# Mean value with respect to a function f

1. Jul 9, 2015

### Akorys

1. The problem statement, all variables and given/known data
I wanted to ask anyone willing to check my proofs to two problems, so that I may know if I am making any mistakes or assumptions.
Let the function $f$ be continuous and strictly monotonic on the positive real axis and let $g$ denote the inverse of $f$. If $a_1\lt a_2\lt...\lt a_n$ are n given positive real numbers, we define their mean value with respect to f to be the number $M_f$ defined as follows:
$$M_f=g(\frac{1}{n}\sum_{i=1} ^{n} f(a_i))$$
1. Show that $a_1\lt M_f\lt a_n$.
2. If $h(x)=af(x)+b$, show that $M_h = M_f$.

2. Relevant equations
As above, $M_f=g(\frac{1}{n}\sum_{i=1} ^{n} f(a_i))$

3. The attempt at a solution
Solution 1.
Since $f$ is a strictly monotonic function, we must have have either
$$f(a_1) \lt f(a_2) \lt ... \lt f(a_n) \text{ or,}$$$$f(a_1) \gt f(a_2) \gt ... \gt f(a_n)$$
It follows that $$nf(a_1) \lt \sum_{i=1}^{n} f(a_i) \lt nf(a_n) \text{ or,}$$$$nf(a_1) \gt \sum_{i=1}^{n} f(a_i) \gt nf(a_n)$$ for each set of inequalities, respectively. Further, these inequalities imply $$f(a_1) \lt \frac{1}{n} \sum_{i=1}^{n} f(a_i) \lt f(a_n) \text{ or,}$$$$f(a_1) \gt \frac{1}{n}\sum_{i=1}^{n} f(a_i) \gt f(a_n)$$ Now we simply consider each case. First, assume that $f$ strictly increases. Then $g$ also strictly increases, and we may apply $g$ to the first set of inequalities above to derive: $$a_1 \lt M_f \lt a_n$$
Now assume $f$ strictly decreases. Then $g$ strictly decreases as well. If we apply $g$ to the second set of inequalities above, the inequality signs must reverse since the function is decreasing. Therefore, we derive: $$g[f(a_1)] \lt g[\frac{1}{n}\sum_{i=1}^{n} f(a_i)] \lt g[f(a_n)]$$$$a_1 \lt M_f \lt a_n$$
This completes the proof of problem 1.

Solution 2.
If $h(x) = af(x) + b$, then $f(x) = \frac{h(x) - b}{a}$. We may apply $g$ to both sides to get $$x = g(\frac{h(x)-b}{a})$$ Let $s(y)=x$ be the inverse function of $h(x)$, such that $y = h(x)$. Through this we attain: $$s(y) = g(\frac{y - b}{a})$$ Now we may consider $M_h$. $$M_h = s(\frac{1}{n}\sum_{i=1}^{n} h(a_i))$$$$M_h = g(\frac{\frac{1}{n}\sum_{i=1}^{n} h(a_i) - b}{a})$$ We may write $b = \frac{1}{n}\sum_{i=1}^{n}b$, which allows us to work with both terms together in the numerator. Combine the denominator into the sum to make the proof clearer. Through this, we may derive: $$M_h = g(\frac{1}{n}[\sum_{i=1}^{n}\frac{ h(a_i) - b}{a}])$$ Since $f(x) = \frac{h(x) - b}{a}$ for each $x = a_1, x = a_2, ..., x=a_n$, the sum reduces to $$M_h = g(\frac{1}{n}\sum_{i=1}^{n} f(a_i) = M_f$$ This is true by definition, and thus completes the proof.

2. Jul 10, 2015

### JonnyG

Looks fine to me.

3. Jul 10, 2015

### Akorys

Thank you for checking it!