From a Fixed-Point Question to a Zero
A fixed point is an input that a function leaves unchanged. For example, if an object’s position is represented by \(x\), and a rule sends it to a new position \(f(x)\), then a fixed point is a position \(c\) for which the rule returns the same position: \(f(c)=c\).
At first, this equation may not look like a root-finding problem. But moving everything to one side gives \(f(c)-c=0\). The function \(g(x)=f(x)-x\) measures the difference between the output of \(f\) and the input. A fixed point of \(f\) is exactly a zero of \(g\).
The key is to check the sign of this difference at the endpoints of an interval. If the difference has opposite signs at the two endpoints, continuity and the Intermediate Value Theorem guarantee that it is zero somewhere between them.
The Fixed-Point Criterion
Let \(a<b\), and suppose \(f\) is continuous on the closed interval \([a,b]\). Define the displacement function \(g\) by \(g(x)=f(x)-x\). Since both \(f(x)\) and \(x\) are continuous, their difference \(g\) is continuous on \([a,b]\).
The condition says that the endpoint displacements have opposite signs, or that at least one endpoint displacement is zero. In inequalities, the strict opposite-sign cases are \(f(a)>a\) and \(f(b)<b\), or \(f(a)<a\) and \(f(b)>b\).
Set \(g(x)=f(x)-x\), so \(g(x)=0\) is equivalent to \(f(x)=x\).
Confirm that \(g\) is continuous on the whole interval and that \(g(a)\) and \(g(b)\) have opposite signs or include zero.
A continuous function with endpoint values on opposite sides of zero must equal zero at some point \(c\in[a,b]\). Then \(f(c)=c\).
Why the Argument Works
Here is the full proof. If \(g(a)=0\), then \(f(a)-a=0\), so \(f(a)=a\), and \(a\) itself is a fixed point. If \(g(b)=0\), the same reasoning gives the fixed point \(b\).
Otherwise, the hypothesis \(g(a)g(b)\le 0\) implies \(g(a)g(b)<0\). Thus one endpoint value of \(g\) is negative and the other is positive. Because \(g\) is continuous on \([a,b]\), the Intermediate Value Theorem says that \(g\) takes every value between \(g(a)\) and \(g(b)\). In particular, it takes the value zero: there is some \(c\in[a,b]\) with \(g(c)=0\). By the definition of \(g\),
Therefore \(c\) is a fixed point. The idea is to track whether \(f(x)\) lies above or below \(x\). If the ordering switches between the endpoints, continuity forces at least one point where the two values agree.
A Useful Special Case
A common way to obtain the endpoint inequalities is to know that \(f\) maps the interval into itself. This means that \(f(x)\in[a,b]\) for every \(x\in[a,b]\). In particular, \(f(a)\ge a\) and \(f(b)\le b\). Therefore
The fixed-point criterion applies. This gives a useful consequence of the theorem: every continuous function from a closed interval into that same interval has at least one fixed point. For the criterion itself, however, it is enough to verify the endpoint inequalities; it is not necessary to prove that every output of \(f\) lies in \([a,b]\).
Worked Examples
Worked Example: An Affine Function on an Interval
Show that \(f(x)=2-\frac{x}{2}\) has a fixed point in \([0,2]\), and find it.
First, \(f\) is continuous because it is an affine function. Calculate the endpoint displacements:
The displacement function is \(g(x)=2-\frac{x}{2}-x=2-\frac{3x}{2}\). Its endpoint values have opposite signs, so the theorem guarantees a zero in \([0,2]\). To find that zero, solve:
Check by substitution:
Since \(\frac{4}{3}\in[0,2]\) and \(f(\frac{4}{3})=\frac{4}{3}\), it is a fixed point.
Worked Example: A Nonlinear Function
Show that \(f(x)=1-x^2\) has a fixed point in \([0,1]\), and determine its value.
The polynomial \(f\) is continuous. At the endpoints,
Thus the theorem guarantees a fixed point. To calculate it, solve \(f(x)=x\):
Using the quadratic formula gives
The root \(\frac{-1-\sqrt{5}}{2}\) is negative, so it is not in \([0,1]\). The other root is \(c=\frac{-1+\sqrt{5}}{2}\). Since \(2<\sqrt{5}<3\), it follows that \(0<c<1\).
Check the fixed-point equation using \(c^2+c-1=0\), which follows from the quadratic equation above:
Therefore \(c=\frac{-1+\sqrt{5}}{2}\) is a fixed point in the required interval.
Worked Example: A Function That Maps an Interval into Itself
Consider \(f(x)=\frac{x+1}{3}\) on \([0,1]\). First verify that its outputs stay in the interval. If \(0\le x\le1\), then \(1\le x+1\le2\), so
In particular, \(f(x)\in[0,1]\). The endpoint displacements are
The function is continuous, so the criterion guarantees a fixed point. Solve the equation directly:
Check by substitution:
The value \(\frac{1}{2}\) belongs to \([0,1]\) and satisfies the defining equation, as required.
A Common Sign-Checking Mistake
The sign test applies to \(g(x)=f(x)-x\), not directly to \(f(x)\). Confusing these functions can lead to a false conclusion. For example, for \(f(x)=2-\frac{x}{2}\) on \([0,2]\), both endpoint outputs are positive:
It would be incorrect to conclude that the fixed-point criterion cannot be used because these two values have the same sign. The relevant values are the displacements:
These have opposite signs, so the theorem applies. Indeed, the fixed point is \(\frac{4}{3}\). Always subtract the input from the output before checking endpoint signs.
What the Theorem Guarantees
The criterion is an existence argument. It guarantees that at least one point in the interval satisfies \(f(c)=c\); it does not, by itself, tell you the value of that point. In examples, an equation may also be solvable directly, but in other settings the theorem can establish existence without providing an explicit formula.
For a fixed-point proof, check three things carefully:
- The interval is closed and has endpoints \(a<b\).
- The function \(f\), and therefore \(g(x)=f(x)-x\), is continuous on the whole interval.
- The endpoint displacements have opposite signs or one is zero.
When these conditions hold, the conclusion follows from the Intermediate Value Theorem. The central move is simple but important: instead of asking directly where \(f\) equals its input, study the continuous difference between them.
Check Your Understanding
For each question, identify the displacement function and use the endpoint information carefully.
- State the equation that characterizes a fixed point of \(f\).
- For \(f(x)=3-x\) on \([1,3]\), calculate the endpoint displacements and determine whether the criterion guarantees a fixed point.
- Why is continuity of \(f\) on the entire interval needed in the fixed-point criterion?
- If \(f\) maps \([a,b]\) into \([a,b]\), what inequalities follow for \(f(a)-a\) and \(f(b)-b\)?
- What is wrong with checking only whether \(f(a)\) and \(f(b)\) have opposite signs?
