Let f:[0,1]->[0,1] be a continuous function. Prove that f has at least one fixed point: an a in A such that f(a)=a. Is the same true for discontinuous functions?

Jared Lowe

Jared Lowe

Answered question

2022-11-15

I'm trying to solve the below exercise.

Let f : [ 0 , 1 ] [ 0 , 1 ] be a continuous function. Prove that f has at least one fixed point: an a A such that f ( a ) = a. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if f ( 0 ) = 0 or f ( 1 ) = 1, the theorem is proved. Suppose not. Define g ( x ) = f ( x ) x, which is a continuous function from [ 0 , 1 ] to [ 0 , 1 ]. Then since f ( 0 ) 0, g ( 0 ) 0, so g ( 0 ) > 0. Since f ( 1 ) 1, g ( 1 ) < 0. By the intermediate value theorem, there exists a [ 0 , 1 ] such that g ( a ) = 0. So f ( a ) a = 0, so f ( a ) = a.
I am fairly sure that the result is not true for discontinuous functions, largely because I requiblack continuity of g to invoke the intermediate value theorem. I am having trouble finding a counterexample, however. Do I define a function piece-wise, with jumps at x = 0 or x = 1, to try to break the intermediate value theorem?

Answer & Explanation

dilettato5t1

dilettato5t1

Beginner2022-11-16Added 25 answers

Counterexamples abound: draw some pictures to find some. For one, try
f ( x ) = { 1 , x [ 0 , 1 ) 0 , x = 1.
Rigoberto Drake

Rigoberto Drake

Beginner2022-11-17Added 2 answers

The clue is in your proof. You defined g ( x ) = f ( x ) x, and found an x-intercept of g. Graphically speaking, a fixed point of the function f is any point where the graph of f crosses the line y = x. So all you have to do is draw a graph that does not cross this line.

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