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.
Best Answer
There are two cases: