Why Recursive Limits Lead to Fixed Points
A recurrence defines each term using the one before it. For example, a rule of the form \(a_{n+1}=g(a_n)\) says to apply the same function \(g\) repeatedly. If the resulting sequence settles toward a limit, it is natural to ask what that limit must satisfy.
The key observation is that, for a convergent sequence, shifting the index by one does not change its limit. Continuity also ensures that applying \(g\) to terms approaching a point produces values approaching \(g\) of that point. Together, these facts give an equation for any possible limit.
Recurrences and Fixed Points
Let \(D\) be a subset of the real numbers and let \(g:D\to D\). A sequence \((a_n)\) follows the recurrence defined by \(g\) if \(a_n\in D\) and
for every \(n\in\mathbb{N}\). A fixed point of \(g\) is a number \(L\in D\) such that
A fixed point stays unchanged when \(g\) is applied. It is important to distinguish this property of the function from convergence of a particular sequence: a function can have fixed points even when a sequence generated by it does not converge.
Continuity at \(L\) is also important. In this setting, saying that \(g\) is continuous at \(L\) means that inputs in \(D\) approaching \(L\) have outputs approaching \(g(L)\). In particular, if \(a_n\to L\), continuity gives \(g(a_n)\to g(L)\). The continuity result for functions of convergent sequences gives this implication.
The Necessary Equation
The theorem gives a necessary condition, not a guarantee. Its proof uses the recurrence, the fact that shifting a convergent sequence does not change its limit, and continuity at the proposed limit.
Because \(a_n\to L\), the sequence \(a_{n+1}\) also tends to \(L\). Removing the first term does not change the limit.
The recurrence gives \(a_{n+1}=g(a_n)\). Since \(a_n\to L\) and \(g\) is continuous at \(L\), \(g(a_n)\to g(L)\).
The same sequence \(a_{n+1}=g(a_n)\) tends both to \(L\) and to \(g(L)\). A convergent sequence has only one limit, so \(L=g(L)\).
For completeness, uniqueness of a sequence limit follows from the triangle inequality. If a sequence had distinct limits \(A\) and \(B\), take \(\varepsilon=|A-B|/3\). Eventually a term \(x_n\) would satisfy both \(|x_n-A|<\varepsilon\) and \(|x_n-B|<\varepsilon\). Then \(|A-B|\leq |A-x_n|+|x_n-B|<2|A-B|/3\), a contradiction. Thus the two limits in the proof must be equal.
Worked Examples: Finding Candidate Limits
Worked Example: A square-root recurrence
Suppose \(a_1=1\) and \(a_{n+1}=\sqrt{2+a_n}\). Find the limit that would be required if this sequence converges.
Every term is nonnegative: \(a_1=1\geq 0\), and a square root is nonnegative whenever it is defined. If the sequence converges to \(L\), then \(L\geq 0\). The function \(g(x)=\sqrt{2+x}\) is continuous for \(x\geq 0\), so the theorem requires
Both sides are nonnegative, so squaring preserves equivalence here:
The algebraic roots are \(L=2\) and \(L=-1\). The second violates \(L\geq 0\), and also fails the original equation because \(-1\neq\sqrt{1}=1\). Checking \(L=2\) in the original equation gives \(\sqrt{2+2}=2\), so it is a valid fixed-point candidate. This calculation identifies a necessary candidate; it does not prove that the sequence converges to it.
Worked Example: A recurrence with no limit
Let \(a_1=0\) and \(a_{n+1}=2-a_n\). The defining function is \(g(x)=2-x\), which is continuous. Its fixed-point equation is
Checking the candidate gives \(g(1)=2-1=1\). But the actual terms are
The sequence alternates between \(0\) and \(2\). Its odd-indexed terms are all \(0\), while its even-indexed terms are all \(2\); these subsequences have different limits. Therefore the original sequence does not converge. The fixed point \(1\) is a candidate required of any limit, not evidence that this particular sequence has one.
Worked Example: A rational recurrence
Let \(a_1=1\) and \(a_{n+1}=1/(1+a_n)\). The terms are positive, and \(g(x)=1/(1+x)\) is continuous for \(x\neq -1\), including every nonnegative input. If the sequence converges to \(L\), then \(L\geq 0\) and
Since \(L\geq 0\), \(1+L\neq 0\), so multiplying by \(1+L\) is valid:
The quadratic formula gives
The root \((-1-\sqrt{5})/2\) is negative, so it cannot be the limit of this positive sequence. The remaining candidate is \(L=(-1+\sqrt{5})/2\). Substitution verifies it: the quadratic equation gives \(L(1+L)=1\), and \(1+L\neq 0\), hence \(L=1/(1+L)\). This verifies the fixed-point equation, not convergence of the sequence.
Why Continuity Is Required
The theorem depends on continuity at the limit. Without it, the recurrence can have a convergent sequence whose limit is not a fixed point. Here is a direct example.
Worked Example: A discontinuous recurrence
Let \(D=\{0\}\cup\{1/n:n\in\mathbb{N}\}\). Define \(g:D\to D\) by \(g(0)=1\) and \(g(1/n)=1/(n+1)\). Set \(a_n=1/n\). Then, for every \(n\in\mathbb{N}\),
Also \(a_n=1/n\to 0\), but \(g(0)=1\neq 0\). Thus the sequence converges to a value that is not a fixed point of \(g\). The reason the theorem does not apply is that \(g\) is not continuous at \(0\): along inputs \(1/n\to 0\), the outputs \(g(1/n)=1/(n+1)\to 0\), not \(g(0)=1\).
Checking the recurrence and limit confirms the example: each term \(1/n\) maps to the next term \(1/(n+1)\), the limit is \(0\), and the function’s value at that limit is \(1\). This shows why continuity is a condition of the fixed-point conclusion, not an optional detail.
A Fixed-Point Calculation Is Not a Convergence Proof
Solving \(L=g(L)\) is often useful because it reduces a question about a recursive sequence to an algebra problem. But the theorem only says that if the sequence converges, its limit must be among the solutions. It does not say that the sequence converges, nor that every fixed point is approached from a chosen starting value.
The alternating example makes the logical gap concrete. Its fixed-point equation has the solution \(L=1\), but its terms keep switching between \(0\) and \(2\). The calculation “\(L=2-L\), so \(L=1\), therefore \(a_n\to1\)” is wrong: the equation was derived under the assumption that a limit exists, and solving it does not establish that assumption. The correction is to treat \(1\) only as a necessary candidate, then separately investigate whether the sequence actually converges.
Another useful caution is to check the original equation after algebraic manipulation. Squaring, multiplying, or cancelling can introduce candidates that do not satisfy the starting equation, or may be invalid under certain conditions. For example, squaring \(L=\sqrt{2+L}\) without first considering the sign can obscure the requirement \(L\geq0\). Substitute proposed roots back into the original fixed-point equation and check that they lie in the relevant domain.
Why the Fixed-Point Equation Matters
Recursive definitions can make direct limit calculations difficult because each term depends on the preceding term. The fixed-point equation gives a compact necessary condition that can narrow the possibilities before a separate convergence argument is attempted. It also exposes conditions that must not be skipped: continuity at the proposed limit and compatibility with the sequence’s domain and possible values.
When the equation has several solutions, it does not choose among them by itself. Information about the actual sequence may rule out some candidates, as nonnegative terms ruled out negative roots in the examples. But identifying a remaining candidate still does not prove that the sequence approaches it. The existence of a limit and the value of that limit are distinct questions.
Check Your Understanding
Answer each question using the assumptions stated in the tutorial.
- If \(a_{n+1}=g(a_n)\), \(a_n\to L\), and \(g\) is continuous at \(L\), what equation must \(L\) satisfy?
- Why does the shifted sequence \(a_{n+1}\) have the same limit as \(a_n\)?
- For the recurrence \(a_{n+1}=2-a_n\) with \(a_1=0\), what is the fixed point, and why does the sequence fail to converge?
- Why must proposed roots of a fixed-point equation be checked in the original equation?
- Which hypothesis of the fixed-point theorem fails in the discontinuous recurrence example?
