(last updated: 3:37:04 PM, November 11, 2020)
Definition 4.4.3 A function \(f: D\subset\mathbb R \rightarrow\mathbb R\) is uniformly continuous on a set \(E\subset D\) if and only if for any given \(\epsilon>0\) there exists \(\delta>0\) such that \(|f(x)-f(t)|<\epsilon\) for all \(x,t\in E\) satisfying \(|x-t|<\delta\). If \(f\) is uniformly continuous on its domain \(D\), we simply say that \(f\) is uniformly continuous.
Example:
Version 1: \(f(x)=x^3\) on \([0,2)\)
Let \(t<x\) then \(|f(x)-f(t)|=f(x)-f(t)=x^3-t^3=(x-t)(x^2+xt+t^2)\). We know that \(x^2+xt+t^2\) is maximized at \(x=t=2\). Even though \(2\) is not in the domain, we can always get arbitrarily close to it, and we just want any upper bound on \(x^2+xt+t^2\), so we plug in \(x=t=2\) to get \(x^2+xt+t^2 \leq 12\) for all \(x,t\in [0,2)\). Thus \(|f(x)-f(t)|\leq 12|x-t|\) for all \(x,t\in [0,2)\). Now given \(\epsilon>0\) choose \(\delta=\frac\epsilon{12}\). Then whenever \(|x-t|<\delta\) and \(x,t\in [0,2)\) we have that \(|f(x)-f(t)|\leq 12|x-t|<12\delta=\epsilon\). So we have shown this funciton satisfies the definition of uniform continuity on the interval \([0,2)\).
Version 2: \(f(x)=x^3\) on \(\mathbb R\)
This situation is different though. Again, assuming \(t<x\) we have that \(|f(x)-f(t)|=f(x)-f(t)=x^3-t^3=(x-t)(x^2+xt+t^2)\), but the factor on the right is now unbounded. We won’t be able to pick a \(\delta\) that works for the entire domain.
Once an \(\epsilon>0\) is given, suppose we think we have found a \(\delta>0\) that works and that \(|x-t|<\delta\). Now pick \(M\in\mathbb R\) so that \(3M^2>2\frac\epsilon\delta\). Since \(\epsilon\) and \(\delta\) are both positive, we can always choose such an \(M\). Now choose any \(t>M\) and set \(x=t+\frac\delta2\), so that \(|x-t|=\frac\delta2<\delta\). However we have that \[\begin{aligned} |f(x)-f(t)|&=(x-t)(x^2+xt+t^2) &\\ &=\frac\delta2(x^2+xt+t^2) &\quad \text{(since $x=t+\frac\delta2$)}\\ &>\frac\delta2 \cdot 3t^2 &\quad \text{(since $x>t$)}\\ &>\frac\delta2 \cdot 3M^2 &\quad \text{(since $t>M$)}\\ &>\frac\delta2 \cdot 2\frac\epsilon\delta &\quad \text{(since $3M^2>2\frac\epsilon\delta$)}\\ &=\epsilon \end{aligned}\] This will be true regardless of what we set \(\delta\) as. Notice though that as \(\delta\) gets smaller (for a fixed \(\epsilon\) value) the cutoff \(M\) we choose will get larger and larger. In order to find \(x\) and \(t\) values that are close together but give far apart function values, we need to go very far on in the domain.
Since given a fixed \(\epsilon\), we cannot find a \(\delta\) that makes the uniform continuity definition hold, we say this funciton is not uniformly continuous.
For the function \(f(x)=x^3\) on \(\mathbb R\), it is not a problem that it is an unbounded function, but that the variation between nearby \(x\) values is unbounded. Making \(x,t\) close together doesn’t necessarily make the function values close together since we can just move out towards \(\infty\) on the domain and make \(f(x)-f(t)\) as large as we like.
Example: Consider the function \(f(x)=\sqrt x\) on \([0,\infty)\). This function is in fact uniformly continuous on its domain.
Consider the inequality for \(x>t\geq 0\) gives \(\sqrt{x} - \sqrt{t} \leq \sqrt{x-t}\). Now, given \(\epsilon>0\), let \(\delta=\epsilon^2\). Then for \(x,t\in [0,\infty)\) with \(|x-t|<\delta\) we have that \(|f(x)-f(t)|=|\sqrt{x} - \sqrt{t}| \leq \sqrt{|x-t|}<\sqrt\delta=\epsilon\). This shows that \(f\) is uniformly continuous.
Remark 4.4.4 To show that \(f\) is not uniformly continuous. Pick sequences \(x_n\) and \(t_n\) in the domain that satisfy \(|x_n-t_n|\leq \frac1n\) for all \(n\in\mathbb N\) but that \(|f(x_n)-f(t_n)|\geq\epsilon\) is eventually true for \(n\) sufficiently large.
The key thing in the above remark is that \(|x_n-t_n|\) must converge to zero so that it is eventually below any possible choice for \(\delta\), thus the use of \(\frac1n\) can be replaced by any sequence converging to zero.
Example:
Let’s revisit \(f(x)=x^3\) on \(\mathbb R\). Let \(x_n=n+\frac1n\) and \(t_n=n\). Then \(|x_n-t_n|=\frac1n\) which converges to zero (and thus will be eventually always smaller than any \(\delta\)). But \(|f(x_n)-f(t_n)|=(n+\frac1n)^3-n^3=n^3+3n^2\frac1n+3n\frac1{n^2}+\frac1{n^3}-n^3 > 3n\). And we have that \(3n>\epsilon\) is eventually true.
We have thus show that given any \(\epsilon>0\) and any \(\delta>0\) there will be an \(n\in\mathbb N\) such that \(|x_n-t_n|=\frac1n<\delta\) and \(|f(x_n)-f(t_n)|\geq\epsilon\), thus this function is not uniformly continuous.
Theorem (not in text) If function \(f\) is uniformly continuous on interval \((a,b)\), uniformly continuous on interval \((b,c)\), and continuous at \(x=b\), then \(f\) is uniformly continuous on interval \((a,b)\cup \{b\}\cup(b,c)=(a,c)\).
Theorem 4.4.6 If \(D\subset\mathbb R\) is a closed and bounded set, and a function \(f:D\rightarrow\mathbb R\) is continuous, then \(f\) is uniformly continuous.
Theorem 4.4.7 If a function \(f:(a,b)\rightarrow\mathbb R\) is uniformly continuous, then \(f (a^+)\) and \(f (b^-)\) are both finite.
Corollary 4.4.8 If a function \(f:(a,b)\rightarrow\mathbb R\) is uniformly continuous if and only if \(f\) can be continuously extended to \([a,b]\).
Theorem 4.4.9 If a function \(f:[a,\infty)\rightarrow\mathbb R\) is continuous with \(\displaystyle \lim_{x\rightarrow\infty}f(x)\) then \(f\) is uniformly continuous on \([a,\infty)\).
Definition 4.4.10 A function \(f : D\subset\mathbb R \rightarrow\mathbb R\) is a Lipschitz function if and only if there exists \(L > 0\), called a Lipschitz constant, such that \[|f(x)-f(t)| \leq L|x - t|\] for all \(x\) and \(t \in D\).
Theorem 4.4.11 If a function is a Lipschitz function, then it is uniformly continuous.
Essentially what Theorem 4.4.11 says is that if \[\frac{|f(x)-f(t)|}{|x-t|}\] is bounded on the domain of the function, of course with \(x\neq t\), then \(f\) is uniformly continuous.
\[ \diamond \S \diamond \]