We know that every bounded sequence of real numbers has a convergent subsequence. For a sequence of functions, say , the notion of boundedness can be stated as: there exists a constant such that for every and for all . Such a sequence is called **uniformly bounded**.

Once we fix some point , the boundedness assumption provides a subsequence which converges at that point. But since different points may require different subsequences, it is not obvious whether we can pick a subsequence which converges for all . (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 requires , which is stronger than for every ). Indeed, does the job. This sequence is uniformly bounded (by ) and converges pointwise to a discontinuous function such that and elsewhere. Any subsequence has the same pointwise limit and since 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 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 defined by

or by a magic one-line formula if you prefer: .

Let be the sequence of the iterates of , that is and . This means , , , and so on.

By construction, the sequence is uniformly bounded (). It is somewhat similar to the example in that we have increasingly rapid oscillations. But the proof that has no pointwise convergent subsequence is elementary. It is based on the following observations.

(A) Suppose that is a sequence such that for each . Then the number satisfies if and if .

The proof of (A) amounts to summing a geometric series. Incidentally, observe that are the digits of in base .

(B) For as above we have . In other words, shifts the ternary digits of to the left. As a consequence, shifts them to the left by places.

(C) Given any subsequence , let if for some , and otherwise. By part (B), which means the first ternary digit of is . By construction, this digit is when is even, and when is odd. By part (A) we have when is even, and when is odd. Thus, does not converge. This completes the proof.

## Remarks

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

The function magnifies each half of by the factor of , and maps it onto .

The important part of the formula for is that when . The rest of it could be some other continuous extension to .

A similar example could be based on the **tent map** , whose graph is shown below.

However, in case of the tent map it is more convenient to use the sequence of **even** iterates: , , and so on.

Indeed, since when , one can simply replace base with base in all of the above computations and arrive at the same conclusion, except the orbit of will now be jumping between the intervals and .

The tent map is conjugate to the logistic map , shown below. This conjugacy is where .

The conjugacy shows that the th iterate is . Therefore, the sequence of the even iterates of has no pointwise convergent subsequence. This provides an example with smooth functions, even polynomials.

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