Let $f$ be a real-valued function on $[0,1]$ with a continuous second derivative. Assume that $f(0)=0, f'(0)= 1, f''(0)\neq 0$, and $0Let $f$ be a real-valued function on $[0,1]$ with a continuous second derivative. Assume that $f(0)=0, f'(0)= 1, f''(0)\neq 0$, and $0<f'(x)<1 $ for all $x\in(0, 1]$. Let $x_1, x_2,\ldots$ be a sequence with $0< x_1 \leq 1$ and with
$$x_n = f\left(\dfrac{x_1+x_2+\cdots+x_{n-1}}{n-1}\right)$$ for $n \geq 2$. Prove that $\displaystyle\lim_{n\to\infty}x_n\ln n = −2/f'' (0)$.
I tried to use fixed point theorem by converting it to form $x_n=\phi(x_{n+1})$ but it is not coming. Please help regarding this.
 A: $\bullet\ $ First, let's prove that $f(x) < x$ for every $x \in (0,1]$.
Let $x \in (0,1]$. By the mean value theorem, there exists $c \in (0,x)$ such that $f(x)-f(0)=f'(c)x$, i.e. such that $f(x)=f'(c)x$. Because $f'(c) \in (0,1)$ by assumption, one deduce that $f(x) < x$.
$\bullet\ $ Let's prove now that $(x_n)_{n \geq 1}$ is decreasing.
By induction, we shall prove that for every $n \geq 1$, one has $x_n < x_{n-1} < ... < x_2 < x_1$.
For $n=1$, the statement is empty.
For $n=2$, one has $x_2 = f(x_1) < x_1$ by the first point.
Let's suppose that the statement is true for an integer $n \geq 2$. Then one has $x_n < ... < x_2 < x_1$. In particular, one deduces that
$$\dfrac{x_1 +\ ... +\ x_n}{n} \leq \dfrac{x_1 +\ ... +\ x_{n-1}}{n-1}$$
But $f$ is increasing over $[0,1]$, so applying $f$ to this inequality leads directly to $x_{n+1} < x_n$, so we have
$$x_{n+1} < x_n < ... < x_1$$
which is the property for the $n+1$-step.
This proves that $(x_n)_{n \geq 1}$ is decreasing.
$\bullet\ $ In particular, the sequence is decreasing and bounded below by $0$, so it converges to a limit $l \in [0,1]$. But by Cesàro's theorem, one has also
$$\lim_{n \rightarrow +\infty} \dfrac{x_1 +\ ... +\ x_n}{n} = l$$
so because $x_n = f \left(\dfrac{x_1 + ... + x_{n-1}}{n}\right)$, by continuity, one has $l=f(l)$. But $0$ is the only fixed point of $f$ on $[0,1]$ (again, because $f(x) < x$ if $0 < x \leq 1$), so $$\lim_{n \rightarrow +\infty} x_n = 0$$
$\bullet\ $ For all $n \geq 1$, let $$y_n = \dfrac{x_1 +\ ... +\ x_n}{n}$$
so that $(y_n)$ also tends to $0$ (Cesàro). By Taylor-Young formula, one has
\begin{align*} x_{n+1} = f(y_{n}) & = f(0)+y_nf'(0)+\dfrac{y_n^2}{2}f''(0) + o(y_n^2) \\ &= y_n+\dfrac{y_n^2}{2}f''(0) + o(y_n^2) \quad \quad \quad \quad \quad \quad (*) \end{align*}
so because $x_{n+1} = (n+1)y_{n+1}-ny_n $, this rewrites as $$(n+1)y_{n+1}-ny_n =y_n+\dfrac{y_n^2}{2}f''(0) + o(y_n^2)$$
i.e. $$y_{n+1} = y_n + \dfrac{y_n^2}{2(n+1)}f''(0) + o\left(\dfrac{y_n^2}{n+1}\right)$$
One deduces that \begin{align*} y_{n+1}^{-1}-y_n^{-1} &= y_n^{-1} \left( 1+\dfrac{y_n}{2(n+1)}f''(0) + o\left(\dfrac{y_n}{n+1}\right)\right)^{-1}-y_n^{-1} \\ &=   -\dfrac{1}{2(n+1)}f''(0) + o\left(\dfrac{1}{n+1}\right) \end{align*}
Now one can sum this : the LHS telescopes, and because $\displaystyle{\sum_{k=1}^n \dfrac{1}{k} \sim \ln(n)}$, one gets $$y_n^{-1} = -\dfrac{1}{2}f''(0)\ln(n) + o\left(\ln(n)\right) $$
i.e.  $y_n \ln(n) \sim -\dfrac{2}{f''(0)}$. But $(*)$ shows that $x_n \sim y_n$, so you finally get
$$\boxed{\lim_{n \rightarrow +\infty} x_n \ln(n) = -\dfrac{2}{f''(0)}}$$
