Let $\delta>0$,$f(x)=\frac{1}{5}x+x^{1+\delta}$.Let $x_0>0$,define $x_n=f(x_{n-1})$ for all $n \in \mathbb{N}$.How to show that $\{5^nx_n\}_{n=1}^{\infty}$ is a bounded sequence for all sufficiently small $x_0$?It seems quite obvious that this is true but I can't prove it rigorously.Can someone help me with this?
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if you write $x_n$ explicitly ,you see that it is bounded by the sum of two geometric series.. – Halil Duru Feb 19 '13 at 23:56
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This is not true in general. But as you say, for $x_0$ small enough, it is true.
Take $$ x_0\leq \left( \frac{4}{5}\right)^{1/\delta}. $$ Then a quick study of $f$ and an induction show that $x_n$ is decreasing and bounded below by $0$, hence it converges to the unique fixed point of $f$, that is $$ \lim_{n\rightarrow+\infty}x_n=0. $$
It follows that $$ x_n=x_{n-1}\left(\frac{1}{5}+x_{n-1}^\delta \right)\leq rx_{n-1} $$ with $$ r=\frac{1}{5}+x_0^\delta . $$
An easy induction shows that $$ x_n\leq r^nx_0\leq r^n $$ for all $n$.
Now observe that $\lim_{x\rightarrow 0^+}\left(\frac{1}{5}+x^\delta\right)^{1+\delta}<\frac{1}{5}$.
So we take $x_0>0$ small enough to have $$ 5r^{1+\delta}<1. $$
Now for all $n$, we have $$ 0\leq 5^nx_n-5^{n-1}x_{n-1}=5^nx_{n-1}^{1+\delta}\leq 5(5r^{1+\delta})^{n-1}. $$
Since the geometric series $\sum (5r^{1+\delta})^{n-1}$ converges, $\sum 5^nx_n-5^{n-1}x_{n-1}$ converges by comparison.
So the sequence $$ 5^kx_k=x_0+\sum_{n=1}^k 5^nx_n-5^{n-1}x_{n-1} $$ converges.
In particular, the sequence $5^kx_k$ is bounded.
Julien
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