Suppose $a<b$ and $f:[a,b] \to [a,b]$ be continous. Suppose that $x \neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.
So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) \geq a$ that $g(a)=f(a)-a \geq 0$. Similarly, because $f(b) \leq b$ then $g(b)=f(b)-b \leq 0$.
Can we just use the fact that because $g(x)$ is continuous, $0 \in [g(b),g(a)]$, the IVT says there exists $c \in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.
How do we show that $c$ is in $(x,y)$??
We know that $g(x)=f(x)-x=y-x \neq 0$
and $g(y)=f(y)-y=x-y \neq 0$ but we don't know that those are in $(a,b)$?
Best Answer
Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.
Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.
Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.