Let $f : [0,1] \to [0,1]$ continuous and $(u_n)$ s.t. $\forall n, u_{n+1}=f(u_n)$. Show that $(u_n)$ converges $\iff u_{n+1} - u_n \to 0$

Question

Would you have a hint for that:

Let $f : [0,1] \to [0,1]$ continuous. Let $(u_n)$ s.t. $\forall n, u_{n+1}=f(u_n)$.

Show that $(u_n)$ converges $\iff u_{n+1} - u_n \to 0$.

I need to prove there is only one accumulation point. I can prove the set of all those points is an interval, so if there are more than one, there is an infinity actually.

(Also, there exists at least one $x$, $f(x) = x$.)

Answer 1

I think you've done most of the work already, thanks to this.

I need to prove there is only one accumulation point. I can prove the set of all those points is an interval, so if there are more than one, there is an infinity actually.

You can also show that, if $\ell$ is an accumulation point, then $f(\ell)=\ell$. Indeed, if $(u_{n_p})$ converges to $\ell$, then $f(u_{n_p})$ converges to $f(\ell)$ by continuity, but $f(u_{n_p}) = u_{n_p+1}$ gets arbitrarily close to $u_{n_p}$.

Now, assume that the set of accumulation points is non-trivial. Then it's a closed interval $[a,b]$ with $a<b$. You know that $f(x) = x$ on this interval. Take an interior point $x_0$. How can the sequence $(u_n) = (f^n (u_0))$ converge to $x_0$? Can you get a contradiction?

Answer 2

Hint: If $lim_nu_{n+1}-u_n=0$, Use the fact that $[0,1]$ is compact, you can extract a sequence $x_{n_p}$ which converges towards $l$ since $x_n\in [0,1]$, show that $u_n$ converges By using the fact that $f(l)=l$ and since $f$ is continuous, the sequences $f^m(x_{n_p})$ converges towards $l$. There is one point of accumulation since $lim_nf(x_{n_p})=lim_nx_{n_p}$ since $lim_nu_{n+1}-u_n=0$. You deduce that for every $m, (x_{{n_p}+m})$ has the limit of $(x_{n_p})$.

If $(x_{n_q})$ converges towards $l'$, $(x_{n_q})=$ the sequence $(f^m(x_{n_p})=x_{{n_p}+m}))$ have an infinite number of same terms, so $l=l'$.

Answer 3

Let $(u_n)$ be defined as in the problem and assume that $u_{n+1} - u_n \to 0$ as $n\to\infty$.

Lemma 1. Any subsequential limit of $(u_n)$ is a fixed point of $f$.

Proof. If $u_{n_k} \to \ell$, then $f(l) - l = \lim_{k\to\infty} \left( f(u_{n_k}) - u_{n_k} \right) = 0$.

Lemma 2. Let $L$ be the set of subsequential limits of $(u_n)$. Then $L$ is an interval, allowing degenerate case.

Proof. It suffices to prove that whenever $a, b$ with $a<b$ lie in $L$, the entire interval $[a, b]$ lies in $L$. Fix $c \in [a, b]$. Then for each $\epsilon > 0$ with $\epsilon < \frac{1}{2}(b-a)$, pick $N$ and $m_a, m_b \geq N$ such that

• $\lvert u_{n+1} - u_n \rvert < \epsilon$ for all $n \geq N$,
• $\lvert u_{m_a} - a \rvert < \epsilon$ and $\lvert u_{m_b} - b \rvert < \epsilon$.

By our choice of $\epsilon$, it follows that $m_a \neq m_b$ and that there exists $n'$ between $m_a$ and $m_b$ such that $\lvert u_{n'} - c \rvert < \epsilon$:

$\hspace{9em}$

(Green bands represent $\epsilon$-neighborhoods.) Finally, fix a sequence $\epsilon_k \downarrow 0$ and apply the previous observation to $\epsilon_k$ to choose an increasing $n'_k$'s with $\lvert u_{n'_k} - c \rvert < \epsilon_k$. Then $u_{n'_k} \to c$ and hence $c \in L$.

Proposition. $(u_n)$ has at most one limit point, hence converges.

Proof. If $(u_n)$ has more than one limit point, then there exist $a < b$ such that $[a, b] \subseteq L$. But this means that $f(x) = x$ on $[a, b]$, which leads the contradiction that $(u_n)$ will stabilize with some constant value in $(a, b)$.

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p.s. While writing my answer, user @D. Thomine also posted the same idea.

Answer 4

If $u_n$ converges it is clear that $u_{n+1}-u_n\to 0$.

Now suppose $u_{n+1}-u_n\to 0$. If $u_{n+1}=u_n$ for some $n$ then $u_k=u_n$ for all $k>n$ and we are done. Hence we may assume $$\tag0u_{n+1}\ne u_n\qquad \text{for all }n.$$ Then it follows that $u_n\ne u_m$ for $n\ne m$ as we cannot have $u_{n+1}-u_n\to 0$ with an eventually periodic (but not eventually constant) sequence.

Lemma 1. Let $(v,w)\subset [0,1]$ be an open interval where $|f(x)-x|>\epsilon$ for all $x\in(v,w)$. Then almost all $u_n$ are $\le v$ or almost all $u_n$ are $\ge w$.

Proof. From $u_{n+1}-u_n\to0$, we see that there are only finitely many $u_n\in(v,w)$. Hence apart from finitely many initial terms, $\{u_n\}_n$ cannot skip across $(v,w)$ and must stay on one side of it. $\square$

Lemma 2. Assum $(0)$. Let $(v,w)\subset [0,1]$ be an open interval where $f(x)=x$ for all $x\in(v,w)$. Then almost all $u_n$ are $\le v$ or almost all $u_n$ are $\ge w$.

Proof. Because of $(0)$, the sequence must avoid $(v,w)$, and as in the proof of lemma 1, it eventually cannot skip across the interval any more. $\square$

Let $A=\overline{\{\,u_n\mid n\in\Bbb N\,\}}$, $B=\{\,x\in[0,1]\mid f(x)=x\,\}$ and $C=A\cap B$. Note that $C$ is precisely the set of accumulation points of $\{u_n\}_n$, hence certainly not empty. Assume $C$ has at least two points, say $c_1,c_2\in C$ with $c_1<c_2$. Then we cannot have $[c_1,c_2]\subseteq B$ because of lemma 2. Hence there exists $d\in(c_1,c_2)$ with $f(d)\ne d$. Then $|f(x)-x|>\frac12|f(d)-d|$ in an open interval around $d$ and we arrive at a contradiction from lemma 1. We conclude that $C$ has at most, hence exactly one point, which then is the limit of the $u_n$.