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 $s_0 = 1 - \sum_{i=1}^n s_i$


I note that

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


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

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


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 $g^{(k+1)}(\cdot) \equiv f$.

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

1 Answer 1


Replicating one proof for the mean value theorem of integration

$\newcommand{\bx}{\mathbf x}\newcommand{\bs}{\mathbf s}$ 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 $f(\xi)$, $$ f(\xi)=c=n!\int_{S_n}f(\bs^T\bx)\,d\bs. $$


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