# is this correct?

• Apr 6th 2010, 04:14 PM
tn11631
is this correct?
suppose that f: [a,b]->[a,b] is continuous. Prove that there is at least one fixed point in [a,b], that is, x such that f(x)=x

This is what I have does it make sense?

Since f is continuous, the function g(x) = f(x) - x is the difference of two continuous functions, hence g is continous on [a,b]

If f(a) = a, then x=a is the fixed point. If not, then f(a) is in (a,b]. So f(a) > a, and g(a) > 0.

If f(b) = b, then x=b is the fixed point. If not, then f(b) is in [a,b). So f(b) < b, and g(b) < 0.

since g is continuous, and g(b) < 0 < g(a), the intermediate value theorem assures the existence of a point x = z in [a,b] at which g(z) = 0. but then f(z) - z = 0, so f(z) = z and z is a fixed point. (with x replaced with z)
• Apr 6th 2010, 04:18 PM
Drexel28
Quote:

Originally Posted by tn11631
suppose that f: [a,b]->[a,b] is continuous. Prove that there is at least one fixed point in [a,b], that is, x such that f(x)=x

This is what I have does it make sense?

Since f is continuous, the function g(x) = f(x) - x is the difference of two continuous functions, hence g is continous on [a,b]

If f(a) = a, then x=a is the fixed point. If not, then f(a) is in (a,b]. So f(a) > a, and g(a) > 0.

If f(b) = b, then x=b is the fixed point. If not, then f(b) is in [a,b). So f(b) < b, and g(b) < 0.

since g is continuous, and g(b) < 0 < g(a), the intermediate value theorem assures the existence of a point x = z in [a,b] at which g(z) = 0. but then f(z) - z = 0, so f(z) = z and z is a fixed point. (with x replaced with z)

Parfait!