Let f be a continuous function on the closed interval [0,1] with a range also contained...

Let f be a continuous function on the closed interval [0,1] with a range also contained in [0,1]. Prove that f that there exists an x in [0,1] such that f(x)=x. Is the same explanation still valid if f is not continuous?

Expert Solution

INTERMEDIATE VALUE THEOREM: Suppose that f : [a,b] |R be continuous such that f(a).f(b) < 0 . Then, there is an x in (a,b) such that f(x)=0.

f : [0,1] [0,1] is continuous.
If f(0)=0 or f(1)=1, we are done.
Assume therefore that f(0) 0 and f(1) 1.
Since f(x) is in [0,1] for all x in [0,1], hence f(0) > 0 and f(1) < 1.

Consider the function g : [0,1] [0,1] defined by, g(x)=f(x)-x for all x in [0,1].
Since f is continuous and the identity function i(x)=x is continuous, hence g is continuous.
Observe that:
g(0) = f(0) - 0 = f(0) > 0
g(1) = f(1) - 1 < 0 as f(1) < 1.
Thus, g(0) > 0 and g(1) < 0 which implies that g(0).g(1) < 0.
By the Intermediate Value Theorem, there exists an x in (0,1) such that g(x) = f(x)-x = 0 and hence, f(x)=x.

No. The claim is not true if f is not continuous. If f is not continuous, then f may not satisfy the Intermediate Value Property.
For example, consider the function f : [0,1] [0,1] defined by,
f(x) = { 1 if x is in [0,1/2]
0 if x is in (1/2,1]
Then, there is no x in [0,1] such that f(x)=x. The reason is simply that f is not continuous(because it is discontinuous at 1/2)

Colby Messinger answered 2 years ago

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