18. Uniform continuity; Banach Contraction Principle; Sets of Discontinuity PDF
Uniform continuity
Uniformly continuous mappings
Uniformly continuous mappings. Let \((X,\rho_X)\) and \((Y,\rho_Y)\) be two metric spaces and \(f:X \to Y\). We say that \(f\) is uniformly continuous on \(X\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[\rho_Y(f(x),f(y))<\varepsilon\] for all \(x,y \in X\) for which \[\rho_X(x,y)<\delta.\]
Remark.
Uniform continuity is a property of a function on a set, whereas continuity can be defined at a single point.
Remarks
Remark 1.
If \(f\) is continuous on \(X\) then for each \(\varepsilon>0\) and \(p \in X\) there is \(\delta>0\) such that \(\rho_X(x,p)<\delta\) implies \(\rho_Y(f(x),f(p))<\varepsilon\).
Thus \(\delta>0\) depends on \(p \in X\) and \(\varepsilon>0\).
Remark 2.
If \(f\) is uniformly continuous on \(X\) then for each \(\varepsilon>0\) there is \(\delta>0\) such that for all \(x,y \in X\) if \(\rho_X(x,y)<\delta\) then \(\rho_X(f(x),f(p))<\varepsilon\).
Thus \(\delta>0\) depends only on \(\varepsilon>0\), but is uniform for all \(x,y \in X\).
Remark 3.
Uniform continuity implies continuity.
Continuity on compact spaces becomes uniform
Theorem. Let \(f\) be a continuous mapping of a compact metric space \((X,\rho_X)\) into a metric space \((Y,\rho_Y)\). Then \(f\) is uniformly continuous on \(X\).
Proof. Let \(\varepsilon>0\) be given.
Since \(f\) is continuous we can associate to each point \(p \in X\) a positive number \(\delta_p>0\) such that if \(q \in B(p,\delta_p)\), then \(\rho_Y(f(p),f(q))<\frac{\varepsilon}{2}\).
Observe that \[X \subseteq \bigcup_{p \in X}B\left(p,\frac{\delta_p}{2}\right).\]
Since \(X\) is compact there are \(p_1,p_2,\ldots,p_n \in X\) so that \[X \subseteq \bigcup_{k=1}^nB\left(p_k,\frac{\delta_{p_k}}{2}\right).\]
Set \[\delta=\frac{1}{2}\min\left(\delta_{p_1},\ldots,\delta_{p_n}\right)>0.\]
Let \(p,q \in X\) be such that \(\rho_X(p,q)<\delta\), then there is \(1 \leq m \leq n\) such that \(p \in B\left(p_m,\frac{\delta_{p_m}}{2}\right).\) Hence \[\rho_X(q,p_m) \leq \rho_X(q,p)+\rho_X(p_m,p) \leq \delta+\frac{\delta_{p_m}}{2}<\delta_{p_m}.\]
Thus we conclude \[\rho_Y(f(p),f(q)) \leq \rho_Y(f(p),f(p_m))+\rho_Y(f(p_m),f(q))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\] This completes the proof. $$\tag*{$\blacksquare$}$$
Example
Exercise. Let \(f(x)=\frac{1}{\sqrt{x}}\). Determine if it is uniformly continuous on \([1,2]\).
Solution. The interval \([1,2]\) is compact and the function \(f\) is continuous at every point of \([1,2]\). Hence, by the previous theorem, it is uniformly continuous.$$\tag*{$\blacksquare$}$$
Exercise. Let \[f(x)=\begin{cases}\frac{1}{x} \text{ if }x\neq 0,\\ 0 \text{ if }x=0. \end{cases}\] Determine if it is uniformly continuous on \([1,2]\).
Solution. The interval \([1,2]\) is compact and the function \(f\) is continuous at every point of \([1,2]\). Hence, by the previous theorem, it is uniformly continuous.$$\tag*{$\blacksquare$}$$
Exercise. Let \[f(x)=\begin{cases}\frac{1}{x} \text{ if }x\neq 0,\\ 0 \text{ if }x=0. \end{cases}\] Determine if it is uniformly continuous on \([0,1]\).
Solution. Let us consider \(a_n=\frac{1}{n}\). Then \(\lim_{n \to \infty}a_n=0\), but \[\lim_{n \to \infty}f(a_n)=\lim_{n \to \infty}n \neq f(0)=0,\] so \(f\) is not continuous at the point \(0\), so it is not uniformly continuous.$$\tag*{$\blacksquare$}$$
Exercise. Show that the function \[f(x)=\begin{cases}\frac{1}{x} \text{ if }x\neq 0,\\ 0 \text{ if }x=0, \end{cases}\qquad \text{ is not uniformly continuous on $(0, 1)$.}\]
Solution. It can be checked that \(f\) is continuous on \((0,1)\).
Suppose that \(f\) is uniformly continuous, then for every \(\varepsilon>0\) there is \(\delta>0\) such that for every \(x, y\in(0, 1)\) if \(|x-y|<\delta\) then \[|f(x)-f(y)|<\varepsilon.\]
We will use this condition with \(\varepsilon=1\) and \(x=\frac{1}{n}\) and \(y=\frac{1}{n+1}\).
This leads to a contradiction, since if \(\frac{1}{n}<\delta\), then we see that \[|x-y|=\frac{1}{n(n+1)}<\delta \quad \text{implies}\quad 1=|n-{n+1}|=|f(x)-f(y)|<1.\]
Banach contraction principle
Lipschitz mapping
Lipschitz mapping. Let \((X,\rho)\) be a metric space. We say that \(\phi:X \to X\) is a Lipschitz mapping of \(X\) into itself with the Lipschitz constant \(C_{\phi}>0\) if it satisfies \[\rho(\phi(x),\phi(y)) \leq C_{\phi}\rho(x,y)\quad \text{ for all }\quad x,y \in X.\]
Remark. Every Lipschitz mapping is uniformly continuous.
Example 1. Let \(X=\mathbb{R}\) and \(\phi(x)=ax+b\), then \(\phi\) is a Lipschitz map with \(C_{\phi}=|a|\) since \[|\phi(x)-\phi(y)|=|a||x-y| \quad \text{ for all }\quad x,y \in \mathbb{R}.\]
Lipschitz mappings - examples
Example 2. Let \((X,\rho)\) be a metric space and \(\varnothing \neq E \subseteq X\).
Define the distance from \(x \in X\) to \(E\) by setting \[\rho_E(x)=\inf\{\rho(x,z)\;:\;z \in E\}.\]
Sometimes we write \(\rho_E(x)=\rho(x,E)\).
Example
One can easily verify that \(\rho(x,E)=0\) iff \(x \in {\rm cl\;}(E)\).
Moreover, \[{\color{blue}|\rho(x,E)-\rho(y,E)| \leq \rho(x,y),}\] thus \(X \ni x \to \rho(x,E)\) is Lipschitz with the Lipschitz constant \(1\).
Indeed, for any \(z\in E\) we have \[\rho(x,E) \leq \rho(x,z) \leq \rho(x,y)+\rho(y,E),\] so \[\rho(x,E)-\rho(y,E) \leq \rho(x,y).\] By symmetry \(\rho(y,E)-\rho(x,E) \leq \rho(x,y),\) and we are done.$$\tag*{$\blacksquare$}$$
Contraction
Contraction. Let \((X,\rho)\) be a metric space. Suppose that \(\phi:X \to X\) and there is \(c \in (0,1)\) such that \[\rho(\phi(x),\phi(y)) \leq {\color{red}c}\rho(x,y)\quad \text{ for all }\quad x,y \in X\] then \(\phi\) is said to be a contraction of \(X\) into \(X\).
Remark. In other words, contractions \(\phi:X \to X\) are Lipschitz maps with the Lipschitz constants \[{\color{red}L_{\phi}<1.}\]
The Banach contraction principle
The Banach contraction principle. If \((X,\rho)\) is a complete metric space and if \(\phi\) is a contraction of \(X\) into \(X\), then there exists one and only one \(x \in X\) such that \({\color{teal}\phi(x)=x}\)
Idea of the proof
Proof of the uniqueness. If there are \(x,y \in X\) so that \(x \neq y\) and \(\phi(x)=x\) and \(\phi(y)=y\), then \[0<\rho(x,y)=\rho(\phi(x),\phi(y)) \leq c\rho(x,y)<\rho(x,y),\] which is a contradiction. Thus we must have \(x=y\). $$\tag*{$\blacksquare$}$$
Remark. The Banach contraction principle says that \(\phi\) has a unique fixed point.
Proof of the existence. The existence of a fixed point of \(\phi:X \to X\) is the essential part of the proof. The proof furnishes a construction method for looking for the fixed point.
Pick \(x_0\) arbitrarily and consider \[\begin{gathered} x_1=\phi(x_0), \qquad x_2=\phi(x_1)=\phi^2(x_0), \qquad x_3=\phi(x_2)=\phi^3(x_0), \ldots\\ {\color{blue} x_{n+1}=\phi(x_{n})=\phi^{n+1}(x_0) \quad \text{ for } \quad n \in \mathbb{N}.} \end{gathered}\]
Observe that \(\rho(x_{n+1},x_n)=\rho(\phi(x_n),\phi(x_{n-1})) \leq c\phi(x_n,x_{n-1}).\)
Thus inductively we obtain \(\rho(x_{n+1},x_n) \leq c^n \rho(x_1,x_0) \text{ for }n \in \mathbb{N}.\)
If \(n<m\) it follows that \[\begin{aligned} \rho(x_n,x_m) &\leq \sum_{j=n+1}^{m}\rho(x_j,x_{j+1}) \leq \left(c^n+c^{n+1}+\ldots+c^{m-1}\right)\rho(x_1,x_0)\\&\leq c^n(1+c+c^2+\ldots)\rho(x_0,x_1)=\frac{c_n}{1-c}\rho(x_0,x_1)\ _{\overrightarrow{n \to \infty}}\ 0. \end{aligned}\]
Thus the sequence \((x_n)_{n \in \mathbb{N}}\) is a Cauchy sequence in \(X\).
But \(X\) is complete metric space so there is \(x \in X\) such that \[\lim_{n \to \infty}x_n=x \quad \text{ for some }\quad x\in X.\]
Since \(\phi:X \to X\) is continuous, thus \[\phi(x)=\lim_{n \to \infty}\phi(x_n)= \lim_{n \to \infty}x_{n+1}=x. \qquad \tag*{$\blacksquare$}\]
Example
Exercise. Consider \(f:[1,\infty) \to [1,\infty)\) defined by \[f(x)=\frac{x}{4}+\frac{1}{4x}.\] Prove that \(f\) has an unique fixed point.
Solution. Note that \(f\) is a contraction. We have \[|f(x)-f(y)|=\left|\frac{x}{4}+\frac{1}{4x}-\frac{y}{4}-\frac{1}{4y}\right| \leq \frac{1}{4}|x-y|+\frac{1}{4}\left|\frac{1}{x}-\frac{1}{y}\right| \leq \frac{|x-y|}{2}.\]
By the Banach contraction principle, \(f\) has a unique fixed point.$$\tag*{$\blacksquare$}$$
Exercise. Prove that there is a unique \(x \geq 0\) such that \[x=\sqrt{2+x}.\]
Solution. Consider \(f:[0,\infty) \to [0,\infty)\) defined by \[f(x)=\sqrt{x+2}.\] By the Banach contraction principle, it suffices to prove that \(f\) is a contraction. Indeed, \[\left|\sqrt{x+2}-\sqrt{y+2}\right|=\frac{|x-y|}{\sqrt{x+2}+\sqrt{2+y}} \leq \frac{1}{2\sqrt{2}}|x-y|.\tag*{$\blacksquare$}\]
Exercise. Prove that the sequence \[\sqrt{2},\ \sqrt{2+\sqrt{2}},\ \sqrt{2+\sqrt{2+\sqrt{2}}},\ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}},\ldots\] converges.
Solution. As in the proof of the Banach contraction principle, the sequence defined by \[x_0=\sqrt{2}\quad \text{ and } \quad x_{n+1}=\sqrt{x_n+2} \quad \text{ for $n\in\mathbb N$}\] converges to the unique solution of \[x=\sqrt{2+x}.\] Since \(f(x)=\sqrt{x+2}\) is a contraction we are done. $$\tag*{$\blacksquare$}$$
Discontinuities
Discontinuities
Discontinuities. If \(x\) is a point in the domain of a function \(f\) at which \(f\) is not continuous we say that
\(f\) is discontinuous on \(X\),
or \(f\) has a discontinuity at \(x\in X\).
Definition. Let \(f:(a,b)\to \mathbb R\). Consider any \(x\) such that \(a< x <b\).
We write \(f(x+)=q\) if \(f(t_n) \ _{\overrightarrow{n \to \infty}}\ q\) for all sequences \((t_n)_{n \in \mathbb{N}}\) in \({\color{blue}(x,b)}\) such that \(t_n \ _{\overrightarrow{n \to \infty}}\ x\).
Similarly, \(f(x-)=q\) if \(f(t_n) \ _{\overrightarrow{n \to \infty}}\ q\) for all sequences \((t_n)_{n \in \mathbb{N}}\) in \({\color{red}(a,x)}\) such that \(t_n \ _{\overrightarrow{n \to \infty}}\ x\).
It is clear that for any \(x \in (a,b)\) \(\lim_{t \to x}f(t)\) exists iff \(f(x+)=f(x-)=\lim_{t \to x}f(t)\).
\(f(x+)\) and \(f(x-)\) - picture
Discontinuity of first and second kind
Let \(f:(a,b)\to \mathbb R\) be given.
Discontinuity of the first kind. If \(f\) is discontinuous at a point \(x\) and if \(f(x+)\) and \(f(x-)\) exist, then \(f\) is said to have discontinuity of the first kind or simple discontinuity at \(x\).
Discontinuity of the second kind. Otherwise the discontinuity is said to be of the second type.
Remark. There are two ways in which a function can have a simple discontinuity:
either \(f(x+) \neq f(x-)\),
or \(f(x+)=f(x-)\neq f(x)\).
\(f(x+) \neq f(x-)\)
\(f(x+)=f(x-)\neq f(x)\)
Continuous from the left and from the right
Continuous from the left. If \(f(x-)=f(x)\) for all \(x \in (a,b)\) then we say that \(f\) is continuous from the left.
Continuous from the right. If \(f(x+)=f(x)\) for all \(x \in (a,b)\) then we say that \(f\) is continuous from the right.
Integer part
Integer part \[\lfloor x\rfloor=\max\{n \in \mathbb{Z}:n \leq x\}\]
\[\text{\color{blue} is continuous from the right.}\].
Fractional part
Fractional part \[\{x\}=x-\lfloor x\rfloor\]
\[\text{\color{blue} is also continuous from the right.}\].
Examples involving characteristic function of \(\mathbb{Q}\)
Characteristic function of \(\mathbb{Q}\). The function \[f(x)=\begin{cases} 1 \text{ if }x \in \mathbb{Q},\\ 0 \text{ if }x \in \mathbb{R} \setminus \mathbb{Q}. \end{cases}\] has a discontinuity of the second kind at every point \(x\) since neither \(f(x+)\) nor \(f(x-)\) exists.
Characteristic function of \(\mathbb{Q}\) times linear function. Define \[f(x)=\begin{cases} x \text{ if }x \in \mathbb{Q},\\ 0 \text{ if }x \in \mathbb{R} \setminus \mathbb{Q}. \end{cases}\] Then \(f\) is continuous at \(x=0\), and \(f\) has a discontinuity of the second kind at every other point \(x\) since neither \(f(x+)\) nor \(f(x-)\) exists.
Example
An example of a function with a simple discontinuity at \(x=0\) that is continuous at every other point is given by the following formula \[f(x)=\begin{cases} x+2 &\text{ if }x <-2,\\ -x-2 &\text{ if }x \in [-2,0),\\ x+2 &\text{ if }x\ge 0. \end{cases}\]
Monotonically increasing and decreasing functions
Monotonically increasing (and decreasing) function. Let \(f:(a,b) \to \mathbb{R}\), then \(f\) is said to be monotonically increasing on \((a,b)\) if \(a<x<y<b\) implies \(f(x) \leq f(y)\). If \(f(x) \geq f(y)\) we obtain the definition of a monotonically decreasing function.
Theorem. Let \(f\) be a monotonically increasing on \((a,b)\). Then \(f(x+)\) and \(f(x-)\) exist at every point at \(x\in (a,b)\). More precisely, \[\sup_{a <t<x}f(t)=f(x-) \leq f(x) \leq f(x+) \leq \inf_{x<t<b}f(t).\]
Furthermore, if \(a<x<y<b\) then \(f(x+) \leq f(y-)\). Analogous result remains true for monotonically decreasing functions.
The set \[E=\{f(t)\;:\; a <t<x\}\] is bounded by \(f(x)\) hence \(A=\sup E \in \mathbb{R}\) and \(A \leq f(x)\).
We have to show \(f(-x)=A\).
Let \(\varepsilon>0\) be given. Since \(A=\sup E\) there is \(\delta>0\) such that \(a<x-\delta<x\) and \(A-\varepsilon<f(x-\delta) \leq A\). Since \(f\) is monotonic \[f(x-\delta) \leq f(t) \leq A \quad \text{ for }\quad t \in (x-\delta,x).\]
Thus \(A-\varepsilon<f(t) \leq A\), so \[|f(x)-A|<\varepsilon \quad \text{ for }\quad t \in (x-\delta,x).\]
Thus \(A=f(x-)\). In a similar way we prove \(f(x+)=\inf_{x<t<b}f(t)\).
Next if \(a<x<y<b\), then
\[f(x+)=\inf_{x<t<b}f(t)=\inf_{x<t<y}f(t).\]
Similarly \[f(y-)=\sup_{a<t<y}f(t)=\sup_{x<t<y}f(t).\]
Thus \[f(x+)=\inf_{x<t<y}f(t) \leq \sup_{x<t<y}f(t) = f(y-).\qquad \tag*{$\blacksquare$}\]
Corollary. Monotonic functions have no discontinuities of the second kind.
Theorem
Theorem. Let \(f:(a,b) \to \mathbb{R}\) be monotonic. Then the set of points of \((a,b)\) of which \(f\) is discontinuous is at most countable.
Proof. Wlog we may assume that \(f\) is increasing.
Let \(E\) be the set of points at which \(f\) is discontinuous.
With every point \(x \in E\) we associate a rational number \(r(x) \in \mathbb{Q}\) such that \[f(x-) <r(x)<f(x+),\] so \(r:E \to \mathbb{Q}\).
Since \(x_1<x_2\) implies \(f(x_1+) \leq f(x_2-)\) we see that \(r(x_1) \neq r(x_2)\) if \(x_1 \neq x_2\).
We have established that the function \(r:E \to \mathbb{Q}\) is injective, thus \[{\rm card\;}(E) \leq {\rm card\;}(\mathbb{Q})={\rm card\;}(\mathbb{N}). \qquad \tag*{$\blacksquare$}\]
