Prove the monotonicity of a sequence with respect to the first term A sequence $x_n$ is defined as follows: $x_0=0, x_1>0$ and $e^{-{x_i}^k}$=$\frac{\int_{x_{i-1}}^{x_{i+1}}e^{-x^{k}}\,\mathrm{d}x}{x_{i+1}-x_{i-1}}, i\ge1$. Here, $0< k\le1$ is a fixed constant.
It is easy to see that $x_n$ can be uniquely determined by $x_1$. Moreover, I want to show that $\forall n\ge2$,$x_n$ is an increasing function of ${x_1}$. I can prove the case when $k=1$ by taking derivatives on each side and using induction(the induction hypothesis is $\frac{\mathrm{d}x_{n}}{\mathrm{d}x_{1}}\ge \frac{\mathrm{d}x_{n-1}}{\mathrm{d}x_{1}}$), but I cannot prove the general case. Can anyone offer some help?
 A: The function $f(t) = e^{-t^k}$ is continuous and strictly decreasing, and that is in fact everything we need to prove the claim.
First we define a function $H : \Bbb R^2 \to \Bbb R$ as
$$
\begin{align}
 H(a, b) &= \frac{1}{b-a} \int_a^b f(t) \,dt  \quad \text{if } a \ne b \, ,\\
 H(a, a) &= f(a) \, .
\end{align}
$$
It is a straightforward calculation to show that $H$ is strictly increasing in the first argument, and strictly decreasing in the second argument.  (Remark: This also follows from the fact that $H(a, b)$ is the slope of the secant line intersecting the graph of the concave function $x \mapsto \int_0^x f(t) \,dt$ at $x=a$ and $x = b$.)
How let $(x_n)$, $(y_n)$ be two sequences with $x_0 = y_0 = 0$, $ 0 < x_1 < y_1$, and
$$
 f(x_i) = \frac{1}{x_{i+1}-x_{i-1}} \int_{x_{i-1}}^{x_{i+1}} f(t) \, dt \, ,\\
 f(y_i) = \frac{1}{y_{i+1}-y_{i-1}} \int_{y_{i-1}}^{y_{i+1}} f(t) \, dt \, .
$$
for $i \ge 1$. We want to show that $x_i < y_i$ for all $i \ge 1$, and this is done with induction.
It suffices to show that for all $i \ge 1$,
$$
 x_{i-1} \le y_{i-1} \land x_i < y_i \implies x_{i+1} < y_{i+1} \, ,
$$
which is done as follows: If $ x_{i-1} \le y_{i-1}$ and $x_i < y_i$ then
$$
H(y_{i-1}, y_{i+1}) = f(y_i) < f(x_i) = H(x_{i-1}, x_{i+1}) \le H(y_{i-1}, x_{i+1})
$$
because $f$ is strictly decreasing, and because $H$ is increasing in the first argument. So we have 
$$
H(y_{i-1}, y_{i+1}) < H(y_{i-1}, x_{i+1}) \implies x_{i+1} < y_{i+1}
$$
because $H$ is decreasing in the second argument. This finishes the proof.
