“If it requires a uniform, it's a worthless endeavor.”
―George Carlin
Definition4.97.Uniform Continuity.
Let \(X,Y\) be metric spaces and let \(f:X\rightarrow Y\) be a function. We say \(f\) is uniformly continuous if for all \(\varepsilon>0\) exists \(\delta>0\) such that for all \(x,y\in X, d(x,y)<\delta\implies d(f(x),f(y))<\varepsilon\text{.}\)
Remark4.98.
For regular continuity, \(\delta\) depends on both \(\varepsilon\) and \(x\text{,}\) but doesn't depend on \(x\) for uniform continuity. We can choose \(\delta\) uniformly for all \(x\text{.}\)
\(f\) is continuous: Note that for all \(x,y\in\mathbb{R},\left|{x^2-y^2}\right|=\left|{x+y}\right|\cdot\left|{x-y}\right|\text{.}\) Given \(\varepsilon>0,x\in\mathbb{R},\) set \(\delta_x=\min\{{1,\frac{\varepsilon}{2\left|{x}\right|+1}}\}\text{.}\) If \(y\in \mathbb{R}\) and \(\left|{x-y}\right|<\delta_x\text{,}\) then
Note: If we knew \(\inf_{x\in\mathbb{R}}\delta_x>0,\) then we could take \(\delta=\inf_{x\in \mathbb{R}}\delta_x>0\) and we'd have for all \(x,y\in\mathbb{R},\left|{x-y}\right|<\delta\implies\left|{f(x)-f(y)}\right|<\varepsilon\text{.}\) For our choice of \(\delta_x,\inf_{x\in\mathbb{R}}\delta_x=0\text{.}\)
\(f\) is not uniformly continuous. We must show exists \(\varepsilon>0\) such that for all \(\delta>0\) exists \(x,y\in \mathbb{R}\) with \(\left|{x-y}\right|<\delta\) and \(\left|{x^2-y^2}\right|\geq \varepsilon\text{.}\) Set \(\varepsilon=1\text{.}\) Fix \(\delta>0\text{.}\) Let \(x=\frac{1}{\delta},y=x+\frac{\delta}{2}\text{.}\) Then \(\left|{x-y}\right|=\frac{\delta}{2}<\delta\) and \(\left|{x^2-y^2}\right|=\left|{x+y}\right|\cdot\left|{x-y}\right|=9\frac{1}{\delta}+\frac{1}{\delta}+\frac{\delta}{2})\frac{\delta}{2}=1+\frac{\delta^2}{4}>1=\varepsilon\text{.}\)
Example4.101.
\(f:[0,\infty)\rightarrow\mathbb{R}, f(x)=\sqrt{x}\) is uniformly continuous.
\(f:(0,\infty)\rightarrow\mathbb{R},f(x)=\frac{1}{x}\) is not uniformly continuous, but for all \(a>0\text{,}\)\(f_{[a,\infty)}:[a,\infty)\rightarrow\mathbb{R}\) is uniformly continuous.
Theorem4.103.Continuous is Uniform From Compact Spaces.
If \(X,Y\) are metric spaces with \(X\) compact, then any continuous function \(X\rightarrow Y\) is uniformly continuous.
Example4.104.
Let \(f:\mathbb{Q}\rightarrow\mathbb{R}\) be given by \(f(x)=\begin{cases}-1&x<\sqrt{2}\\ 1&x>\sqrt{2}\end{cases}\text{.}\) Then \(f\) is continuous but not uniformly continuous.
Continuity: If \(x\in\mathbb{Q}\text{,}\) fix \(\delta>0\) with either \((x-\delta,x+\delta)\subseteq (-\infty,\sqrt{2})\) or \((x-\delta,x+\delta)\subseteq (\sqrt{2},\infty)\text{.}\) Note that if \(y\in\mathbb{Q}\) with \(\left|{x-y}\right|<\delta\text{,}\) then \(\left|{f(x)-f(y)}\right|=0<\varepsilon\text{.}\)
Not uniformly continuous: For \(\delta>0\text{,}\) let \(x\in(\sqrt{2}-\frac{\delta}{2},\sqrt{2})\cap\mathbb{Q}\) and \(y\in(\sqrt{2},\sqrt{2}+\frac{\delta}{2})\cap \mathbb{Q}\text{.}\) Then \(\left|{x-y}\right|<\delta\) and \(\left|{f(x)-f(y)}\right|<2\text{.}\)
Theorem4.105.Uniform Convergence and Cauchy Sequences.
If \(f:X\rightarrow Y\) is a uniformly continuous function, then for any Cauchy sequence \((x_n)_{n=1}^\infty\subseteq X\text{,}\) the sequence \((f(x_n))_{n=1}^\infty\subseteq Y\) is also Cauchy.
Let \((x_n)_{n=1}^\infty\subseteq X\) be Cauchy. Fix \(\varepsilon>0\text{.}\) Fix \(\delta>0\) so that whenever \(x,x'\in X\) with \(d(x,x')<\delta\) we have \(d(f(x),f(x'))<\varepsilon\) (using that \(f\) is uniformly continuous). Since \((x_n)_{n=1}^\infty\subseteq X\) is Cauchy, exists \(N\in\mathbb{Z}_+\) such that for all \(k,l\geq N, d(x_k,x_l)<\delta\text{,}\)\(d(f(x_k),f(x_l))<\varepsilon\) by the choice of \(\delta\text{.}\)
Remark4.106.
The converse is false! (I think ”converse" here means just continuous) \(f:\mathbb{R}\rightarrow \mathbb{R},f(x)=x^2\) is a counterexample.
If \(f:X\rightarrow Y\) is a function between metric spaces \(X\) and \(Y\text{,}\) then \(f\) preserves Cauchy sequences as in the proposition iff \(f\) is “uniformly continuous on totally bounded sets” (ie for all totally bounded \(A\subseteq X, f_A:A\rightarrow Y\) is uniformly continuous).
Theorem4.108.UMP for Complete Metric Spaces.
Let \(X,Y\) be metric spaces with \(Y\) complete. Let \(A\subseteq X\) and let \(f:A\rightarrow Y\) be uniformly continuous. Then \(\exists!\) continuous function \(\overline{f}:\overline{A}\rightarrow Y\) so the \(\overline{f}(a)=f(a)\) for all \(a\in A\text{.}\) Further, \(\overline{f}\) is uniformly continuous.
Uniqueness (this won't require that \(f\) is uniformly continuous): Suppose \(g,h:\overline{A}\rightarrow Y\) are continuous and \(g(a)=h(a)\) for all \(a\in A\text{.}\) If \(a\in\overline{A}, \exists \,(a_n)_{n=1}^\infty\subseteq A\) such that \(a=\lim_{n\rightarrow\infty}a_n\text{.}\) So
Existence Suppose \(a\in\overline{A}\text{.}\) Let \((a_n)_{n=1}^\infty\subseteq A\) be such that \(a=\lim_{n\rightarrow\infty}a_n\text{.}\) Then \((a_n)_{n=1}^\infty\subseteq A\) is Cauchy. Since \(f\) is uniformly continuous, \((f(a_n))_{n=1}^\infty\subseteq Y\) is Cauchy. Since \(Y\) is complete, \((f(a_n))_{n=1}^\infty\subseteq A\) converges. Define \(\overline{f}:\overline{A}\rightarrow Y\) by \(\overline{f}(a)=\lim_{n\rightarrow\infty}f(a_n)\text{.}\)
\(\overline{f}\) is well defined, ie if \((a_n)_{n=1}^\infty\subseteq A\) and \((b_n)_{n=1}^\infty\subseteq A\) with \(\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n\in\overline{A}\text{,}\) then \(\lim_{n\rightarrow\infty}f(a_n)=\lim_{n\rightarrow\infty}f(b_n)\text{.}\)
Let \(y_1=\lim_{n\rightarrow\infty}f(a_n),y_2=\lim_{n\rightarrow\infty}f(b_n)\text{.}\) To show \(y_1=y_2,\) we'll show for any \(\varepsilon>0,d(y_1,y_2)\leq \varepsilon\text{.}\) Let \(\varepsilon>0\text{.}\) Since \(f\) is uniformly continuous, exists \(\delta>0\) such that if \(a,b,\in A,d(a,b)<\delta\implies d(f(a),f(b))<\varepsilon\text{.}\) If \(c=\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n\in\overline{A},\) then exists \(N\geq 1\) such that for all \(n\geq N,d(a_n,c)<\frac{\delta}{2},d(b_n,c)<\frac{\delta}{2}\text{.}\) Then for all \(n\geq N, d(a_n,b_n)<\delta\text{.}\) So for all \(n\geq N\text{,}\) we have \(d(f(a_n),f(b_n))<\varepsilon\text{.}\) Now, \(d(y_1,y_2)=\lim_{n\rightarrow\infty}d(f(a_n),f(b_n))\leq \varepsilon\text{.}\) Since this holds for all \(\varepsilon>0,\) we have \(y_1=y_2\text{.}\)
\(\overline{f}(a)=f(a)\) for all \(a\in A\text{.}\)
Note that \((a)_{n=1}^\infty\subseteq A\) (constant sequence \(a\)) and \(a=\lim_{n\rightarrow\infty}a\text{.}\) So \(\overline{f}(a)=\lim_{n\rightarrow\infty}f(a)=f(a)\text{.}\)
\(\overline{f}\) is uniformly continuous.
Fix \(\varepsilon>0\text{.}\) Since \(f\) is uniformly continuous, exists \(\delta>0\) such that for all \(a,b\in A, d(a,b)<\delta\implies d(f(a),f(b))<\varepsilon\text{.}\) Now fix \(a,b\in\overline{A}\) with \(d(a,b)<\delta\text{.}\) We'll show that \(d(\overline{f}(a),\overline{f}(b))\leq \varepsilon\text{.}\) Fix sequences \((a_n)_{n=1}^\infty \subseteq A\) and \((b_n)_{n=1}^\infty \subseteq A\) such that \(a=\lim_{n\rightarrow\infty}a_n\) and \(b=\lim_{n\rightarrow\infty}b_n\text{.}\) So \(\overline{f}(a)=\lim_{n\rightarrow\infty}f(a_n)\) and \(\overline{f}(b)=\lim_{n\rightarrow\infty}f(b_n)\text{.}\) Note that \(\lim_{n\rightarrow\infty}d(a_n,b_n)=d(a,b)<\delta\text{.}\) So exists \(N\geq 1\) such that for all \(n\geq N\text{,}\)\(d(a_n,b_n)<\delta\text{.}\) So for all \(n\geq N, d(f(a_n),f(b_n))<\varepsilon\text{.}\) Letting \(n\rightarrow\infty, d(\overline{f}(a),\overline{f}(b))\leq \varepsilon\text{.}\)
