TITLE: A generalized (MacLaurin's) average for functions
QUESTION [18 upvotes]: The average value of a function $y=f(x)$,  on an interval $[a,b]$, is ${1\over {b-a}}\int_a^b f(t)dt$. This of course relates to the arithmetic average. It is easy to see that a corresponding formula for the geometric average is $\exp\left({1\over {b-a}}\int_a^b \ln(f(t))dt\right)$.
There are many other types of averages. In particular the ones motivated by the elementary symmetric polynomials are interesting as they "mix" the function values. My question is: How can we evaluate those averages?
To be specific, consider a real positive continuous function $y=f(x)$ on $[a,b]$. Create a partition of $n$ sub-intervals of width $\Delta x$. Let $Y=(y_1,y_2,\cdots, y_n)$ be the values of the function $f$ at some point in  those intervals. Define the elementary symmetric polynomials $e_k=e_k(Y)$, for $1\le k \le n$, through
$$
\prod_{i=1}^n (t+y_i)= t^n+e_1t^{n-1}+\cdots+e_n.
$$ 
Define the average 
$$
a_k(Y)={\root k \of {{e_k} \over {\left (n \atop k \right )}}}.
$$
Define $A_\alpha(f)$, the $\alpha$-average of $f$ over $[a,b]$, as the limit of $a_k(Y)$ as $n \to \infty$, $\Delta x \to 0$, and $k/n \to \alpha$. Note $\alpha=0$ corresponds to the arithmetic average and $\alpha=1$ is the geometric average.
What do we know about $A_\alpha$ for $0<\alpha <1$? How can we compute it? For example if $f(x)=x$, $[a,b]=[1,2]$, and $\alpha=1/2$ what is $A_\alpha$?
Edit 1:
Some related inequalities are  Maclaurin's  and  Newton's.
Edit 2:
I guess the requirement of continuity can be relaxed to piecewise continuity and still have a unique limit.  Finding $A_\alpha$ for the following function, for a given $m>0$, will also be of interest:
$$f(x)= \cases { 1  & if  $ \quad 0 \le x \le 1/2$  \cr
m & if $ \quad 1/2 < x \le 1$ }.$$

REPLY [2 votes]: Wow, so much inspiration.
Here's a generalization of your question : let us begin with a function $f : ]0,\infty[ \to ]\inf_{x > 0} f(x),\sup_{x > 0} f(x)[$ which is monotone. Let $g : [0,1] \to ]0,\infty[$. Define the $f$-average of $g$ by 
$$
A_f(g) = f^{-1} \left( \int_0^1 f(g(t)) dt \right).
$$
Note that when $f(x) = x$ you get the arithmetic average and when $f(x) = \log(x)$ you get the geometric average. These functions are increasing. However, you can also take $f(x) = 1/x$ and get the harmonic mean. This function is decreasing ; perhaps that is why it behaves differently than the other averages. 
This $f$-average is not completely randomly defined ; it has some properties. In a similar way that arithmetic and geometric averages are defined for finite sets instead of functions, we can define the $f$-mean of $n$ numbers by 
$$
M_f(x_1, \dots, x_n) = f^{-1} \left( \frac 1n \sum_{i=1}^n f(x_i) \right).
$$
Again, taking $f$ as stated before gives us arithmetic, geometric and harmonic means. Now we can see that the $f$-average of the $f$-mean of $g_1(t), \dots, g_n(t)$ is the $f$-mean of the $f$-averages : 
$$
A_f( M_f(g_1,\dots,g_n)) = f^{-1} \left( \int_0^1 f(M_f(g_1,\dots,g_n)(t)) \, dt \right) \\
= f^{-1} \left( \int_0^1 \frac 1n \sum_{i=1}^n f(g_i(t)) \, dt \right) \\
= f^{-1} \left( \frac 1n \sum_{i=1}^n \int_0^1 f(g_i(t)) \, dt \right) \\
= f^{-1} \left( \frac 1n \sum_{i=1}^n f(A_f(g_i)) \right) \\
= M_f(A_f(g_1), \dots, A_f(g_n)).
$$
We can also obtain a generalized QM-AM-GM-HM inequality : 
Theorem. Suppose $f,g$ are twice differentiable, $f' \neq 0 \neq g'$, $f'' \le g''$ and $h(x) = g'(x) / f'(x)$ is injective. Then for all $x_1, \dots, x_n \in ]0,\infty[$, we have $M_f(x_1,\dots,x_n) \le M_g(x_1,\dots,x_n)$. 
The proof is similar to the one with QM-AM-GM-HM except that the facts used for $f$ and $g$ are quite clear now. Let us minimize $M_g(x_1,\dots,x_n)$ subject to $M_f(x_1,\dots,x_n) = C$. It follows from the Lagrange Multipliers that there exists a $\lambda$ such that 
$$
\frac{g'(x_i)}{g'(M_g(x_1,\dots,x_n))} = \frac{\lambda f'(x_i)}{f'(M_f(x_1,\dots,x_n))} \quad \Longrightarrow \quad \frac{g'(x_i)}{f'(x_i)} = \frac{\lambda g'(M_g)}{f'(M_f)} = \frac{g'(x_j)}{f'(x_j)}, \quad i \neq j.
$$
Since we assumed $h(x) = g'(x)/f'(x)$ to be injective, $x_1 = x_2 = \dots = x_n \overset{def}= x$, so it follows that $(x,\dots,x)$ is a minimum and 
$$
M_g(x_1,\dots,x_n) \ge M_g(x,\dots,x) = x = M_f(x,\dots,x) = C = M_f(x_1,\dots,x_n).
$$
Note that letting $f \in \{x^2, x, \log x, 1/x\}$, this proves QM-AM-GM-HM. 
What would really help for computations is if we could find the functions $f_{\alpha}$ corresponding to the $A_{\alpha}$-averages, if such functions exist. This would definitely be the way to go to compute these averages. Would those functions happen to be nicely differentiable, we could also show using this theorem that $A_{\alpha}(f)$ is a decreasing function of $\alpha$.