Fixed Points: Intermediate Value Theorem

Question

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?

$c$ is a fixed point (though not by the fixed point definition). — TheMobiusLoops, Nov 06 '13 at 06:12
Just like what you did, you pick $x\in (-1,1)$, the IVT gives you $x_0$ which might not be $x$. You need to use IVT, but not to $f$. — , Nov 06 '13 at 06:18
I'm tempted to say rotate $\mathbb{R}^2$ by $-\frac{\pi}{4}$ about the origin and then use IVT. EDIT: John's method is better — Tim Ratigan, Nov 06 '13 at 06:20
Possible duplicate of Show that a continuous function has a fixed point — Jigao, Dec 16 '18 at 13:12

score 6 · Accepted Answer · answered Nov 06 '13 at 06:22

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.

Fixed Points: Intermediate Value Theorem

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