- #1

member 731016

- Homework Statement
- Please see below

- Relevant Equations
- Please see below

For this problem,

My proof is

Since ##f'## is increasing then ##x < y <z## which then ##f(x) < f(y) < f(z)##

This is because,

##f''(t) \ge 0## for all t

## \rightarrow \int \frac{df'}{dt} dt \ge \int 0~dt = 0## for all t

##\rightarrow \int df' \geq 0## for all t

##f ' \geq 0## for all t

##\frac{df}{dt} \geq 0## for all t

##\int df \geq \int 0~dt## for all t

##f(t) \geq 0##

Now ##\frac{f(y) - f(x)}{y - x} \geq 0##

##\frac{f(z) - f(y)}{z - y}##

Assume ##y - x = z - y = c##

##\frac{f(y) - f(x)}{c} \geq 0 \implies f(y) - f(x) \geq 0##

##\frac{f(z) - f(y)}{c} \geq 0 \implies f(z) - f(y) \geq 0##

Thus we, consider two cases,

(1) ##f(z) - f(y) \geq f(y) - f(x) \geq 0##

(2) ##f(y) - f(x) \geq f(z) - f(y) \geq 0##

Note that (2) is impossible since ##f(x) < f(y) < f(z)##

##f(y) \geq 0 \implies \frac{f(y) - f(z)}{y - x} \geq \frac{0}{y - x} = 0##

##f(z) \geq 0 \implies \frac{f(z) - f(y)}{z - y} \geq \frac{0}{z - y} = 0##

We can assume that ##z - y = y - x##, since one possible function is ##f(x) = x^n## when ##n \in \mathbb{N}##. Consider case ##n = 1##, then there is a function so that ##f(z) - f(y) \geq f(y) - f(x)## however, for ##n > 1## ##f(z) - f(y) \geq f(y) - f(x)## Of course, we have only considered one case of the polynomial functions and it can be generalized to any increasing function I think.

Does anybody please know where to prove from here?

Thanks!

My proof is

Since ##f'## is increasing then ##x < y <z## which then ##f(x) < f(y) < f(z)##

This is because,

##f''(t) \ge 0## for all t

## \rightarrow \int \frac{df'}{dt} dt \ge \int 0~dt = 0## for all t

##\rightarrow \int df' \geq 0## for all t

##f ' \geq 0## for all t

##\frac{df}{dt} \geq 0## for all t

##\int df \geq \int 0~dt## for all t

##f(t) \geq 0##

Now ##\frac{f(y) - f(x)}{y - x} \geq 0##

##\frac{f(z) - f(y)}{z - y}##

Assume ##y - x = z - y = c##

##\frac{f(y) - f(x)}{c} \geq 0 \implies f(y) - f(x) \geq 0##

##\frac{f(z) - f(y)}{c} \geq 0 \implies f(z) - f(y) \geq 0##

Thus we, consider two cases,

(1) ##f(z) - f(y) \geq f(y) - f(x) \geq 0##

(2) ##f(y) - f(x) \geq f(z) - f(y) \geq 0##

Note that (2) is impossible since ##f(x) < f(y) < f(z)##

##f(y) \geq 0 \implies \frac{f(y) - f(z)}{y - x} \geq \frac{0}{y - x} = 0##

##f(z) \geq 0 \implies \frac{f(z) - f(y)}{z - y} \geq \frac{0}{z - y} = 0##

We can assume that ##z - y = y - x##, since one possible function is ##f(x) = x^n## when ##n \in \mathbb{N}##. Consider case ##n = 1##, then there is a function so that ##f(z) - f(y) \geq f(y) - f(x)## however, for ##n > 1## ##f(z) - f(y) \geq f(y) - f(x)## Of course, we have only considered one case of the polynomial functions and it can be generalized to any increasing function I think.

Does anybody please know where to prove from here?

Thanks!