- #1

Akorys

- 23

- 3

## Homework Statement

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 [itex]f[/itex] be continuous and strictly monotonic on the positive real axis and let [itex]g[/itex] denote the inverse of [itex]f[/itex]. If [itex]a_1\lt a_2\lt...\lt a_n[/itex] are n given positive real numbers, we define their mean value with respect to f to be the number [itex]M_f[/itex] defined as follows:

[tex]M_f=g(\frac{1}{n}\sum_{i=1} ^{n} f(a_i))[/tex]

1. Show that [itex]a_1\lt M_f\lt a_n[/itex].

2. If [itex] h(x)=af(x)+b[/itex], show that [itex]M_h = M_f[/itex].

## Homework Equations

As above, [itex]M_f=g(\frac{1}{n}\sum_{i=1} ^{n} f(a_i))[/itex]

## The Attempt at a Solution

Solution 1.

Since [itex]f[/itex] is a strictly monotonic function, we must have have either

[tex]f(a_1) \lt f(a_2) \lt ... \lt f(a_n) \text{ or,}[/tex][tex]f(a_1) \gt f(a_2) \gt ... \gt f(a_n)[/tex]

It follows that [tex] nf(a_1) \lt \sum_{i=1}^{n} f(a_i) \lt nf(a_n) \text{ or,}[/tex][tex]nf(a_1) \gt \sum_{i=1}^{n} f(a_i) \gt nf(a_n)[/tex] for each set of inequalities, respectively. Further, these inequalities imply [tex] f(a_1) \lt \frac{1}{n} \sum_{i=1}^{n} f(a_i) \lt f(a_n) \text{ or,}[/tex][tex]f(a_1) \gt \frac{1}{n}\sum_{i=1}^{n} f(a_i) \gt f(a_n)[/tex] Now we simply consider each case. First, assume that [itex]f[/itex] strictly increases. Then [itex]g[/itex] also strictly increases, and we may apply [itex]g[/itex] to the first set of inequalities above to derive: [tex]a_1 \lt M_f \lt a_n[/tex]

Now assume [itex]f[/itex] strictly decreases. Then [itex]g[/itex] strictly decreases as well. If we apply [itex]g[/itex] to the second set of inequalities above, the inequality signs must reverse since the function is decreasing. Therefore, we derive: [tex]g[f(a_1)] \lt g[\frac{1}{n}\sum_{i=1}^{n} f(a_i)] \lt g[f(a_n)][/tex][tex]a_1 \lt M_f \lt a_n[/tex]

This completes the proof of problem 1.

Solution 2.

If [itex] h(x) = af(x) + b[/itex], then [itex] f(x) = \frac{h(x) - b}{a}[/itex]. We may apply [itex]g[/itex] to both sides to get [tex] x = g(\frac{h(x)-b}{a})[/tex] Let [itex]s(y)=x[/itex] be the inverse function of [itex]h(x)[/itex], such that [itex] y = h(x) [/itex]. Through this we attain: [tex] s(y) = g(\frac{y - b}{a})[/tex] Now we may consider [itex]M_h[/itex]. [tex] M_h = s(\frac{1}{n}\sum_{i=1}^{n} h(a_i))[/tex][tex]M_h = g(\frac{\frac{1}{n}\sum_{i=1}^{n} h(a_i) - b}{a})[/tex] We may write [itex] b = \frac{1}{n}\sum_{i=1}^{n}b [/itex], 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: [tex] M_h = g(\frac{1}{n}[\sum_{i=1}^{n}\frac{ h(a_i) - b}{a}]) [/tex] Since [itex] f(x) = \frac{h(x) - b}{a}[/itex] for each [itex] x = a_1, x = a_2, ..., x=a_n[/itex], the sum reduces to [tex] M_h = g(\frac{1}{n}\sum_{i=1}^{n} f(a_i) = M_f[/tex] This is true by definition, and thus completes the proof.