ATTRACTIVE PROPERTIES VIA GENERALIZED DERIVATIVES OF CONTINUOUS FUNCTIONS

BY

A. M. BRUCKNER and Z. BUCZOLICH *

ABSTRACT. Let I be a closed interval in R and let x_{0} ∈ I. We obtain a condition on a set E having x_{0} as a point of accumulation that guarantees that any continuous function f : I → I that has x_{0} as a fixed point and posesses a derivative f′_{E}(x_{0}) relative to E at x_{0} with |f′_{E}(x_{0})| < 1 has some point x ≠ x_{0} attracted to x_{0}. We also obtain conditions that guarantee that the point x can be chosen in E.

1. Introduction. Let f be a continuous function mapping a closed interval I into itself and let x_{0} be a fixed point of f. An elementary theorem asserts that if |f′(x_{0})| < 1, then there is a neighborhood U of x_{0} such that for each x ∈ U, the sequence {f^{n}(x)} of iterates of x under f converges to x_{0}. Here f^{0}(x) = x and f^{n}(x) = f^{n−1}(x) n = 1, 2, ....

A variant of this theorem can be obtained by replacing the condition on the derivative by a condition on the difference- quotient: if there is a number d ∈ (0, 1) such that

−d < \frac{f(x) − f(x_{0})}{x − x_{0}} < d  (1)

for all x ∈ U, then lim_{n→∞} f^{n}(x) = x_{0} for all x ∈ U.

Suppose, now, that we replace the full neighborhood U of x_{0} with some other set E. We can ask whether the condition |f′_{E}(x_{0})| < 1 where f′_{E}(x_{0}) denotes the derivative of f with respect to E at x_{0}, or the condition (1) holding for all x ∈ E, still implies some sort of attraction to the fixed point x_{0}. The strength of the conclusion will depend, of course, on the set E. It is, perhaps surprising that porosity conditions, rather than density conditions, are the relevant ones.

* This paper was prepared while the second listed author visited University of California at Davis.

2. Main Results.

Suppose x_{0} is a fixed point of f and 0 < f′(x_{0}) < 1. It is then clear that for x sufficiently close to x_{0} the sequence {f^{n}(x)} approaches x_{0} monotonically. Suppose we weaken the assumption to 0 < f′_{ap}(x_{0}) < 1. This means that there is a set E having x_{0} as a point of density such that

0 < lim_{x→x_{0};x∈E} \frac{f(x) − f(x_{0})}{x − x_{0}} < 1.

If x ∈ E and f^{n}(x) ∈ E for all n = 1, 2, ..., then f^{n}(x) → x_{0}. Perhaps the fact that E has x_{0} as a density point suggests there must be an x attracted to x_{0} whose orbit lies entirely in E. We shall see that this is not necesseraly the case, although there will be a point not necesseraly in E, whose orbit approaches x_{0}. Our goal in this section is to obtain two theorems. The first provides a condition on a set E that guarantees that any continuous function such that 0 < lim_{x→x_{0};x∈E} \frac{f(x)−f(x_{0})}{x−x_{0}} = d < 1 has some point x (not necessarily in E) whose orbit is attracted to the fixed point x_{0} without landing on x_{0}. The second theorem provides a condition that guarantees that if 0 < lim_{x→x_{0};x∈E} \frac{f(x)−f(x_{0})}{x−x_{0}} < 1 then there is a point x ∈ E whose orbit is attracted to x_{0} without landing on x_{0}. Actually each theorem reveals a bit more.

We begin with an example that may be instructive.

Example 2.1. Assume that 0 ∈ int(I), where int(I) denotes the interior of the interval I. Then there exists a continuous function f : I → I and a set E such that f(0) = 0, 0 is a right density point of E, lim_{x↘0;x∈E} \frac{f(x)}{x} = 1/2, and |f^{n}(x)| > δ_{x} > 0 n = 1, 2, ..., for every x ∈ E, x > 0. The notation x ↘ y means that x converges to y from the right.

Verification of Example 2.1. Before turning to the details we want to sketch the idea of our example. We split each interval [2^{−n}, 2^{−n+1}] into 2^{n} subintervals I_{n,m} n = 1, 2, ..., m = 1, 2, ..., 2^{n} and we choose small subintervals J_{n,m} about the endpoints of I_{n,m}. The set E will be (0, 1] \ ∪_{n,m}J_{n,m} and the function f will be constant on each set E ∩ I_{n,m}. These constants will be chosen so that we can obtain lim_{x↘0;x∈E} \frac{f(x)}{x} = 1/2 and each constant will equal a fixed point of f. Hence for any x ∈ E we have f^{k}(x) = f(x), k = 1, 2, ..., that is, no point of E will be attracted to 0.

Put I_{n,m} = [2^{−n} + (m − 1)4^{−n}, 2^{−n} + m4^{−n}] and J_{n,m} = (2^{−n} + (m − 4^{−n})4^{−n}, 2^{−n} + (m + 4^{−n})4^{−n}) for n = 1, 2, ..., m = 1, 2, ..., 2^{n}. Put E = (0, 1] \ ∪_{n,m}J_{n,m} and f(x) = \frac{1}{2}(2^{−n} + m4^{−n}) for x ∈ I_{n,m} ∩ E, n = 1, 2, ..., m = 1, 2, ..., 2^{n}. We also put f(2^{−n} + m4^{−n}) = 2^{−n} + m4^{−n} for n = 1, 2, ..., m = 1, 2, ..., 2^{n}. Otherwise define f arbitrarily so as to be continuous. Then for every x ∈ E there exist n and m such that x ∈ I_{n,m}. Thus f(x) = \frac{1}{2}(2^{−n} + m4^{−n}) = 2^{−n−1} + 2m4^{−n−1} and hence f^{k}(f(x)) = f(x) k = 1, 2, .... As the reader easily can verify 0 is a right density point of E and lim_{x↘0;x∈E} \frac{f(x)}{x} = 1/2.

We remark that it will follow from Theorem 2.1 below that there is one point x not in E whose orbit is attracted to 0.

Lemma 2.1. Suppose that f(0) = 0 ∈ int(I), 0 < f(x) ≤ x for x > 0, f : I → I is continuous, there exists a_{n} ↘ 0 such that \frac{f(a_{n})}{a_{n}} < d ∈ (0, 1), and a_{n+1}/a_{n} > d. We also assume that an interval [c_{0}, b_{0}] is given such that 0 < c_{0} < d·b_{0} ≤ a_{1}, f([c_{0}, b_{0}])∩[c_{0}, b_{0}] ≠ ∅. Then 0 ∈ cl(∪{f^{n}([c_{0}, b_{0}]) : n = 1, 2, ...}).

Proof. Since f is continuous f^{n}([c_{0}, b_{0}]) is an interval denoted by [c_{n}, b_{n}]. From 0 < f(x) ≤ x it follows that 0 ≤ ... ≤ c_{n} ≤ c_{n−1} ≤ ... < c_{0}, and 0 ≤ ... ≤ b_{n} ≤ b_{n−1} ≤ ... ≤ b_{0}. From f([c_{0}, b_{0}]) ∩ [c_{0}, b_{0}] ≠ ∅ it follows that there exist z, w ∈ [c_{0}, b_{0}] such that f(w) = z. Then f^{n+1}(w) = f^{n}(z) ∈ [c_{n+1}, b_{n+1}] ∩ [c_{n}, b_{n}] for every n ≥ 1. Thus ∪_{k=0}^{n}f^{k}([c_{0}, b_{0}]) = [c_{n}, b_{0}].

To obtain a contradiction suppose that lim_{n→∞} c_{n} = γ > 0. Then ∪_{k=0}^{∞}[c_{k}, b_{k}] = ∪_{k=0}^{∞}f^{k}([c_{0}, b_{0}]) ⊃ (γ, b_{0}]. We also have f(γ) = γ since otherwise f(γ) < γ and by the continuity of f there would exist a c_{n} such that f(c_{n}) < γ which is clearly impossible. If there is x ∈ (γ, b_{0}] such that f(x) < γ then from (γ, b_{0}] ⊂ ∪_{k=0}^{∞}[c_{k}, b_{k}] it follows that x ∈ [c_{k}, b_{k}] for a k ∈ N. This would imply that c_{k+1} ≤ f(x) < γ which again contradicts the definition of γ. Therefore f(x) ≥ γ for x ∈ (γ, b_{0}]. Since a_{n+1}/a_{n} > d and b_{0} < a_{1} there exists an n such that γ < a_{n} ≤ γ/d. By the assumption of this lemma f(a_{n}) < d·a_{n}. On the other hand also by the assumption of this lemma γ ≤ c_{0} < d · b_{0} and hence γ/d ∈ (a_{n}, b_{0}]. Thus f(a_{n}) ≥ γ = d · γ/d ≥ d · a_{n}, a contradiction proving lim_{n→∞} c_{n} = 0, which implies 0 ∈ cl(∪{f^{n}([c_{0}, b_{0}]) : n = 1, 2, ...}).

Lemma 2.2. If f(0) = 0, 0 < f(x) ≤ x for x > 0, f is continuous, and there exist 0 < c_{0} < b_{0} such that 0 ∈ cl(∪{f^{n}([c_{0}, b_{0}]) : n = 1, 2, ...}) then there exists y ∈ [c_{0}, b_{0}] such that lim_{n→∞} f^{n}(y) = 0 and f^{n}(y) > 0 for n = 1, 2, ... .

Proof. From the continuity of f^{n} it follows that f^{n}([c_{0}, b_{0}]) is an interval [c_{n}, b_{n}]. Since f(x) ≤ x we have c_{0} ≥ c_{1} ≥ ... ≥ c_{n} ≥ ... > 0, b_{0} ≥ b_{1} ≥ ... ≥ b_{n} ≥ ... > 0. The assumption of this lemma implies that lim_{n→∞} c_{n} = 0. If there exists c_{n} such that c_{n} = c_{n+1} then f([c_{n}, b_{n}]) = [c_{n+1}, b_{n+1}] = [c_{n}, b_{n+1}] ⊂ [c_{n}, b_{n}] and hence f^{m}([c_{n}, b_{n}]) ⊂ f^{m−1}([c_{n}, b_{n}]) ⊂ f([c_{n}, b_{n}]) ⊂ [b_{n}, c_{n}] contradicting the fact that lim_{n→∞} c_{n} = 0. Thus c_{0} > c_{1} > ... > c_{n} > ... > 0. Put F_{n} = f^{−n}([c_{n}, c_{n−1}]) ∩ [c_{0}, b_{0}]. If we have finitely many F′_{n}s say F_{n_{1}}, F_{n_{2}}, ..., F_{n_{m}}, n_{1} > n_{2} > ... > n_{m} then we show that their intersection is nonempty. Choose a z ∈ [c_{n_{m}}, c_{n_{m}−1}) ∩ [c_{n_{m}}, b_{n_{m}}]. Then there exists a u ∈ F_{n_{m}} such that f^{n_{m}}(u) = z. Suppose that there exists a j such that u ∉ F_{n_{j}}, that is, f^{n_{j}}(u) ∉ [c_{n_{j}}, c_{n_{j}−1}]. Thus f^{n_{j}}(u) ∈ (c_{n_{j}−1}, b_{n_{j}}] ⊂ [c_{n_{j}−1}, b_{n_{j}−1}] and hence there exists a w ∈ [c_{0}, b_{0}] such that f^{n_{j}}(u) = f^{n_{j}−1}(w) and hence z = f^{n_{j}+(n_{m}−n_{j})}(u) = f^{n_{j}−1+(n_{m}−n_{j})}(w) = f^{n_{m}−1}(w) ∈ [c_{n_{m}−1}, b_{n_{m}−1}] contradicting to z ∈ [c_{n_{m}}, c_{n_{m}−1}). Thus by compactness there exists a y ∈ ∩_{n=1}^{∞}F_{n}. Since f^{n}(y) ∈ [c_{n}, c_{n−1}] we obtain lim_{n→∞} f^{n}(y) = 0.

Lemma 2.3. Suppose that f(0) = 0, f is continuous, and there exist q_{n}, r_{n} → 0, q_{1} > r_{1} > q_{2} > r_{2} > ... > q_{k} > r_{k} > ... > 0, such that f(q_{k}) > 0, and f(r_{k}) = 0, k = 1, 2, ... then there exists a nonempty perfect set P such that lim_{n→∞} f^{n}(y) = 0, f^{n}(y) > 0, for every y ∈ P, n = 1, 2, ... .

Proof. Put J_{∅} = [r_{1}, q_{1}]. Suppose that s is a finite zero-one sequence of length m ∈ {0, 1, 2, ...} (when m = 0 then s = ∅) and J_{s} = [r_{n(s)}, q_{n(s)}]. Since f(q_{n(s)}) > 0 we can choose n(s0) > n(s1) > n(s) such that J_{s0} = [r_{n(s0)}, q_{n(s0)}] ⊂ [0, f(q_{n(s)})], and J_{s1} = [r_{n(s1)}, q_{n(s1)}] ⊂ [0, f(q_{n(s)})].

Put F_{∅} = J_{∅}. Suppose that s is of length n − 1, the closed set F_{s} is defined, and f^{n−1}(F_{s}) = J_{s}. Then put F_{s0} = f^{−n}(J_{s0}) ∩ F_{s} and F_{s1} = f^{−n}(J_{s1}) ∩ F_{s}. From J_{s0} ⊂ [0, f(q_{n(s)})] it follows that for every x ∈ J_{s0} there exists a z ∈ J_{s} such that f(z) = x. Since f^{n−1}(F_{s}) = J_{s} we can find a w ∈ F_{s} such that f^{n−1}(w) = z. Thus f^{n}(w) = x and w ∈ F_{s0}. This implies that f^{n}(F_{s0}) = J_{s0}. A similar argument shows that f^{n}(F_{s1}) = J_{s1}.

Put H_{m} = ∪{F_{s} : s is a zero-one sequence of length m} and H = ∩_{m=1}^{∞}H_{m}. Since the sets F_{s} are closed and there are 2^{m} zero-one sequences of length m the set H_{m} is closed. Therefore H is also closed.

Suppose that y ∈ H. Then f^{m}(y) ∈ [r_{n(s)}, q_{n(s)}], where s is of length m. It is easy to see that n(s) > m. Thus f^{m}(y) → 0 and f^{m}(y) > 0.

If ϕ is an infinite zero-one sequence then denote by ϕ|m the first m terms of ϕ.

For arbitrary zero-one sequence ϕ choose a y(ϕ) ∈ ∩_{m=1}^{∞}F_{ϕ|m} ∈ H. We remark that y(ϕ) is well defined since F_{ϕ|m} m = 1, 2, ... is a nested sequence of closed sets. If ϕ, ψ are different zero- one sequences then there exists an m such that ϕ|m = ψ|m = s and ϕ|m+1 ≠ ψ|m+1. By symmetry we may assume that ϕ|m+1 = s0 and ψ|m+1 = s1. Then y(ϕ) ∈ F_{s0} and y(ψ) ∈ F_{s1}. Since f^{m+1}(y(ϕ)) ∈ J_{s0}, f^{m+1}(y(ψ)) ∈ J_{s1} and J_{s0} ∩ J_{s1} = ∅ we obtain y(ϕ) ≠ y(ψ). Therefore the cardinality of H is that of the continuum and there exists a nonempty perfect subset P ⊂ H. This proves Lemma 2.3.

It will be convenient to state our results in the language of porosity. Let S be a set, x_{0} a point and h > 0. Let ℓ(h) denote the length of the longest interval in [x_{0}, x_{0} + h] \ S. The right porosity of S at x_{0} is defined as

p_{+}(x_{0}, S) = lim sup_{h→0} \frac{ℓ(h)}{h}.

If p_{+}(x_{0}, S) = 0, we say S is nonporous from the right at x_{0}. If p_{+}(x_{0}, S) = 1 we say S is strongly porous from the right at x_{0}. The notions of left porosity and (bilateral) porosity are now defined in the obvious way.

Theorem 2.1. If x_{0} is a fixed point of the continuous function f and

0 < \frac{f(x) − f(x_{0})}{x − x_{0}} < d < 1

for all x in set E such that p_{+}(E, x_{0}) < 1−d, then there exists y such that lim_{n→∞} f^{n}(y) = x_{0} and for every n, f^{n}(y) ≠ x_{0}.

Proof. Replacing f by f(x + x_{0}) − x_{0} we may assume that x_{0} = 0. The porosity condition now implies the existence of a strictly decreasing sequence {a_{n}} such that a_{n} → 0, 0 < f(a_{n})/a_{n} < d, and a_{n+1}/a_{n} > d.

If there exist x_{n} ↘ 0 such that f(x_{n}) ≤ 0 then using 0 < f(a_{n}) one can choose q_{n}, r_{n} → 0, q_{1} > r_{1} > ... > q_{k} > r_{k} > ... > 0, such that f(q_{k}) > 0, f(r_{k}) = 0 for k = 1, 2, .... In this case applying Lemma 2.3 we can complete the proof of Theorem 2.1.

Thus we may assume that there exists δ > 0 such that 0 < f(x) for x ∈ (0, δ]. Put

f_{1}(x) = \{ \begin{matrix}f(x), & if f(x) ≤ x, and 0 ≤ x < δ; \\ x, & if f(x) > x and 0 ≤ x < δ; \\ f(x), & if x < 0; \\ min(δ, f(δ)), & if δ ≤ x. \end{matrix}

Since a_{n} ↘ 0 there exists m ∈ N such that a_{m} < δ. Put a′_{n} = a_{n+m} for n = 1, 2, ... .

Obviously 0 < f_{1}(x) ≤ x for x > 0. Put b_{0} = a′_{1} and choose a c_{0} ∈ (0, f_{1}(b_{0})]. Then f_{1}([c_{0}, b_{0}]) ∩ [c_{0}, b_{0}] ≠ ∅. Applying Lemma 2.1 for f_{1}, a′_{n} and [c_{0}, b_{0}] we obtain that 0 ∈ cl(∪{f_{1}^{n}([c_{0}, b_{0}]) : n = 1, 2, ...}). Therefore we can apply Lemma 2.2 for f_{1} and hence there exists y ∈ [c_{0}, b_{0}] such that lim_{n→∞} f_{1}^{n}(y) = 0 and f_{1}^{n}(y) > 0 for n = 1, 2, ... . From 0 < f_{1}^{n}(x) ≤ x it follows that 0 < ... ≤ f_{1}^{n+1}(y) ≤ f_{1}^{n}(y) ≤ ... ≤ y. If f_{1}^{n+1}(y) = f_{1}^{n}(y) for an n ∈ N then f_{1}^{ℓ+1}(y) = f_{1}^{n+1+(ℓ−n)}(y) = f_{1}^{n+(ℓ−n)}(y) = f_{1}^{ℓ}(y) for every ℓ ≥ n and this would contradict the fact that lim_{n→∞} f_{1}^{n}(y) = 0. Therefore 0 < f_{1}(f_{1}^{n}(y)) < f_{1}^{n}(y) for n = 0, 1, ... . Since 0 < ... < f_{1}^{n}(y) < ... < y < b_{0} = a_{m+1} ≤ a_{m} < δ we obtain that 0 < f_{1}^{n}(y) < δ and f_{1}(f_{1}^{n})(y) < f_{1}(y) n = 1, 2, ... . From the definition of f_{1} it follows that f^{n}(y) = f_{1}^{n}(y) for n = 1, 2, ... . Thus lim_{n→∞} f^{n}(y) = 0 and this completes the proof of Theorem 2.1.

Example 2.2 below shows that the porosity condition in Theorem 2.1 cannot be weakened.

Example 2.2. Suppose that 0 < ϵ < 1/2. Put f(x) = 2^{−n} for x ∈ [2^{−n}, \frac{2^{−n+1}}{1+ϵ}], n = 1, 2, ... . If \frac{2^{−n+1}}{1+ϵ} < x < 2^{−n+1}, n = 1, 2, ... choose 2^{−n} ≤ f(x) ≤ 2^{−n+1} so that it continuously connects f(\frac{2^{−n+1}}{1+ϵ}) = 2^{−n} and f(2^{−n+1}) = 2^{−n+1}. If x ≤ 0 or x > \frac{2}{1+ϵ} choose f so that f(0) = 0, f(\frac{2}{1+ϵ}) = 1 and f is continuous. Then the porosity of E = \{x : f(x) < (\frac{1}{2} + ϵ)x\} at 0 is no greater than 1/2 since f(\frac{2^{−n+1}}{1+ϵ}) = \frac{1+ϵ}{2}(\frac{1}{1+ϵ} · 2^{−n+1}) < (\frac{1}{2} + ϵ)\frac{2^{−n+1}}{1+ϵ} n = 1, 2, ... . It is also clear that each interval [2^{−n}, 2^{−n+1}] is mapped into itself and hence no point is attracted from the right to 0.

We saw in Example 2.1 that even when E is nonporous at x_{0} and f′_{E}(x_{0}) = 1/2 there may be no point in E whose orbit is attracted from the right to x_{0}. Observe that R \ E was also nonporous from the right at x_{0} in that example. One can rectify this flaw by requiring R \ E to be strongly porous from the right at x_{0}.

Theorem 2.2. Suppose E is nonporous from the right at x_{0} and R \ E is strongly porous from the right at x_{0}. If f is a continuous function having x_{0} as a fixed point and satisfies an inequality

0 < d_{1} < \frac{f(x) − f(x_{0})}{x − x_{0}} < d_{2} < 1

for all x in E then there exists y ∈ E such that lim_{n→∞} f^{n}(y) = x_{0} and f^{n}(y) > 0 for n = 1, 2, ... . In particular, if f′_{E}(x_{0}) exists and 0 < f′_{E}(x_{0}) < 1, the conclusion follows.

Proof. Replacing f by f(x + x_{0}) − x_{0} we can reduce the proof to the case that x_{0} = 0. The assumptions of the theorem imply that there exist c_{0} < b_{0}, and a sequence {a_{n}} converging monotone decreasingly to 0 such that for each n, a_{n} ∈ E, a_{n+1}/a_{n} > d_{2}, [c_{0}, b_{0}] ⊂ E ∩ [0, a_{1}] and \frac{c_{0}}{b_{0}} < d_{1}. Put

f_{1}(x) = \{ \begin{matrix}f(x), & if f(x) ≤ x and 0 ≤ x; \\ x, & if f(x) > x and 0 ≤ x; \\ f(x), & if x < 0. \end{matrix}

Observe that f_{1}(a_{n})/a_{n} < d_{2}, c_{0} < d_{1}b_{0} < d_{2}b_{0} < b_{0} ≤ a_{1} and finally f_{1}(b_{0})/b_{0} > d_{1} > c_{0}/b_{0}. Therefore f_{1}(b_{0}) > c_{0}, and hence f_{1}([c_{0}, b_{0}])∩[c_{0}, b_{0}] ≠ ∅. Thus we can apply Lemma 2.1 and Lemma 2.2 for f_{1} with {a_{n}}, d_{2}, c_{0}, b_{0} and find a point y ∈ [c_{0}, b_{0}] ⊂ E such that lim_{n→∞} f_{1}^{n}(y) = 0, f_{1}^{n}(y) > 0, n = 0, 1, ... . If there exists an n ≥ 1 such that f_{1}^{n}(y) ≠ f^{n}(y) then f_{1}(f_{1}^{n−1}(y)) = f_{1}^{n−1}(y) and hence f_{1}^{m}(y) = f_{1}^{m−n+1}(f_{1}^{n−1}(y)) = f_{1}^{n−1}(y) which contradicts to lim_{n→∞} f_{1}^{n}(y) = 0. Thus f_{1}^{n}(y) = f^{n}(y) and this proves Theorem 2.2.

Example 2.3. In this example we want to show that in Theorem 2.2 the condition 0 < d_{1} < \frac{f(x)−f(x_{0})}{x−x_{0}} < d_{2} < 1 can not be replaced by 0 < \frac{f(x)−f(x_{0})}{x−x_{0}} < 1. Put

[c_{n}, b_{n}] = [\frac{1}{n} \frac{1}{(n + 1)!}, \frac{1}{n} \frac{1}{n!}]

and E = ∪_{n=1}^{∞}[c_{n}, b_{n}]. Then E is non porous from the right at 0 and R \ E is strongly porous from the right at 0. Put f(\frac{b_{n+1}+c_{n}}{2} ) = \frac{b_{n+1}+c_{n}}{2} , n = 1, 2, ..., and f(x) = \frac{b_{n+1}+c_{n}}{2} for x ∈ [c_{n}, b_{n}], n = 1, 2, ... . Otherwise choose f so that it is continuous on R. Then, obviously no point from E is attracted to 0.

3. Results for ”typical” continuous functions.

The previous results show that adequate behavior of the difference quotient \frac{f(x)−f(x_{0})}{x−x_{0}} on sets satisfying certain porosity conditions with respect to a fixed point x_{0} will guarantee certain attractive behavior to x_{0}. But ”most ” continuous functions will not exhibit this behavior with respect to any fixed point [BH]. In this section, we make this statement precise and show that, nonetheless, most continuous functions have the property that to each fixed point x_{0} corresponds a nonempty perfect set P (x_{0}) such that for each y ∈ P (x_{0}), lim_{n→∞} f^{n}(y) = x_{0} with f^{n}(y) ≠ x_{0} for each n = 1, 2, ... .

Let C denote the class of continuous functions mapping I into I. When C is furnished with the sup norm it becomes a complete metric space.

Theorem 3.1. a) Let A consist of those functions f in C such that for each x_{0} ∈ I the set

S(x_{0}) = \{x : |\frac{f(x) − f(x_{0})}{x − x_{0}}| < 1\}

is bilaterally strongly porous at x_{0}. Then A is a residual subset of C.

b) Let B consist of those functions f in C such that to every fixed point x_{0} of f corresponds a nonempty perfect set P (x_{0}) such that lim_{n→∞} f^{n}(y) = x_{0} for y ∈ P (x_{0}) and f^{n}(y) ≠ x_{0} for all n. Then B is a residual subset of C.

Proof. a) In [BH] one finds that for f in some residual subset D of C, the set S = {x : f(x) = g(x)} is bilaterally strongly porous at each point of S whenever f satisfies a Lipschitz condition. If for some x_{0} ∈ I the set S(x_{0}) in part a) of the statement fails to be bilaterally strongly porous at x_{0}, we can argue as in the proof of [BH] Thm. 3.2, that is, we can find a Lipschitz function g that agrees with f on a subset of S(x_{0}) that is also not strongly porous at x_{0}. It follows that D ⊂ A so A is residual in C.

To prove b) assume that I = [a_{0}, b_{0}] and put C_{1} = {f ∈ C : f(a_{0}) ≠ a_{0} and f(b_{0}) ≠ b_{0}}. Obviously C_{1} is a residual subset of C.

Assume that J = [a, b] ⊂ I is given. Denote by M(J) the set of functions f ∈ C_{1} such that the maximum of f on J is achieved at a fixed point of f and this fixed point is different from b. First we show that for a given J the set M(J) is nowhere dense in C_{1}. Let N(f, ϵ) be the sphere in C_{1} that is centered at f and has radius ϵ > 0. We find a sphere N(g, δ) inside N(f, ϵ) whose intersection with M(J) is empty. Without loss of generality, suppose f ∈ M(J) with f(c) = c = max_{x∈J} f(x), a ≤ c < b. We may also assume ϵ < b − c and choose d ∈ (c, b) with d − c < ϵ/2. Let g ∈ N(f, ϵ/2) satisfy d < g(c) < f(c) + ϵ/2, g(x) ≤ c for x ≥ d. This is possible since f ≤ c on J. Let u be the maximum fixed point of g in J. Then c < u < d. Choose δ ∈ (0, ϵ/2) such that g(c) − δ > d and g(x) + δ < d for x ≥ u. If h ∈ N(g, δ) then h(c) > d and h has no fixed points greater than d ∈ J. Thus N(g, δ) ∩ M(J) is empty and N(g, δ) ⊂ N(f, ϵ). It follows that M(J) is nowhere dense in C_{1}.

By applying the above result to each interval with rational endpoints and by applying a similar argument for ”minimum” instead of ”maximum”, we find that the set of functions E in C_{1} that achieve a local extremum at a fixed point is first category in C_{1} and hence these functions form a first category subset of C.

Now according to a theorem of Jarnik [B:p.213], each function in some residual subset S of C has every real number as a derived number at every point. This means that for f ∈ S, x_{0} ∈ I and t ∈ R, there exists a sequence x_{n} converging to x_{0} such that (f(x_{n}) − f(x_{0}))/(x_{n} − x_{0}) converges to t. Let f be in (C_{1} ∩ S) \ E. Applying Lemma 2.3 to f (with respect to x_{0} and f(x_{0}) instead of 0 and f(0) ) we find f in the required set B. Thus the residual set (C_{1} ∩ S) \ E is contained in B so B is residual in C. This concludes our proof.

Finally we remark that using similar techniques to those in Chapter VIII of [B] one can show that for the typical f : I → I the fixed point set is a nowhere dense non-empty perfect set.

References

[B] A.BRUCKNER, Differentiation of Real Functions, Springer Verlag 1978.

[BH] A.BRUCKNER and J.HAUSSERMANN, Strong porosity features of typical continuous functions, Acta Math. Hung. 45 (1-3) 1985, 7- 13.

Department of Mathematics, University of California, Santa Barbara, California, 93106; USA Department of Analysis Eötös Loránd University 1088 Budapest, Múzeum krt. 6-8, Hungary. H-1088

at 11:43:47 3/20/1991