Proof given ##x < y < z## and a twice differentiable function

[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!

 

[Hill]

f′≥0 for all t

This conclusion is incorrect. Consider the counterexample: $f=x^2$.

 

[fresh_42]

This conclusion is incorrect. Consider the counterexample: $f=x^2$.

What do you mean? For $x<z$ we get for $f(t)=t^2$
$$
\dfrac{f(y)-f(x)}{y-x}=\dfrac{y^2-x^2}{y-x}=y+x< z+y=\dfrac{z^2-y^2}{z-y}=\dfrac{f(z)-f(y)}{z-y}
$$
so where is the problem?

 

[Hill]

What do you mean? For $x<z$ we get for $f(t)=t^2$
$$
\dfrac{f(y)-f(x)}{y-x}=\dfrac{y^2-x^2}{y-x}=y+x< z+y=\dfrac{z^2-y^2}{z-y}=\dfrac{f(z)-f(y)}{z-y}
$$
so where is the problem?

The problem is with the OP's conclusion that

f′≥0 for all t

It is what my post (#2) says:

This conclusion is incorrect.

 

[fresh_42]

No, You said ...

This conclusion is incorrect.

... so how in the world could anybody know what you meant by "this", especially if you're not sure anyway?

Consider the counterexample: $f=x^2$.

... which is no counterexample. It is in fact the generic example: one has to consider the zeros of $f''(t)$ and place $x,y,z$ among possible zero(s).

 

[Hill]

how in the world could anybody know what you meant by "this"

By looking at the quote just above it:

I give a counterexample to "this".

 

[fresh_42]

By looking at the quote just above it:

View attachment 346378

I give a counterexample to "this".

My bad (eyesight), I read it as

f''(t) ≥ 0

.

 

[pasmith]

By the mean value theorem, there exist $\zeta \in (x,y)$ and $\eta \in (y, z)$ such that $$\begin{split}f'(\zeta) &= \frac{f(y) - f(x)}{y - x} \\ f'(\eta) &= \frac{f(z) - f(y)}{z - y}. \end{split}$$ Now use the fact that $\zeta < \eta$ and $f'$ is increasing.

 

[SammyS]

My bad (eyesight), I read it as
.

Not entirely your fault. This is what happens when the PF "Insert quotes" feature is used to quote a LaTeX expression and the expression is not edited to be displayed in LaTeX.

You get this:

f′≥0 for all t

rather than this:

$\displaystyle f ' \geq 0$ for all t

(Adding a small space, further clarifies things.)

$\displaystyle f\, ' \geq 0$ for all t