MHB Prove $|P(a)-P(b)|<\dfrac{1}{2}$ for Algebra Challenge Function $P$

[anemone]

Let $P$ be a function defined on $[0, 1]$ such that $P(0)=P(1)=1$ and $|P(a)-P(b)|<|a-b|$, for all $a\ne b$ in the interval $[0, 1]$.

Prove that $|P(a)-P(b)|<\dfrac{1}{2}$.

 

[Albert1]

Let $P$ be a function defined on $[0, 1]$ such that $P(0)=P(1)=1$ and $|P(a)-P(b)|<|a-b|$, for all $a\ne b$ in the interval $[0, 1]$.

Prove that $|P(a)-P(b)|<\dfrac{1}{2}$.

Spoiler

for simplicity , I will set :$P(x)=x(x-1)+1=(x-\dfrac{1}{2})^2+\dfrac {3}{4}$
then $max(P(a)-P(b))\leq 1-\dfrac {3}{4}<\dfrac {1}{2}$
take $a=1,b=\dfrac {1}{2}$

 

[topsquark]

Spoiler

for simplicity , I will set :$P(x)=x(x-1)+1=(x-\dfrac{1}{2})^2+\dfrac {3}{4}$
then $max(P(a)-P(b))\leq 1-\dfrac {3}{4}<\dfrac {1}{2}$
take $a=1,b=\dfrac {1}{2}$

Are you simply demonstrating an example or is there some way this will generalize to all possible P(x)?

-Dan

 

[chisigma]

Let $P$ be a function defined on $[0, 1]$ such that $P(0)=P(1)=1$ and $|P(a)-P(b)|<|a-b|$, for all $a\ne b$ in the interval $[0, 1]$.

Prove that $|P(a)-P(b)|<\dfrac{1}{2}$.

[sp]Because is $|P(a) - P (b)|<|a - b|$ for all $a \ne b$ in [0,1] is also $|P' (x)|< 1$ for all x in [0,1] . But is also $\int_{0}^{1} P' (x)\ d x = P(1) - P(0) = 0$ , so that for all a and b in [0,1] is $ |\int_{a}^{b} P'(x)\ d x| = |P(b) - P(a)| \le \frac{1}{2}$...[/sp]

Kind regards$\chi$ $\sigma$

 

[Albert1]

Are you simply demonstrating an example or is there some way this will generalize to all possible P(x)?

-Dan

In general you can set $P(x)=x^n(x-1)^n+1$
and then find its derivative to get $max(P(x))$ and $min(P(x))$
for $x\in [0,1]$

 

[Albert1]

[sp]Because is $|P(a) - P (b)|<|a - b|$ for all $a \ne b$ in [0,1] is also $|P' (x)|< 1$ for all x in [0,1] . But is also $\int_{0}^{1} P' (x)\ d x = P(1) - P(0) = 0$ , so that for all a and b in [0,1] is $ |\int_{a}^{b} P'(x)\ d x| = |P(b) - P(a)| \le \frac{1}{2}$...[/sp]

Kind regards$\chi$ $\sigma$

very clever solution !
but $ |P(b) - P(a)| =\dfrac {1}{2}$ should be excluded , for $a,b \in [0,1]$ and $a\neq b$
this is the same as using probability method:
$\dfrac { |P(b) - P(a)|}{1-0}<\dfrac { |b - a|}{1-0}<\dfrac {1}{2}$

 

[topsquark]

[sp]so that for all a and b in [0,1] is $ |\int_{a}^{b} P'(x)\ d x| = |P(b) - P(a)| \le \frac{1}{2}$...[/sp]

I'm sorry, [math]\chi ~ \sigma[/math], I can't see how you are getting |b - a| < 1/2 in the last step?

-Dan

 

[Albert1]

I'm sorry, [math]\chi ~ \sigma[/math], I can't see how you are getting |b - a| < 1/2 in the last step?

-Dan

let :$0\leq b=a\pm k\leq 1$

for : $a+k-(a-k)=2k\leq (1-0)=1$

or $k \leq \dfrac {1}{2}$

then : $|b - a| =|a\pm k -a| =|\pm k| \leq \dfrac {1}{2}$

$\therefore |P(b) -P(a)| <|b -a| \leq \dfrac {1}{2}$

 

[anemone]

Sorry for the late reply and thanks Albert for participating, and your explanations and also thanks to chisigma for your solution...

Solution of other:

Spoiler

We consider the following cases:
[TABLE="class: grid, width: 700"]
[TR]
[TD]Case I ($|a-b|\le \dfrac{1}{2}$):

If $|a-b|\le \dfrac{1}{2}$, then

$|P(a)-P(b)|<|a-b|\le\dfrac{1}{2}$, as desired.[/TD]
[TD]Case II ($|a-b|> \dfrac{1}{2}$):

By symmetry, we may assume $a>b$.

Then

$\begin{align*}|P(a)-P(b)|&=|P(a)-P(1)+P(0)-P(b)|\\&\le |P(a)-P(1)|+|P(0)-P(b)|\\&<|a-1|+|0-b|\\&=1-a+b-0\\&=1-(a-b)\\&<\dfrac{1}{2}\end{align*}$

as desired.[/TD]
[/TR]
[/TABLE]