Turn a Fixed-Point Equation into a Sign Test
A point \(c\) is a fixed point of a function \(f\) when \(f(c)=c\). That equation can be difficult to solve directly. A useful alternative is to subtract \(x\) from the function value and study \(g(x)=f(x)-x\): the fixed points of \(f\) are exactly the zeros of \(g\). If \(g\) is continuous and has opposite signs at two points, it must take the value zero somewhere between them.
Throughout, \(\mathbb{R}\) has its standard topology, and every interval \([a,b]\) has the subspace topology inherited from \(\mathbb{R}\). Continuity on an interval means continuity for this topology. The function \(f\) in the test need only be continuous and real-valued on the interval; it is not assumed to map the interval into itself.
If \(f\) is continuous, then \(g\) is continuous. Indeed, the identity function \(x\mapsto x\) is continuous, and at any point \(c\), continuity of \(f\) gives, for any \(\varepsilon>0\), a \(\delta>0\) such that \(|f(x)-f(c)|<\varepsilon/2\) when \(|x-c|<\delta\). Shrinking \(\delta\) if necessary so that \(\delta\leq\varepsilon/2\), we obtain \(|g(x)-g(c)|\leq |f(x)-f(c)|+|x-c|<\varepsilon\). This is the continuity needed for the sign test.
The Fixed-Point Sign Test
The weak inequality includes the cases where one of the two tested points is already fixed. The strict-sign conclusion is stronger: if neither tested point is fixed and the signs differ, the fixed point lies strictly between them. If \(u=v\), the product condition can hold only when \(f(u)-u=0\), so the conclusion still holds, with \(c=u\).
Proof. Set \(g(x)=f(x)-x\), which is continuous by the estimate above. First suppose \(g(u)<0<g(v)\). Define \(S=\{x\in[u,v]:g(x)\leq 0\}\). The set \(S\) is nonempty because \(u\in S\), and it is bounded above by \(v\). By completeness of \(\mathbb{R}\), \(c=\sup S\) exists.
Continuity and \(g(u)<0\) give points of \(S\) strictly larger than \(u\), so \(c>u\). Continuity and \(g(v)>0\) give a left-hand interval ending at \(v\) on which \(g\) is positive. No point of that interval belongs to \(S\), so \(c<v\).
If \(g(c)<0\), continuity gives a neighborhood of \(c\) on which \(g\) remains negative. Since \(c<v\), that neighborhood contains some \(x\in(c,v]\), and then \(x\in S\), contradicting that \(c\) is an upper bound for \(S\). If \(g(c)>0\), continuity gives a neighborhood of \(c\) on which \(g\) is positive. Since \(c>u\) and \(c=\sup S\), there must be some \(s\in S\) in that neighborhood with \(s\leq c\): otherwise a point strictly below \(c\) in the neighborhood would be an upper bound for \(S\). This contradicts positivity there. Thus \(g(c)=0\), with \(u<c<v\).
If instead \(g(u)>0>g(v)\), apply the argument just given to the continuous function \(-g\); its zero is also a zero of \(g\) and lies in \((u,v)\). Finally, if either endpoint value is zero, that endpoint itself is a zero. These cases exhaust \(g(u)g(v)\leq 0\). Since \(g(c)=0\) means \(f(c)=c\), the result follows. \(\square\)
The central idea is a sign change, not a particular formula for \(f\). The proof uses the least-upper-bound property of the real numbers to locate the zero. Equivalently, the sign test is the sign-change consequence of the Intermediate Value Theorem, applied to \(g=f-\operatorname{id}\).
Worked Applications
Worked Example: A Fixed Point at an Endpoint
Let \(f:[0,1]\to\mathbb{R}\) be \(f(x)=2x-1\). Use the sign test on \([0,1]\), and identify the fixed point it guarantees.
Here \(g(x)=f(x)-x=(2x-1)-x=x-1\). At the left endpoint, \(g(0)=0-1=-1\), so \(f(0)-0=-1<0\). At the right endpoint, \(g(1)=1-1=0\), so the product of the endpoint differences is \((-1)\cdot0=0\). The theorem guarantees a fixed point in \([0,1]\); in this case the right endpoint is already one.
Check: \(f(1)=2(1)-1=1\), so \(f(1)=1\). In fact, solving \(f(x)=x\) gives \(2x-1=x\), hence \(x=1\), confirming that this is the only fixed point. The signs are not strictly opposite, so the theorem does not place the fixed point in the open interval.
Worked Example: A Fixed Point Strictly Between Two Test Points
Let \(f:[1,3]\to\mathbb{R}\) be \(f(x)=x^2-5x+8\). Apply the sign test using the endpoints of the interval, then verify the fixed point exactly.
The difference function is \(g(x)=f(x)-x=x^2-6x+8=(x-2)(x-4)\). At \(x=1\), \(g(1)=(1-2)(1-4)=(-1)(-3)=3>0\). At \(x=3\), \(g(3)=(3-2)(3-4)=(1)(-1)=-1<0\). The endpoint differences have strictly opposite signs, so the theorem gives a fixed point in \((1,3)\).
Factoring \(g\) shows its zeros are \(2\) and \(4\), and only \(2\) lies in \((1,3)\). Substitution checks the answer: \(f(2)=2^2-5(2)+8=4-10+8=2\). Thus \(2\) is a fixed point in the tested interval.
Check: the computed endpoint values are \(f(1)=1-5+8=4\) and \(f(3)=9-15+8=2\), so \(f(1)-1=4-1=3\) and \(f(3)-3=2-3=-1\), as used above.
Worked Example: Two Fixed Points Despite Equal Endpoint Signs
Let \(f:[0,1]\to\mathbb{R}\) be \(f(x)=x^2+\frac{3}{16}\). Calculate the endpoint signs and find all the fixed points in the interval.
Subtracting \(x\) gives \(g(x)=x^2-x+\frac{3}{16}=(x-\frac14)(x-\frac34)\). At \(x=0\), \(g(0)=\frac{3}{16}>0\). At \(x=1\), \(g(1)=1-1+\frac{3}{16}=\frac{3}{16}>0\). The signs at the two endpoints agree, so the sign test makes no conclusion.
The factorization nevertheless gives two zeros, \(x=\frac14\) and \(x=\frac34\), both in \([0,1]\). Verify each directly: \(f(\frac14)=(\frac14)^2+\frac{3}{16}=\frac{1}{16}+\frac{3}{16}=\frac14\), and \(f(\frac34)=(\frac34)^2+\frac{3}{16}=\frac{9}{16}+\frac{3}{16}=\frac{12}{16}=\frac34\).
Check: both endpoint function values are \(f(0)=\frac{3}{16}\) and \(f(1)=1+\frac{3}{16}=\frac{19}{16}\). Thus \(f(0)-0=\frac{3}{16}\) and \(f(1)-1=\frac{19}{16}-1=\frac{3}{16}\), confirming the equal positive signs.
What the Test Does—and Does Not—Say
The condition is sufficient for existence: a zero difference at one test point or opposite signs at two test points guarantees a fixed point between them. It is not a necessary condition for a fixed point somewhere in the interval. The last example has two fixed points even though the differences at \(0\) and \(1\) are both positive.
Common mistake. From \(g(0)>0\) and \(g(1)>0\), one might incorrectly conclude that \(g(x)>0\) throughout \([0,1]\), and therefore that \(f\) has no fixed point there. In the last example, this would assert that \(x^2-x+\frac{3}{16}\) never vanishes. But the exact factorization \(g(x)=(x-\frac14)(x-\frac34)\) shows \(g(\frac14)=g(\frac34)=0\). The endpoint signs alone do not determine the signs in between.
There is a second limitation: the sign test guarantees at least one fixed point, not exactly one. To establish uniqueness, one needs additional information beyond the endpoint signs. Also, this result does not assume \(f([a,b])\subseteq[a,b]\); it concerns a real-valued continuous function and a sign change of \(f(x)-x\). Requiring a function to map the interval into itself is a different hypothesis used in other fixed-point results.
Check Your Understanding
For each question, begin by forming the difference function \(g(x)=f(x)-x\).
- For \(f:[0,2]\to\mathbb{R}\), \(f(x)=3-x\), apply the sign test at \(0\) and \(2\). Is the guaranteed fixed point an endpoint or an interior point?
- Suppose \(f:[a,b]\to\mathbb{R}\) is continuous and \(f(a)-a<0<f(b)-b\). What does the test guarantee, and where must the fixed point lie?
- Can equal positive values of \(f(u)-u\) and \(f(v)-v\) rule out a fixed point between \(u\) and \(v\)? Explain using the worked quadratic example.
- Why does the sign test still apply when \(f\) does not map \([a,b]\) into itself?
- If \(u=v\) and \((f(u)-u)(f(v)-v)\leq0\), what must be true about \(u\)?