Elementary proof of fixed point theorem.

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
ArcanaNoir
Messages
778
Reaction score
4
I'd like someone to check this proof out for violations of math law. It seems like hackery to me, but then so does a lot of what my professor says, so maybe it's not. If it's flawed, just tell me why. Please don't suggest other ways to prove this, that's my job this weekend.

Thanks :)

p.s. This is my first post using [itex]\LaTeX[/itex], since my discovery that it was there all along (used to see the raw text).

Homework Statement


[itex]f: [0,1] \to [0,1][/itex] and [itex]f[/itex] is continuous [itex]\Rightarrow \exists x \in [0,1] : f(x)=x[/itex]

Homework Equations


Anything elementary calculus. Trying to keep it as simple as possible.

The Attempt at a Solution



let [itex]g(x)=x[/itex]
Then [itex]g(0) \le f(x) \le g(1)[/itex]
or, [itex]g(a) \le f(x) \le g(b)[/itex] where [itex][a,b]=[0,1][/itex]

[tex]\lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x[/tex]
therefore
[tex]\lim_{x\to x}{f(x)} = x[/tex]

Since [itex]f(x)[/itex] takes on the same value as [itex]g(x)[/itex] at some point [itex]x[/itex], [itex]f(x)=g(x)=x[/itex] at this point.

So, for some [itex]x \in [0,1], f(x)=x[/itex]
 
Last edited:
on Phys.org
ArcanaNoir said:
I'd like someone to check this proof out for violations of math law. It seems like hackery to me, but then so does a lot of what my professor says, so maybe it's not. If it's flawed, just tell me why. Please don't suggest other ways to prove this, that's my job this weekend.

Thanks :)

p.s. This is my first post using [itex]\LaTeX[/itex], since my discovery that it was there all along (used to see the raw text).

Homework Statement


[itex]f: [0,1] \to [0,1][/itex] and [itex]f[/itex] is continuous [itex]\Rightarrow \exists x \in [0,1] : f(x)=x[/itex]


Homework Equations


Anything elementary calculus. Trying to keep it as simple as possible.


The Attempt at a Solution


This "proof" is flawed. Comments on each line below.

let [itex]g(x)=x[/itex]
Then [itex]g(0) \le f(x) \le g(1)[/itex]
No point renaming x to something else. This just says 0<f(x)<1
or, [itex]g(a) \le f(x) \le g(b)[/itex] where [itex][a,b]=[0,1][/itex]

And this just says the same thing.

[tex]\lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x[/tex]
But a and b are constants. But anyway, x is presumably known to be continuous.

therefore
[tex]\lim_{x\to x}{f(x)} = x[/tex]
Strange way to write a limit and f(x) = x doesn't follow from anything you have done.
Since [itex]f(x)[/itex] takes on the same value as [itex]g(x)[/itex] at some point [itex]x[/itex], [itex]f(x)=g(x)=x[/itex] at this point.

So, for some [itex]x \in [0,1], f(x)=x[/itex]

You aren't close. You said no suggestions. Do you want a hint?
 
No thanks on the hint. I guess it's back to the drawing board. Thanks for the input :)