If $\lim_{n \rightarrow \infty}\sum_{k=0}^np_{k}=\infty$ show that $f$ has a fixed point

Question

We have $(p_{n})_{n\geq0}$ a sequence of strictly positive real numbers and $a$ a real number and the continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that the sequence $$\left(\frac{\sum_{k=0}^np_{k}f^{(k)}(a)}{\sum_{k=0}^np_{k}}\right)_{n\geq1}$$ is bounded, where $f^{(k)}$ is the composite of $k$ times of $f$ (and $f^{0}=\text{id}_{\mathbb{R}})$. If $$\lim_{n \rightarrow \infty}\sum_{k=0}^np_{k}=\infty$$ show that $f$ has a fixed point.
I need to say that this problem looked ugly when I first saw it. So if the series of $p_{n}$ diverges to $\infty$ and knowing that the "big" sequence is bounded then $$\lim_{n\rightarrow \infty}\sum_{k=0}^np_{k}f^{(k)}(a)$$ is $0$ or $+/-\infty$. It is a fixed point problem right? So we can take the continuous function $g:R\rightarrow R$, $g(x)=f(x)-x$. So we need to show somehow this function $g$ it is 0 at some point. If $$\lim_{n\rightarrow \infty}\sum_{k=0}^np_{k}f^{(k)}(a)=-\infty$$ then there exists $f^{(k)}(a)<0$ right?

Answer 1

I'll do a proof by contradiction: I assume that $f$ has no fixed point. By $f$'s continuity we can say that W.L.G. :\begin{equation}f(x)>x,\;\;\forall x\in \mathbb R.\end{equation} Now $\{f^k(a)\}$ cannot be bounded, if not, we have that $\{f^k(a)\}\subset[-R,R]$ and exists the minimum of the function $g(x)=f(x)-x\ge m$ (by https://en.wikipedia.org/wiki/Extreme_value_theorem): $$f^k(a)=f(f^{k-1}(a))\ge f^{k-1}(a)+m\ge \ldots\ge a+km $$ This is a contradiction by the $f'$s boundedness. So there exists a subsequence $f^{k_h}(a)\stackrel{\text{W.L.G.}}{\to}+\infty$. Let $M\in \mathbb R$, will exists $h_0>0$ s.t. $f^{k_{h_0}}(a)>M$. Now by the first inequality we know that: $$f^{k_{h_0}+n}(a)>f^{k_{h_0}+n-1}(a)>\ldots>f^{k_{h_0}}(a)>M.$$Hence: $$\frac{\sum_{k=1}^N p_kf^k(a)}{\sum_{k=1}^N p_k}=\frac{\sum_{k=1}^{h_0-1} p_kf^k(a)}{\sum_{k=1}^N p_k}+\frac{\sum_{k={h_0}}^N p_kf^k(a)}{\sum_{k=1}^N p_k}> $$ $$>\frac{\sum_{k=1}^{h_0-1} p_kf^k(a)}{\sum_{k=1}^N p_k}+M\frac{\sum_{k={h_0}}^N p_k}{\sum_{k=1}^N p_k}=$$ $$=C_{h_0} +M\frac{\sum_{k={h_0}}^N p_k}{\sum_{k=1}^{h_0-1} p_k+\sum_{k=h_0}^N p_k}\stackrel{N\to\infty}{\to}C_{h_0}+M.$$ To conclude I chose $0<M=M'-\inf_N \frac{\sum_{k=1}^N p_kf^k(a)}{\sum_{k=1}^N p_k}$ and and I observe that $$\liminf_N \frac{\sum_{k=1}^N p_kf^k(a)}{\sum_{k=1}^N p_k}>C_{h_0}+M>M',$$ and by the generality of $M'>0$ we get a contradiction because the LHS must be bounded.

Answer 2

This is heavily inspired by Bongo's answer.

Consider the sequence $(f^{(n)}(a))_{n \ge 0}$, i.e. $$ a, f(a), f(f(a)), f(f(f(a))), \ldots $$

Case 1: The sequence $(f^{(n)}(a))$ is not monotone, i.e. there are indices $m$ and $n$ such that $f^{(m+1)}(a) > f^{(m)}(a)$ and $f^{(n+1)}(a) < f^{(n)}(a)$.

Setting $x_1 = f^{(m)}(a)$ and $x_2 = f^{(n)}(a)$ we have $f(x_1) - x_1 > 0$ and $f(x_2) - x_2 < 0$. The intermediate value theorem then implies that $f(x^*) - x^* = 0$ for some $x^*$ between $x_1$ and $x_2$. That $x^*$ is a fixed point of $f$.

Case 2: The sequence $(f^{(n)}(a))$ is monotone (increasing or decreasing). Then $$ L = \lim_{n \to \infty} f^{(n)}(a) $$ exists in $\Bbb R \cup \{ -\infty, +\infty \}$. If we can show that $L$ is finite then $$ f(L) = f\left(\lim_{n \to \infty} f^{(n)}(a)\right) = \lim_{n \to \infty} f^{(n+1)}(a) = L $$ implies that $L$ is a fixed point of $f$.

In order to show that $L$ is finite we use the general form of the Stolz-Cesàro theorem: $$ L = \liminf_{n \to \infty} \frac{p_n f^{(n)}(a)}{p_n} \le \liminf_{n \to \infty} \frac{\sum_{k=0}^n p_k f^{(k)}(a)}{\sum_{k=0}^n p_k} \le M $$ and $$ L = \limsup_{n \to \infty}\frac{p_n f^{(n)}(a)}{p_n} \ge \limsup_{n \to \infty} \frac{\sum_{k=0}^n p_k f^{(k)}(a)}{\sum_{k=0}^n p_k} \ge -M \, , $$ where $M$ is a bound for the absolute value of $ \frac{\sum_{k=1}^n p_k f^{(k)}(a)}{\sum_{k=1}^n p_k} $. So $L$ is finite and therefore a fixed point of $f$.

Answer 3

Let $f\colon \mathbb{R}\to \mathbb{R}$ continuous, without fixed points. Then the function $f(x)-x$ has constant sign, $>0$, or $< 0$. Two cases:

1. $f(x) > x$ for all $x$. Then for all $a \in \mathbb{R}$, $(f^n(a))_{n\ge 0}$ is strictly increasing with limit $+\infty$. (if limit were finite $l$ then $f(l)= l$, contradiction).

2. $f(x) < x$ for all $x$. Then for all $a\in \mathbb{R}$, $(f^n(a))_{n\ge 0}$ is strictly decreasing with limit $-\infty$.

Now, to finish the problem, use Cesaro-Stolz.