n\to\infty}a_n) = \lim_{n\to\infty} f(a_n) = \lim_{n\to\infty} a_{n+1} = a$$
• Since $f(x) < x$ for $x > 0$, $a$ cannot be positive and hence it must be $0$.

Please note that in above claims, we have not used the fact the right derivative of $f$ at $0$ exists and equal to $\frac12$. By definition of right derivative, for any $\epsilon > 0$, there exists a $\delta(\epsilon) > 0$ such that:

$$\left|\frac{f(x) - f(0)}{x - 0} - \frac12\right| < \epsilon \quad\text{ whenever }\quad x \in (0,\delta(\epsilon))$$

Take $\epsilon$ as $\frac14$ and notice $f(0) = 0$, this implies:

$$f(x) \le \frac34 x \quad\text{ whenever }\quad x \in [ 0, \delta(\frac14) )$$

By Claim 2, $a_n \to 0$ as $n \to \infty$. This means there is an integer $N$ such that

$$a_n < \delta(\frac14) \quad\text{ whenever }\quad n \ge N$$

For such $n$, we have $$a_{n+1} = f(a_n) \le \frac34 a_n \quad\implies\quad a_{n} \le \left(\frac34\right)^{n-N} a_N$$

As a consequence, for any integer $M > N$, the partial sums of the original series we care is bounded from above by a number independent of $M$.

$$\sum_{k=0}^{M} a_k = \sum_{k=0}^{N-1} a_k + \sum_{k=N}^{M} a_k \le \sum_{k=0}^{N-1} a_k + a_N \sum_{k=N}^{M} \left(\frac34\right)^{k-N} \le \sum_{k=0}^{N-1} a_k + 4a_N$$

Being a series of non-negative numbers, this implies $\displaystyle \sum_{k=0}^{\infty} a_k$ converges.

For right continuity (we have the right derivative) we have that $f(0)=0$. Thus the case $a_0=0$ trivially converges. Otherwise we have that the succession is strictly monotone (thus it has a limit).

There exists a neighbourhood of $0$, $V=(0,a)$ such that, $\forall h \in V$ we have $\vert \frac{f(h)-f(0)}{h} \vert =\vert \frac{f(h)}{h} \vert \leq \frac{1}{2}+\epsilon = M$.

Now let $a_0=h\in (0,1]$. If there is a V, as above, such that $h \in V$ and $M < \frac{1}{h}$ the series converges using the comparison test with the geometric series, because: $\vert a_1 \vert = \vert f(h) \vert \leq M h \Rightarrow \vert a_n \vert \leq (M h)^n$ (with $\vert M h \vert < 1$).

If no such V exists then $M\geq h$ and the series doesn't converge because $\vert a_n \vert \geq (M h)^n \geq 1$

• Since the right limit of the derivative at $0$ is $1/2$, there is some neighborhood $[0,\epsilon)$ in which $f(x) < (2/3)x$ (the $2/3$ is arbitrary; anything larger than $1/2$ can be used). That estimate should allow you to prove that the series will converge. – Carl Mummert Sep 9 '13 at 13:40

I am getting the followings for your given problem without using the right hand differentiability at $x = 0$, at not much required for getting convergence of the series $\sum a_n$.

As $f :[0,1] \rightarrow [0,1]$ is a continuous function all the value of $a_n$ will be in $[0,1]$. So $\{a_n\}$ is a bounded positive term sequence.

Now $a_{n+1} = f(a_n) < a_n$ as $f(x) < x$. So we can conclude that $\{a_n\}$ is a monotone sequence. A monotone bounded sequence is always convergent. You can prove it and so $\{a_n\}$ is convergent, though it is not your question which is about the convergence of $\sum a_n$.

As the sequence $\{a_n\}$ is monotone we shall say $a_n < a_0$ and as all the terms are positive we shall say $[a_n]^\frac{1}{n} < [a_0]^\frac{1}{n}$. Let $a_0 < 1$ then $\lim_{n \rightarrow \infty} a_n < 1$ because of monotonicity of $\{a_n\}$. Then by root text we can say $\sum a_n$ is convergent.

• In this case, root test doesn't work because $\lim_{n\to\infty} a_0^{\frac{1}{n}} = 1$. Another thing is by this argument, the series corresponds to $a_n = \frac{1}{n+2}$ ( generated by $f(x) = x/(x+1)$ with $a_0 = \frac12$ ) converges. This is clearly not true. – achille hui Sep 9 '13 at 14:17
• @achillehui Thank you for instructions. What will be correct argument? – Dutta Sep 10 '13 at 2:50
• The correct argument consists of two parts. 1) show $a_n$ converges to 0 ($a_n$ is strictly monotonic decreasing + bounded from below + $f$ continuous is enough to prove this) 2) Since $a_n \to 0$, aside from finitely many terms, you can apply the argument in Carl Mummert's comment to bound $a_n$ from above by a geometric series. By comparison test, you are done. – achille hui Sep 10 '13 at 4:54
• I have understood. Please write it down as an answer. – Dutta Sep 10 '13 at 6:33