Validity of this proof that any continuous function with domain and range in [0,1] must have a fixed point.

Question

The following proof was given in a solutions manual to a question asking to prove that a continuous function with domain and range in $[0,1]$ must have a fixed point:

Consider the function $F(x) = f(x) - x$, where $f$ is any continuous function with domain $[0,1]$ and range in $[0,1]$. We shall prove that f has a fixed point. Now if $f(0) = 0$ then we are done: $f$ has a fixed point (the number $0$), which is what we are trying to prove. So assume $f(0) \neq 0$. For the same reason we can assume that $f(1) \neq 1$. Then $F(0) = f(0) > 0$ and $F(1) = f(1) - 1 < 0$. So by the Intermediate Value Theorem, there exists some number $c$ in the interval $(0,1)$ such that $F(c) = f(c) - c = 0$. So $f(c) = c$, and therefore $f$ has a fixed point.

However my issue with the above proof is that $F(1) = f(1) - 1$ is not always less than zero (as stated above) as in the case of function $y = x$ at $(1,1)$ the function $F(1) = 1 - 1 = 0$

likewise $F(0) = f(0) - 0$ is not always greater than zero (as stated above) as in the case of function $y = x$ at $(0,0)$ the function $F(0) = 0 - 0 = 0$

I know they explicitly asked us to assume that $f(0) \neq 0$ and $f(1) \neq 1$ but in the case of a proof how can we work we choose to ignore the points that seem to me can negate the proof at $(1,1)$ and $(0,0)$, especially when we are dealing with a continuous function? I agree with the rest of the proofs and understand it and even acknowledge that by following their assumptions, the proof is flawless, but what I want to understand is how it is justified. Thank you.

Answer 1

There are two cases:

1. if either $f(0) = 0$ or $f(1) = 1$: you already have a fixed point! This is what you are missing.
2. otherwise you can use the intermediate values theorem.

Answer 2

The point is that if $f(1)=1$ means $1$ is a fixed point of $f$. So in that case further argument is unnecessary. If $0$ and $1$ are not fixed points, then a further argument is required. So the proof is working only with this case - $0$ and $1$ are already dealt with.

Answer 3

The whole point of the proof is splitting the case into three scenarios:

• If $F(1)$ is not less than zero, then $f(1)=1$ and a fixed point exists.
• If $F(0)$ is not positive, then $f(0)=0$ and a fixed point exists.
• If $F(0)$ is positive and $F(1)$ is negative, then a fixed point exists by the IVT.

Note: When you get to point three, you are actually proving the statement:

Say $f:[0,1]\to[0,1]$ is a function for which $0$ and $1$ are not fixed points. Then there exists such a value $c\in(0,1)$ that $f(c)=c$.