# Can one average “close” smooth functions?

Suppose $M$ is a connected, smooth, second-countable manifold.

Let $U \subset M^n$ be some neighbourhood of the diagonal. We will call a function $a: U \times \Delta_n \rightarrow M$ an "averaging operator of order n" if $a|_{U \times v_i} = \pi_i$, where $v_i$ denotes the i-th vertex of $\Delta_n$.

Intuitively, $a$ should tell us how to take a weighted average of "close" functions.

Does every $M$ admit a smooth averaging operator of order $n$, for some $U$ and all $n$?

(I am trying to prove that smooth functions are dense in $Hom(M,N)$ without embedding $N$ into any $\mathbb{R}^n$ and the above result would greatly help me.)

-
 $\Delta_n$ is standard $n$-simplex? – Paul Nov 16 '11 at 8:27 Yes, it is the standard n-simplex as a submanifold of R^n. – Piotr Pstragowski Nov 16 '11 at 19:08 I would remark that this question has been posted even on MathOverflow. There, until now it has received three interesting answers. Cf. mathoverflow.net/questions/81105/… – Giuseppe Nov 18 '11 at 19:11