Prove that a function has a fixed point

Amer

Active member
Let F be a continuous function from [a,b] onto [a,b] prove that F has a fixed point
in the interval [a,b]

it is clear for me by drawing the product of [a,b]x[a,b] any line which pass through all the image should intersect with the diagonal but i cant make a mathematical proof.

I tried by looking at
x-f(x) and suppose it is a postive or negative function and reach a contradiction

any hints

Evgeny.Makarov

Well-known member
MHB Math Scholar
You should use the intermediate value theorem saying that if g(a) <= 0 and g(b) >= 0 where g is some continuous function on [a, b], then g(c) = 0 for some c ∈ [a, b]. Looking at your picture, can you construct such a function g from F and the diagonal?

---------- Post added at 22:50 ---------- Previous post was at 22:46 ----------

I tried by looking at x-f(x) and suppose it is a postive or negative function and reach a contradiction
There is no need to suppose. Since a ≤ f(x) ≤ b for all x ∈ [a, b] it is clear that x - f(x) changes sign on [a, b].

• Amer

Amer

Active member
suppose that F dose not intersect with the diagonal that means x- F(x) is either positive or negative
let g(x) = x - F(x)
suppose that it is positive g(a) = a - F(a)

0 ≤ a- F(a)
F(a) ≤ a
F(a) = a fixed point

now taking other case, is it ok ? thanks again

Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
suppose that F dose not intersect with the diagonal that means x- F(x) is either positive or negative
let g(x) = x - F(x)
suppose that it is positive g(a) = a - F(a)
Do you mean, "Suppose that g(a) = a - F(a) is positive"?

0 ≤ a- F(a)
"Positive" means strictly greater than zero.

now taking other case
What do you mean by this?

I think you have the right idea, but it has to be expressed properly.

• Amer

Amer

Active member
ok I come up with the complete proof
let g(x) = x - F(x)
since F(x) is onto there exist c in [a,b] such that F(c) = b, if c=b then we are finished
if c<b then g(c) = c - F(c) = c- b < 0

same there exist e in [a,b] such that F(e) = a, if e=a we are finished if e> a

g(e) = e - F(e) = e - a > 0

since g(x) is continuous function by intermediate theorem g(x) has a zero in [a,b]

Evgeny.Makarov

Well-known member
MHB Math Scholar
This is correct; however, F: [a, b] -> [a, b] has a fixpoint even if it is not onto, just continuous. Why didn't you use the hint given above?
Since a ≤ f(x) ≤ b for all x ∈ [a, b] it is clear that x - f(x) changes sign on [a, b].
We have g(a) = a - f(a) ≤ 0 and g(b) = b - f(b) ≥ 0. Then apply the IVT.

CaptainBlack

Well-known member
Let F be a continuous function from [a,b] onto [a,b] prove that F has a fixed point
in the interval [a,b]

it is clear for me by drawing the product of [a,b]x[a,b] any line which pass through all the image should intersect with the diagonal but i cant make a mathematical proof.

I tried by looking at
x-f(x) and suppose it is a postive or negative function and reach a contradiction

any hints
As wıth other posts consıder $$g(x)=x-f(x)$$. If $$g(a)g(b) \le 0$$ you are done (eıther usıng the ıntermedıate value theorem of usıng the bısectıon method whıch converges to a zero of a contınuous functıon on an ınterval wıth thıs property or trıvıally ıf zero). So you need only show that $$g(a)$$ and $$g(b)$$ cannot both be of the same sıgn.

CB

Last edited:
• Amer

Amer

Active member
This is correct; however, F: [a, b] -> [a, b] has a fixpoint even if it is not onto, just continuous. Why didn't you use the hint given above?We have g(a) = a - f(a) ≤ 0 and g(b) = b - f(b) ≥ 0. Then apply the IVT.
if it is just continuous ?
I used the hint, maybe not