Folded letters and uniformly bounded function sequences

We know that every bounded sequence of real numbers has a convergent subsequence. For a sequence of functions, say {f_n\colon [0, 1]\to \mathbb R}, the notion of boundedness can be stated as: there exists a constant {M} such that {|f_n(x)|\le M} for every {n} and for all {x\in [0, 1]}. Such a sequence is called uniformly bounded.

Once we fix some point {x\in [0, 1]}, the boundedness assumption provides a subsequence {\{f_{n_k}\}} which converges at that point. But since different points may require different subsequences, it is not obvious whether we can pick a subsequence {\{f_{n_k}\}} which converges for all {x\in [0, 1]}. (Such a subsequence is called pointwise convergent.)

It is easy to give an example of a uniformly bounded sequence with no uniformly convergent subsequence (uniform convergence {f_n\to f} requires {\sup |f_n-f|\to 0}, which is stronger than {f_n(x) \to f(x)} for every {x}). Indeed, {f_n(x) = x^n} does the job. This sequence is uniformly bounded (by {M=1}) and converges pointwise to a discontinuous function {g} such that {g(1)=1 } and {g(x)=0} elsewhere. Any subsequence {f_{n_k}} has the same pointwise limit {g} and since {g} is not continuous, the convergence cannot be uniform.

But what would be an example of a uniformly bounded sequence of continuous functions with no pointwise convergent subsequence? In Principles of Mathematical Analysis Rudin gives {f_n(x) = \sin nx} as such an example but then uses the Lebesgue Dominated Convergence Theorem to prove the non-existence of pointwise convergent subsequences. I do not want to use the DCT.

The simplest example I could think of is based on the letter-folding function {F\colon [0, 1]\to [0, 1]} defined by

{\displaystyle F(x) = \begin{cases} 3x, & x\in [0, 1/3] \\ 2 - 3x, & x\in [1/3, 2/3] \\  3x - 2, & x \in [2/3, 1]  \end{cases} }

or by a magic one-line formula if you prefer: {F(x) = 3x - 1 - |3x-1| + |3x - 2|}.

Letter-folding function

Let {\{f_n\}} be the sequence of the iterates of {L}, that is {f_0(x) = x} and {f_{n+1}(x) = F(f_n(x))}. This means {f_1 = F}, {f_2 = F\circ F}, {f_3 = F\circ F \circ F}, and so on.

The second and third iterates of F

By construction, the sequence {\{f_n\}} is uniformly bounded ({M=1}). It is somewhat similar to the example {\{\sin nx\}} in that we have increasingly rapid oscillations. But the proof that {\{f_n\}} has no pointwise convergent subsequence is elementary. It is based on the following observations.

(A) Suppose that {\{a_j\}} is a sequence such that {a_j\in \{0, 2\}} for each {j \in \mathbb N}. Then the number {\displaystyle x = \sum_{j=1}^\infty \frac{a_j}{3^j}} satisfies {x\in [0, 1/3]} if {a_1=0} and {x\in [2/3, 1]} if {a_1 = 2}.

The proof of (A) amounts to summing a geometric series. Incidentally, observe that {a_1, a_2, \dots} are the digits of {x} in base {3}.

(B) For {x} as above we have {\displaystyle F(x) = \sum_{j=1}^\infty \frac{a_{j+1}}{3^j}}. In other words, {F} shifts the ternary digits of {x} to the left. As a consequence, {f_n} shifts them to the left by {n} places.

(C) Given any subsequence {\{f_{n_k}\}}, let {a_j = 2} if {j = n_{2k} + 1} for some {k}, and {a_j=0} otherwise. By part (B), {\displaystyle f_{n_k}(x) = \sum_{j=1}^\infty \frac{a_{j + n_k}}{3^j} } which means the first ternary digit of {f_{n_k}(x)} is {a_{n_k + 1}}. By construction, this digit is {2} when {k} is even, and {0} when {k} is odd. By part (A) we have {f_{n_k}(x) \ge 2/3} when {k} is even, and {f_{n_k}(x) \le 1/3} when {k} is odd. Thus, {\{f_{n_k}(x)\}} does not converge. This completes the proof.

Remarks

The set of all points {x} of the form considered in (A), (B), (C), i.e., those with all ternary digits {0} or {2}, is precisely the standard Cantor set {C}.

The function {F} magnifies each half of {C} by the factor of {3}, and maps it onto {C}.

The important part of the formula for {F} is that {F(x) = 3x\bmod 1} when {x\in C}. The rest of it could be some other continuous extension to {[0, 1]}.

A similar example could be based on the tent map {T(x) = 1 - |2x-1|}, whose graph is shown below.

The tent map

However, in case of the tent map it is more convenient to use the sequence of even iterates: {T\circ T}, {T\circ T\circ T\circ T}, and so on.

The second and fourth iterates of T

Indeed, since {T(T(x)) = 4x \bmod 1} when {x\in [0, 1/4]\cup [1/2, 3/4]}, one can simply replace base {3} with base {4} in all of the above computations and arrive at the same conclusion, except the orbit of {x} will now be jumping between the intervals {[0, 1/4]} and {[1/2, 3/4]}.

The tent map is conjugate to the logistic map {L(x) = 4x(1-x)}, shown below. This conjugacy is {L\circ \varphi = \varphi\circ T} where {\varphi(x) = (1 - \cos \pi x)/2}.

Logistic map

The conjugacy shows that the {n}th iterate {L^{n}} is {\varphi\circ T^n \circ \varphi^{-1}}. Therefore, the sequence of the even iterates of {L} has no pointwise convergent subsequence. This provides an example with smooth functions, even polynomials.

The second and fourth iterates of L

One could use the sequence of all iterates (not just the even ones) in the constructions based on {T} and {L}, it just takes a bit of extra work that I prefer to avoid.