Proof about a fixed point

[Ozma]

The problem is the following: let [MATH]f:[0,1] \to [0,1][/MATH] a continuous function. Show that [MATH]f[/MATH] has a fixed point in [MATH][0,1][/MATH]; that is, there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=x_0[/MATH].

My textbook solves this problem like this: "The function [MATH]g[/MATH] defined by [MATH]g(x)=f(x)-x[/MATH] with [MATH]x \in [0,1][/MATH] is continuous, and [MATH]g(0)=f(0) \geq 0[/MATH], and [MATH]g(1)=f(1)-1 \leq 0[/MATH]. Since [MATH]g[/MATH] has the intermediate value property, there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]g(x_0)=0[/MATH]."

I've tried to solve it myself, in the following way: by hypothesis [MATH]f[/MATH] is continuous in [MATH][0,1][/MATH], so [MATH]f[/MATH] satisfies the intermediate value theorem. So for all [MATH]\lambda \in [0,1][/MATH] there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=\lambda[/MATH]. Since [MATH]\lambda \in [0,1][/MATH] is arbitrary and [MATH]x_0 \in [0,1][/MATH], choosing [MATH]\lambda = x_0[/MATH] it follows that there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=x_0[/MATH].

Is my approach correct? Or is somehow wrong and I should've done as the textbook? Thank you.

 

[Dr.Peterson]

How do you conclude this:

So for all [MATH]\lambda \in [0,1][/MATH] there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=\lambda[/MATH].

There are certainly continuous functions f and numbers λ such that there is no such [MATH]x_0 \in [0,1][/MATH]; for example, what if [MATH]f(x) = 0[/MATH]?

I suspect you are either misunderstanding the intermediate value theorem, or assuming something like [MATH]f(0) = 0[/MATH] and [MATH]f(1) = 1[/MATH].

 

[lev888]

The problem is the following: let [MATH]f:[0,1] \to [0,1][/MATH] a continuous function. Show that [MATH]f[/MATH] has a fixed point in [MATH][0,1][/MATH]; that is, there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=x_0[/MATH].

My textbook solves this problem like this: "The function [MATH]g[/MATH] defined by [MATH]g(x)=f(x)-x[/MATH] with [MATH]x \in [0,1][/MATH] is continuous, and [MATH]g(0)=f(0) \geq 0[/MATH], and [MATH]g(1)=f(1)-1 \leq 0[/MATH]. Since [MATH]g[/MATH] has the intermediate value property, there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]g(x_0)=0[/MATH]."

I've tried to solve it myself, in the following way: by hypothesis [MATH]f[/MATH] is continuous in [MATH][0,1][/MATH], so [MATH]f[/MATH] satisfies the intermediate value theorem. So for all [MATH]\lambda \in [0,1][/MATH] there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=\lambda[/MATH]. Since [MATH]\lambda \in [0,1][/MATH] is arbitrary and [MATH]x_0 \in [0,1][/MATH], choosing [MATH]\lambda = x_0[/MATH] it follows that there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=x_0[/MATH].

Is my approach correct? Or is somehow wrong and I should've done as the textbook? Thank you.

I don't think your approach works. For any [MATH]\lambda \in [0,1][/MATH] there exists [MATH]x_0 \in [0,1][/MATH] such that [MATH]f(x_0)=\lambda[/MATH]. But, for each [MATH]\lambda[/MATH] there will be a different [MATH]x_0[/MATH]. Sure, you can make one of the [MATH]\lambda[/MATH]'s [MATH]x_0[/MATH], but the corresponding value in the domain will not necessarily be [MATH]x_0[/MATH].

 

[lev888]

How do you conclude this:

There are certainly continuous functions f and numbers λ such that there is no such [MATH]x_0 \in [0,1][/MATH]; for example, what if [MATH]f(x) = 0[/MATH]?

I suspect you are either misunderstanding the intermediate value theorem, or assuming something like [MATH]f(0) = 0[/MATH] and [MATH]f(1) = 1[/MATH].

Does [MATH]f:[0,1] \to [0,1][/MATH] mean that the range is [0,1]?

 

[Ozma]

@Dr.Peterson: Ok, maybe I get where is the mistake. The fact that the codomain of [MATH]f[/MATH] is [MATH][0,1][/MATH] doesn't mean that [MATH]f(0)=0[/MATH] and [MATH]f(1)=1[/MATH], and the intermediate value theorem only assures that a continuous function [MATH]g[/MATH] in an interval [MATH][a,b][/MATH] assumes all the values between [MATH]g(a)[/MATH] and [MATH]g(b)[/MATH]. So, when using this theorem, I must always refer to the image of [MATH]g[/MATH] (which I don't know here), not its codomain.

@lev888: I intuitively get why you are right and that in general the corresponding value in the domain will be different from [MATH]x_0[/MATH], but how can I deduce this from the logic of the statements? I mean, I know that [MATH]x_0[/MATH] depends on [MATH]\lambda[/MATH] because [MATH]\lambda[/MATH] is quantified before [MATH]x_0[/MATH], but from what can I deduce that for different [MATH]\lambda [/MATH]s I get different [MATH]x_0[/MATH]s?

Thanks to both of you for the answers.

 

[lev888]

... but from what can I deduce that for different [MATH]\lambda [/MATH]s I get different [MATH]x_0[/MATH]s?

From the definition of function:
"In mathematics, a function is a binary relation between two sets that associates to each element of the first set exactly one element of the second set." (from wikipedia)

Having the same x for different values of y violates the "exactly one" part of the definition.