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

  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1717374255292.png

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!
 
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  • #2
ChiralSuperfields said:
f′≥0 for all t
This conclusion is incorrect. Consider the counterexample: ##f=x^2##.
 
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  • #3
Hill said:
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?
 
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  • #4
fresh_42 said:
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
ChiralSuperfields said:
f′≥0 for all t
It is what my post (#2) says:
Hill said:
This conclusion is incorrect.
 
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  • #5
No, You said ...
Hill 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?
Hill said:
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).
 
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  • #6
fresh_42 said:
how in the world could anybody know what you meant by "this"
By looking at the quote just above it:

1717386526684.png


I give a counterexample to "this".
 
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  • #7
Hill said:
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
.
 
Last edited:
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  • #8
By the mean value theorem, there exist [itex]\zeta \in (x,y)[/itex] and [itex]\eta \in (y, z)[/itex] such that [tex]\begin{split}f'(\zeta) &= \frac{f(y) - f(x)}{y - x} \\
f'(\eta) &= \frac{f(z) - f(y)}{z - y}.
\end{split}[/tex] Now use the fact that [itex]\zeta < \eta[/itex] and [itex]f'[/itex] is increasing.
 
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  • #9
fresh_42 said:
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:
ChiralSuperfields said:
f′≥0 for all t

rather than this:
ChiralSuperfields said:
##\displaystyle f ' \geq 0## for all t

(Adding a small space, further clarifies things.)
ChiralSuperfields said:
##\displaystyle f\, ' \geq 0## for all t
 
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