Uniform continuity (sequence dual)
Uniform continuity: 2/3
ε-δ vs. sequence formulation
Set adherence and function convergence have both an "epsilon-delta" formulation and a sequence formulation. We covered propositions that mapped between these equivalences.
Uniform convergence also has this duality. Below, uniform convergence is formulated in terms of sequences. The sequences equivalence formulations below look very similar to the formulation of equivalent Cauchy sequences, but they are not necessarily Cauchy, and that is their distinction.
Equivalent sequences
Let \( (a_n)_{n=m}^{\infty} \) and \( (b_n)_{n=m}^{\infty} \) be sequences of reals. We say that \( (a_n)_{n=m}^{\infty} \) and \( (b_n)_{n=m}^{\infty} \) are equivalent sequences iff for any real \( \varepsilon > 0 \) there exists an integer \( N \ge m \) such that \( |a_k - b_k| < \varepsilon \) for all \( k \ge N \).
This formulation can also be split up by using intermediate definitions for ε-close sequences and eventual ε-close sequences.
The above definition can be succinctly formulated using sequence limits also.
Equivalent sequences (via limits)
Let \( (a_n)_{n=m}^{\infty} \) and \( (b_n)_{n=m}^{\infty} \) be sequences of reals. We say that \( (a_n)_{n=m}^{\infty} \) and \( (b_n)_{n=m}^{\infty} \) are equivalent sequences iff \( \lim_{n \to \infty}(a_n - b_n) = 0\).
Now on to uniform convergence phrased using equivalent sequences.
Uniform convergence (via equivalent sequences)
Let \( X \subseteq \mathbb{R} \) be a set and let \( f: X \to \mathbb{R} \) be a function. Then the following two statements are logically equivalent:
- f is uniformly continuous.
- For any two equivalent sequences \( (a_n)_{n=m}^{\infty} \) and \( (b_n)_{n=m}^{\infty} \) whose elements come from \( X \), the sequences \( (f(a_n))_{n=m}^{\infty} \) and \( (f(b_n))_{n=m}^{\infty} \) are also equivalent sequences.
Useful properties
Three useful properties concerning uniformly continuous functions are:
- they map Cauchy sequences to Cauchy sequences.
- they map bounded sets to bounded sets (the image of a bounded set under a continuously convergence function is a bounded set).
- If a fuction \( f \) has a closed interval domain and is continuous then it is also uniformly continuous.
The last of these is proved by Tao; the others are exercises.
Mapping of sequences
Function convergence and sequence convergence
The duality of function convergence and sequence convergence means that if \( f \) is a continuous function, then \( f \) maps convergent sequences to convergent sequences.
Uniform function convergence and pairs of equivalent sequences
Connecting the two
To see the connection between the two ideas above, observer that a sequence \( (a_n)_{n=0}^{\infty} \) converges to \( x \) if and only if the sequence \( (a_n)_{n=0}^{\infty} \) is equivalent to \( (x)_{n=0}^{\infty} \).
Example
\( f(x) := \frac{1}{x} \)
Consider the function \( f : (0, 2) \to \mathbb{R} \) defined by \( f(x):= \frac{1}{x} \). We can confirm that the two sequences \( (\frac{1}{n})_{n=1}^{\infty} \) and \( (\frac{1}{2n})_{n=1}^{\infty} \) are equivalent sequences in \( (0, 2) \). However, the sequences \( (f(\frac{1}{n}))_{n=1}^{\infty} \) and \( (f(\frac{1}{2n}))_{n=1}^{\infty} \) (which are \( (n)_{n=1}^{\infty} \) and \( (2n)_{n=1}^{\infty} \) respectively) are not equivalent sequences, so \( f \) cannot be uniformly continuous.
\( f(x) := x^2 \)
The function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x):= x^2 \) is a continuous function, but, it turns out, it is not uniformly continuous. In some sense, the continuity gets worse approaching infinity.
To confirm this, consider the sequences \( (n)_{n=1}^{\infty} \) and \( (n + \frac{1}{n})_{n=1}^{\infty} \). They are equivalent; however, the sequences \( (f(n))_{n=1}^{\infty} \) and \( (f(n + \frac{1}{n}))_{n=1}^{\infty} \) are not. Evaluating \( f \), the sequences become \( (n^2)_{n=1}^{\infty} \) and \( (n^2 + 2 + \frac{1}{n})_{n=1}^{\infty} \) respectively. The former, \( (n^2)_{n=1}^{\infty} \), never becomes 2-close to the latter, \( (n^2 + 2 + \frac{1}{n})_{n=1}^{\infty} \). Thus, it can be concluded that \( f \) is not uniformly continuous.