1.2 Uniform Continuity
Recall the definition of (standard) continuity:
In this definition, the ‘distance’ \(\delta\) away from \(c\) which ensures \(f(x)\) remains within a distance \(\epsilon\) of \(f(c)\) depends on two things: the choice of \(\epsilon\), and where you are in the domain \(D\), i.e. your choice of \(c\). If instead, your choice of \(\delta\) remains the same no matter where you are in \(D\), then \(f\) is said to be uniformly continuous. An example of this definition is seen in Figure 1.1.3
Figure 1.1: An example of a uniformly continuous function. Here, for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that we can translate a rectangle of width \(2\delta\) and height \(2\epsilon\) along the function without penetrating the top or bottom edges of the rectangle
Furthermore, from Definition 1.4, we see by fixing the value of \(y\), we deduce that uniform continuity implies standard continuity! In fact, when the function domain is compact (i.e. think \([a,b]\)), the reverse also holds true:
Let \(f:[a,b] \to \mathbb{R}\). Then \(f\) is continuous if and only if it is uniformly continuous.
1.2.1 Uniform Continuity and Differentiability
You may remember that if a function \(f:I \to \mathbb{R}\) is differentiable on an open interval \(I \subseteq \mathbb{R}\), then it is continuous on \(I\). However, we cannot strengthen this result in the way you might expect. Namely, it is not true that differentiability implies uniform continuity.
To see why, consider \(f:\mathbb{R} \to \mathbb{R}\) given by \(f(x) = x^2\). We know that the derivative function is \(f':\mathbb{R} \to \mathbb{R}\) given by \(f'(x) = 2x.\) However, f is not uniformly continuous on \(\mathbb{R}\).
To prove this, we consider the negation of the definition, i.e. we seek \(\epsilon_0 >0\) such that for all \(\delta > 0\), there exists \(x,y \in \mathbb{R}\) such that \(\lvert x - y \rvert < \delta\), and \(\lvert f(x) - f(y) \rvert \geq \epsilon_0.\)
Try \(\epsilon_0 = 1.\) Then \[\begin{align*} \lvert f(x) - f(y) \rvert \geq 1 &\Leftrightarrow \lvert x - y \rvert \lvert x + y \rvert \geq 1. \end{align*}\]
Looking only at non-negative values of \(x,y\) (which we can do since we are searching for \(x\) and \(y\) in this problem), our constraint \(\lvert x - y \rvert < \delta\) suggests we try setting \(y = x + \frac{\delta}{2}.\) In this case \[\lvert f(x) - f(y) \rvert \geq 1 \Leftrightarrow \frac{\delta}{2}\left(2x + \frac{\delta}{2}\right) \geq 1,\] from which we obtain the requirements \[x \geq 0 \quad \text{and} \quad x \geq \frac{1}{\delta} - \frac{\delta}{4}.\] Hence, taking \[x = \max\left\lbrace 0 , \frac{1}{\delta} - \frac{\delta}{4}\right\rbrace \quad \text{and} \quad y = x + \frac{\delta}{2},\] yields \[\lvert x - y \rvert < \delta \quad \text{and} \quad \lvert f(x) - f(y) \rvert \geq 1.\] So, we have found an \(\epsilon_0>0\) such that for any positive \(\delta\), we have found \(x,y\) with \(\lvert x - y \rvert < \delta\), and \(\lvert f(x) - f(y) \rvert \geq \epsilon_0.\) This shows that \(f\) is not uniformly continuous.
However, all hope is not lost. In fact, using the Mean Value Theorem, we can recover a result linking differentiability and uniform continuity!
Let \(f:[a,b] \to \mathbb{R}\) be continuous on \([a,b]\) and differentiable on \((a,b).\) If \(f\) is differentiable on \((a,b)\) with bounded derivative, i.e. \(\exists L > 0\) such that \(\lvert f'(x) \rvert < L \;\;\forall x \in (a,b)\), then \(f\) is uniformly continuous.
1.2.2 Other forms of Continuity
Whilst less relevant to this course, there are versions of continuity which are stronger still! The first we will mention here is known as Hölder continuity.
Ok, this definition looks a little complicated, so a visual such as Figure 1.2 is probably quite welcome here for some geometric intuition.
Figure 1.2: An example of a Hölder continuous function. In this case, there exists constants \(L\) and \(\alpha\) such that we can translate a double parabola along the function so that the function remains within the shaded areas.
You’ve already shown in a previous problem sheet that if \(D\) is an interval, and \(\alpha > 1\), then the only \(\alpha\)-Hölder continuous functions are constant. Another important class of Hölder continuous functions occurs when \(\alpha = 1.\) This is a case you’re also likely to have come across in the problem sheets:
Again, this is something we can visualise (see Figure 1.3). Quite handily, if we introduce \(\delta = \epsilon/L\), we see that a Lipschitz continuous function satisfies the definition of uniform continuity, and so is also continuous!
Figure 1.3: An example of a Lipschitz continuous function. In this case, we can translate a double cone along the function, so that the function remains in the shaded area.