f and g

[grobber]

Let f,g:[a,b]->[a,b] be continuous functions s.t. f(g(x))=g(f(x)) for all x in the domain and f is monotonous. Prove that there is a c in [a,b] s.t. f(c)=g(c)=c.

[Valentin Vornicu]

nice one. vietnam undergrad 2003.

first of all each of f and g have a fixed point. if f is increasing then let g(d)=d be a fixed point of g. Then f(g(d))=g(f(d)) => f(d) = g(f(d)) which means that also f(d) is a fixed point. In this way we create a sequence (f⁽ⁿ⁾ (d) )_n which is a bounded sequence, and also monotonous (since f is increasing), thus convergent. it's limit is the point c we are looking for.

the case when f is decreasing is even easier: consider a random fixed point of f, f(c)=c. Then g(c)=f(g(c)) then g(c) is also a fixed point, and if g(c)<c then g(c) = f(g(c)) > f(c) =c which is a contradiction! again a contradiction arises when we try g(c)>c. Thus only one option remains valid: g(c)=c=f(c) and we are done.

hope all of you know why a continuous function f:[a,b]->[a,b] has a fixed point

[christi]

How??

[Valentin Vornicu]

well consider the auxiliar function g(x)=f(x)-x. Obviously it's continous, and g(a)=f(a)-a \geq 0 and g(b)=f(b)-b \leq 0 thus it must also take the intermediate value 0 ...

maybe your "how" was directed to another thing?

[amfulger]

[Valentin Vornicu]

f is decreasing ( that is f(x)<f(y) when x>y) and g(c) I supposed it is smaller than c which means that f(g(c))>f(c). nu equality involved.

[amfulger]

So what you mean is that you only consider the situations f is increasing (x>y => f(x)>f(y))and f is decreasing (x>y =>f(x)<f(y)).
Then I suppose that you consider that the monotonous functions are injective. In Romania we use strictly monotonous for this. So what does the monotonous in the problem text stand for?

[Valentin Vornicu]

in Romanian you use the word "monoton". In English monotonous means either increasing or decreasing, that is (f(x)-f(y))(x-y) > 0 or (f(x)-f(y))(x-y) <0 for any x,y. I belive that you have an English problem here

[grobber]

I don't think monotonous means strict inequalities. I also think that we should use "strictly monotonous" if we're referring to monotonous and injective functions.

[Valiowk]

Well, if you see this link and that link, you'll understand where the confusion comes from.

Personally, I can't even remember what I learnt monotone to be during training sessions... (this is what query forms are for...)

[liyi]

In high school books, 'increasing' and 'decreasing' mean strict inequalities.
But in the most math analysis books, it allows equality. So we should use 'strictly increasing(decreasing)'