# Showing $\exists \xi : \frac{f(\xi)}{n!} = \int \cdots \int_{S_n} f(s_0 x_0 + \cdots + s_n x_n) d s_1 \cdots d s_n$ for unit simplex $S_n$

## Problem

Suppose we have $$f : \mathbb{R} \to \mathbb{R}$$ continuous on $$I\left( x_0, \cdots, x_n \right)$$, where $$I(\cdot)$$: the smallest interval containing $$x_0, \cdots, x_n$$.

Show $$\exists \xi \in I\left( x_0, \cdots, x_n \right)$$, such that

$$\frac{f(\xi)}{n!} = \int \cdots \int_{S_n} f\left(s_0 x_0 + s_1 x_1 + \cdots + s_n x_n\right) d s_1 \cdots d s_n$$

where $$S_n := \left\{ (s_1, \cdots, s_n): s_i \ge 0, \sum_{i=1}^n s_i \le 1 \right\}$$ : the unit simplex in $$\mathbb{R}^n$$, and $$s_0 = 1 - \sum_{i=1}^n s_i$$

## Try

I note that

\begin{align} f\left( \left(1-\sum_{i=1}^n s_i\right) x_0 + s_1 x_1 + \cdots + s_n x_n\right) &= f\left( x_0 + s_1 (x_1 - x_0) + \cdots s_n (x_n - x_0) \right) \\[8pt] \end{align}

thus

$$\int \cdots \int_{S_n} (RHS) d s_n \cdots d s_1 = \int_0^{s_1} \cdots \int_0^{1-\sum_{j=1}^{n-2} s_j} \frac{1}{x_n-x_0} \left[ F\left(x_0 + s_1 (x_1 - x_0) + \cdots s_n (x_n - x_0)\right) \right]_0^{1-\sum_{j=1}^{n-1}s_j} d s_{n-1} \cdots s_1$$

However, in this way I do not see any proceedings. How should I proceed? Any help will be appreciated.

## Backgrounds

Actually raising this problem is due to reading from a wiki article where the remainder term is represented

$$R(x) = g[x_0,\ldots,x_k,x] \ell(x) = \ell(x) \frac{g^{(k+1)}(\xi)}{(k+1)!}, \quad \quad x_0 < \xi < x_k$$

without any explanations. Here $$g^{(k+1)}(\cdot) \equiv f$$.

• Do you at least assume that $f$ is continuous? This doesn't seem to work if $f$ is not. Sep 23, 2019 at 17:45
• @WETutorialSchool Oh I missed the point, thank you Sep 23, 2019 at 22:12

$$\newcommand{\bx}{\mathbf x}\newcommand{\bs}{\mathbf s}$$ As $$\bs\mapsto f(\bx^T\bs)$$ is continuous over the compact $$S_n$$, resp. $$f$$ over $$I(\bx)$$, there are points $$x_\max$$ and $$x_\min\in I(\bx)$$ so that $$f(x_\min)\le f(\bs^T\bx)\le f(x_\max) ~~~~\forall \bs\in S_n.$$ where $$\bs=(s_0,s_1,...,s_n)$$ and $$\bx=(x_0,x_1,...,x_n)$$. Then inserting this inequality into the integral gives $$f(x_\min)\le c=n!\int_{S_n}f(\bs^T\bx)\,d\bs\le f(x_\max) ,$$ using that $$\text{vol}(S_n)=\int_{S_n}1\cdot d\bs=\frac1{n!}$$. By the intermediate value theorem for continuous functions, there exists a $$\xi$$ between $$x_\min$$ and $$x_\max$$ so that the value $$c$$ in the middle is exactly $$f(\xi)$$, $$f(\xi)=c=n!\int_{S_n}f(\bs^T\bx)\,d\bs.$$