For a function $f:D\rightarrow \mathbb{R}$, a solution of the equation
$\hspace{150pt}$$f(x)=x$, for $x\in D$
is called a fixed point of $f$. A fixed point corresponds to a point at which the graph of the function $f$ intersects the line $y=x$. If $f:[-1,1]\rightarrow \mathbb{R}$ is continuous, $f(-1)>-1$, and $f(1)<1$, show that $f:[-1,1]\rightarrow \mathbb{R}$ has a fixed point.
By the intermediate value theorem, since $f$ is continuous on $[-1,1]$, if $c$ is a number strictly between $f(-1)$ and $f(1)$ then there is a point $x_0$ in the open interval $(-1,1)$ at which $f(x_0)=c$.
Now, $c=x$ on $(-1,1)$, so for any $x$ such that $-1 < x < 1$, there exists an $x_0\in(-1,1)$ such that $f(x_0)=x$.
I'm not really sure what to do about the endpoints though. Any suggestions?
Best Answer
Consider the function $g\colon [-1,1] \to \mathbb R$ defined by $g(x) = f(x) - x$. Notice that since $f(x)$ and $x$ are continuous on $[-1,1]$, we know that $g$ is also continuous on $[-1,1]$. Furthermore, observe that: \begin{align*} g(-1) &= f(-1) - (-1) > -1 + 1 = 0 \\ g(1) &= f(1) - 1 < 1 - 1 = 0 \end{align*} Hence, since $g(1) < 0 < g(-1)$, it follows by the Intermediate Value Theorem that there exists some $c \in (-1,1)$ such that $g(c) = 0 \iff f(c) - c = 0 \iff f(c) = c$. So $c$ is a fixed point of $f$, as desired.