17. Continuous functions; Continuous functions on compact and connected sets PDF
Limits
Limits. Let \((X,\rho_X)\), \((Y,\rho_Y)\) be metric spaces. Suppose \(E \subseteq X\) and \(f:E \to Y\) and \(p\) is a limit point of \(E\). We write \[f(x) \ _{\overrightarrow{x \to p}}\ q \quad \text{ or }\quad \lim_{x \to p}f(x)=q.\] if there is a point \(q \in X\) satisfying the following \(\varepsilon\)-\(\delta\) condition:
For every \(\varepsilon>0\) there exists \(\delta>0\) such that \[\rho_Y(f(x),q)<\varepsilon\] for all points \(x \in E\) for which \(0<\rho_X(x,p)<\delta\).
Special case
If \(X=Y=\mathbb{R}\) then
\[\rho_X(x,y)=\rho_Y(x,y)=|x-y|\] and the condition reads as follows:
Limit. For every \(\varepsilon>0\) there exists \(\delta>0\) such that for all \(x \in E\) if \[0<|x-p|<\delta,\] then \[|f(x)-q|<\varepsilon.\]
Theorem
Theorem (Characterizations of Continuity). Let \((X,\rho_X)\), \((Y,\rho_Y)\) be metric spaces and \(E \subseteq X\), \(f:X \to Y\), and \(p\in X\) be as in the previous definition. Then
\(\lim_{x \to p}f(x)=q\) iff
\(\lim_{n \to \infty}f(p_n)=q\) for every sequence \((p_n)_{n \in \mathbb{N}}\) in \(E\) such that \(p_n \neq p\) and \(\lim_{n \to \infty}p_n =p\).
Proof (A)\(\Longrightarrow\)(B). Suppose that (A) holds. Choose \((p_n)_{n \in \mathbb{N}}\) like in condition (B). Let \(\varepsilon>0\) be given, then there exists \(\delta>0\) such that \[\rho_Y(f(x),q)<\varepsilon \quad \text{ if }\quad x \in E \quad \text{ and }\quad 0<\rho_X(x,p)<\delta.\] Also there exists \(N \in \mathbb{N}\) such that \(n \geq N\) implies \(0<\rho_X(p_n,p)<\delta\). Thus we also have \(\rho_Y(f(p_n),q)<\varepsilon\) for \(n\ge N\) showing that (B) holds. $$\tag*{$\blacksquare$}$$
Proof (B)\(\Longrightarrow\)(A). Conversely suppose (A) is false. Then there exists some \(\varepsilon>0\) such that for every \(\delta>0\) there exists a point \(x \in E\) (depending on \(\delta\)) for which \[\rho_Y(f(x),q) \geq \varepsilon \quad \text{ but } \quad 0<\rho_X(x,p)<\delta.\] Taking \(\delta_n=\frac{1}{n}\) for each \(n \in \mathbb{N}\) we thus find a sequence \((p_n)_{n \in \mathbb{N}}\) in \(E\) satisfying \(\lim_{n \to \infty}p_n =p\) but \[\rho_Y(f(p_n),q) \geq \varepsilon.\] thus (B) is false as desired. $$\tag*{$\blacksquare$}$$
Remark. It was possible to choose the sequence \((p _n)_{n \in \mathbb{N}}\) in \(E\) in one step thanks to the Axiom of Choice. Without assuming the Axiom of Choice the previous theorem is not provable.
Theorem. Suppose that \((X,\rho_X)\) is a metric space, and \(E \subseteq X\), and \(p\) is a limit point of \(E\). Let \(f,g:E \to \mathbb{R}\) be functions such that \[\lim_{x \to p}f(x)=A\quad \text{ and }\quad \lim_{x \to p}g(x)=B.\]
Then
\(\lim_{x \to p}(f+g)(x)=A+B\),
\(\lim_{x \to p}(f\cdot g)(x)=A\cdot B\),
\(\lim_{x \to p}\left(\frac{f}{g}\right)(x)=\frac{A}{B}\) if \(B \neq 0\) and \(g(x) \neq 0\) for \(x \in E\).
Continuous function
Continuous at the point \(p\). Suppose that \((X,\rho_x)\) and \((Y,\rho_Y)\) are metric spaces, \(E \subseteq X\), \(p \in E\) and \(f:E \to Y\). The function \(f\) is said to be continuous at point \(p\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
\[\rho_Y(f(x),f(p))<\varepsilon\] for all points \(x \in E\) for which \[\rho_X(x,p)<\delta.\]
Continuous function. If the function \(f:E \to Y\) is continuous at every point of \(E\) then \(f\) is said to be continuous on \(E\).
Special case
If \(X=Y=\mathbb{R}\) then
\[\rho_X(x,y)=\rho_Y(x,y)=|x-y|\] and the function \(f:E\to \mathbb{R}\) is said to be continuous at point \(p\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
\[|f(x)-f(p)|<\varepsilon\] for all points \(x \in E\) for which \[|x-p|<\delta.\]
Example
Example. Let us define \(f:\mathbb{R} \to \mathbb{R}\) by \[f(x)=\begin{cases} 1 \text{ if }x \in \mathbb{Q},\\ 0 \text{ if }x \not\in \mathbb{Q}. \end{cases}\] Determine if \(f\) is continuous or not at the point \(0\).
Solution. Let us consider the sequence \((a_n)_{n \in \mathbb{N}}\), where \(a_n=\sqrt{2}/n\). Then \(\lim_{n \to \infty}a_n=0\) and \(a_n \not\in \mathbb{Q}\), so \(f(a_n)=0\). Then \[\lim_{n \to \infty}f(a_n)=0 \neq 1 =f(0),\] so \(f\) is not continuous at point \(0\).$$\tag*{$\blacksquare$}$$
Example. Let us define \(f:\mathbb{R}^2 \to \mathbb{R}\) by \[f(x,y)=\begin{cases} 1 \text{ if }x+y \in \mathbb{Q},\\ 0 \text{ if }x+y \not\in \mathbb{Q}. \end{cases}\] Determine if \(f\) is continuous or not at the point \((0,0)\).
Solution. Let us consider the sequence \((a_n)_{n \in \mathbb{N}}\), where \(a_n=(0,\sqrt{2}/n)\). Then \(\lim_{n \to \infty}a_n=(0,0)\) and \(0+\sqrt{2}/n \not\in \mathbb{Q}\), so \(f(a_n)=0\). Then \[\lim_{n \to \infty}f(a_n)=0 \neq 1 =f(0,0),\] so \(f\) is not continuous at point \((0,0)\).$$\tag*{$\blacksquare$}$$
Remark
Remark. If \(p\) is an isolated point of \(E\) then our definition implies that every function \(f\) which has \(E\) as its domain is continuous at \(p\). For, no matter which \(\varepsilon>0\) we choose, we can pick \(\delta>0\) so that the only point \(e \in E\) for which \[\rho_X(x,p)<\delta\] is \(x=p\), then \[\rho_Y(f(x),f(p))=0<\varepsilon.\]
Fact. In the situation of the definition of continuity assume also that \(p\) is a limit point of \(E\). Then \(f\) is continuous at \(p\) iff \(\lim_{x \to p}f(x)=f(p).\)
Proof. It is obvious if we compare two previous definitions.$$\tag*{$\blacksquare$}$$
Theorem
Theorem. Suppose that \((X,\rho_X)\), \((Y,\rho_Y)\), and \((Z,\rho_Z)\) are metric spaces, let \(E \subseteq X\) and \(f:E \to Y\) and \(g:f[E] \to Z\) be given and define \(h:E \to Z\) by \[{\color{blue}h(x)=g(f(x)), \quad x \in E.}\] If \(f\) is continuous at a point \(p \in E\) and \(g\) is continuous at the point \(f(p)\), then \(h\) is continuous at \(p\). In other words \[\lim_{x\to p}h(x)=\lim_{x\to p}g(f(x))=g(f(p))=h(p).\]
Let \(\varepsilon>0\) be given.
Since \(g\) is continuous at \(f(p)\) there is \(\eta>0\) such that \[\rho_Z(g(y),g(f(p)))<\varepsilon\quad \text{ if }\quad \rho_Y(y,f(p))<\eta \quad \text{ and }\quad y \in f[E].\]
Since \(f\) is continuous at \(p\), there is \(\delta>0\) such that \[\rho_Y(f(x),f(p))<\eta \quad \text{ if }\quad \rho_X(x,p)<\delta \quad \text{ and }\quad x \in E.\]
If follows that \[\rho_Z(h(x),h(p))=\rho_Z(g(f(x)),g(f(p)))<\varepsilon\] if \(\rho_X(x,p)<\delta\) and \(x \in E\). Thus \(h\) is continuous at \(p \in E\).$$\tag*{$\blacksquare$}$$
Example
Example. Assume that \(f:\mathbb{R}^2 \to (0, \infty)\) is continuous for all \((x,y) \in \mathbb{R}^2\). Prove that \(h(x,y)=\sqrt{f(x,y)}\) is continuous.
Solution. Let us note that the function \(g:(0, \infty) \to (0, \infty)\) defined by \[g(x)=\sqrt{x}\] is continuous. We have \[h=g \circ f,\] so \(h\) is continuous by the previous theorem. $$\tag*{$\blacksquare$}$$
Theorem
Theorem. A mapping \(f\) of a metric space \((X,\rho_X)\) into a metric space \((Y,\rho_Y)\) is continuous on \(X\) iff \(f^{-1}[V]\) is open in \(X\) for every open set \(V\) in \(Y\).
Proof. Suppose that \(f\) is continuous on \(X\) and \(V \subseteq Y\) is open.
We have to show that \(f^{-1}[V]\) is open in \(X\). Let \(p \in f^{-1}[V]\). Since \(V\) is open \(B_{\rho_Y}(f(p),\varepsilon) \subseteq V\) for some \(\varepsilon>0\).
Since \(f\) is continuous at \(p \in X\) there is \(\delta>0\) such that \[\rho_{Y}(f(x),f(p))<\varepsilon \quad \text{ if }\quad \rho_X(x,p)<\delta.\] Thus \[B_{\rho_X}(p,\delta) \subseteq f^{-1}[V]=\{x \in X\;:\;f(x) \in V\}.\]
Conversely, suppose \(f^{-1}[V]\) is open in \(X\) for any open \(V \subseteq Y\).
Fix \(p \in X\) and \(\varepsilon>0\) and consider \[{\color{blue}V=B_{\rho_Y}(f(p),\varepsilon)}\] which is open thus \(f^{-1}[V]\) is open, hence there is \(\delta>0\) so that \(B_{\rho_X}(p,\delta) \subseteq f^{-1}[V]\).
Thus if \(\rho_X(x,p)<\delta\), then \(x \in f^{-1}[V]\), hence \[f(x) \in V=B_{\rho_Y}(f(p),\varepsilon) \quad \iff \quad \rho_Y(f(x),f(p))<\varepsilon. \qquad \tag*{$\blacksquare$}\]
Corollary. A mapping \(f:X \to Y\) between metric spaces \((X,\rho_X)\) and \((Y,\rho_Y)\) is continuous iff \(f^{-1}[C]\) is closed in \(X\) for any closed set \(C\) in \(Y\).
Proof. A set is closed iff its complement is open. We are done by invoking the previous theorem, since \(f^{-1}[E^c]=(f^{-1}[E])^c\) for every open set \(E \subseteq Y\). $$\tag*{$\blacksquare$}$$
Example
Example. Let \(f:\mathbb{R} \to \mathbb{R}\) be continuous and \(a \in R\). Prove that the set \[A=\{x \in \mathbb{R}\;:\;f(x)>a\}\] is open.
Solution: We have \[\{x \in \mathbb{R}\;:\;f(x)>a\}=f^{-1}[(a,\infty)]\] and \((a,\infty)\) is open in \(\mathbb{R}\), so by the previous theorem, \(A\) is open.$$\tag*{$\blacksquare$}$$
Example. Prove that the set \[A=\{(x,y) \in \mathbb{R}\;:\;\sqrt{x^2+y^2}<1\}\] is open in \(\mathbb{R}^2\) with the Euclidean metric.
Solution: Let us consider a continuous function \(f:\mathbb{R}^2 \to \mathbb{R}\) defined by \[f(x,y)=\sqrt{x^2+y^2}.\] Moreover, by the previous theorem \[A=\{(x,y) \in \mathbb{R}\;:\;f(x,y)<1\}=f^{-1}[B(0,1)]\] is open since \(B(0,1)\) is an open unit ball in \(\mathbb{R}^2\).$$\tag*{$\blacksquare$}$$
Theorem
Theorem. Let \(f,g:X \to \mathbb{R}\) be two continuous functions on a metric space \((X,\rho_X)\). Then \(f+g\), \(f\cdot g\), and \(\frac{f}{g}\) are continuous. In the last case we assume \(g(x) \neq 0\) for all \(x \in X\).
Example 1. Every polynomial \[p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\] is a continuous function on \(\mathbb{R}\).
Example 2. The exponential function \(f(x)=e^x\) is continuous as we have shown that for any \((a_n)_{n \in \mathbb{N}}\) so that \(\lim_{n \to \infty}a_n=a\) one has \(\lim_{n \to \infty}e^{a_n}=e^a.\)
Examples
Example 3. \(f(x)=|x|\) is continuous on \(\mathbb{R}\) since \(|f(x)-f(y)| \leq |x-y|\).
Example 4. \(f(x)=\lfloor x\rfloor=\max\{n \in \mathbb{Z}\;:\; n \leq x\}\) is NOT continuous at any \(x \in \mathbb{Z}\).
Example 5. \(f(x)=x^{\alpha}\) for any \(\alpha \in \mathbb{R}\) is continuous on \((0,\infty)\).
Example 6. If \(f,g:X \to \mathbb{R}\) are continuous then \(\max\{f,g\}\) and \(\min\{f,g\}\) are continuous as well. Indeed,
\[\max\{f,g\}=\frac{f+g+|f-g|}{2},\qquad \min\{f,g\}=\frac{f+g-|f-g|}{2}.\]
Continuity and compactness
Continuity and compactness
Bounded function. A mapping \(f:E \to \mathbb{R}\) is said to be bounded if there is a number \(M>0\) such that \[|f(x)| \leq M\quad \text{ for all }\quad x \in E.\]
Theorem (4.4.1). Suppose that \(f\) is a continuous mapping of a compact metric space \((X,\rho_X)\) into a metric space \((Y,\rho_Y)\). Then \(f[X]\) is compact in \(Y\).
Let \((V_{\alpha})_{\alpha \in A}\) be an open cover of \(f[X]\), i.e. \[f[X] \subseteq \bigcup_{\alpha \in A}V_{\alpha}.\] Since \(f\) is continuous then each set \(f^{-1}[V_{\alpha}]\) is open in \(X\). Since \(X\) is compact and \[X \subseteq \bigcup_{\alpha \in A}f^{-1}[V_{\alpha}]\] thus there are \(\alpha_1,\alpha_2,\ldots,\alpha_n \in A\) so that \[X \subseteq \bigcup_{j=1}^n f^{-1}[V_{\alpha_j}].\] Since \(f[f^{-1}[E]] \subseteq E\) we have \[f[X] \subseteq f\big[\bigcup_{j=1}^{n}f^{-1}[V_{\alpha_j}]\big] \subseteq \bigcup_{j=1}^{n}V_{\alpha_j}. \qquad \tag*{$\blacksquare$}\]
Corollary
Corollary. If \(f:X \to \mathbb{R}\) is continuous on a compact metric space \((X,\rho_{X})\) then \(f[X]\) is closed and bounded in \(\mathbb{R}\). Specifically, \(f\) is bounded.
Theorem. Suppose \(f:X \to \mathbb{R}\) is continuous on a compact metric space \((X,\rho_X)\) and \[M=\sup_{p \in X}f(p)\quad \text{ and }\quad m=\inf_{p \in X}f(p).\] Then there are \(p,q\) such that \[f(p)=M \quad \text{ and }\quad f(q)=m.\]
Proof. \(f[X] \subseteq \mathbb{R}\) is closed and bounded. Thus \(M\) and \(m\) are members of \(f[X]\) and we are done. $$\tag*{$\blacksquare$}$$
Theorem
Theorem. Suppose \(f\) is continuous injective mapping of a compact metric space \(X\) onto a metric space \(Y\). Then the inverse mapping \(f^{-1}\) defined on \(Y\) by \[f^{-1}(f(x))=x, \ \ \ x \in X\] is a continuous mapping of \(Y\) onto \(X\).
Proof. The inverse \(f^{-1}:Y \to X\) is well defined since \(f:X \to Y\) is one-to-one and onto. It suffices to prove that \(f[V]\) is open in \(Y\) for every open set \(V\) in \(X\). Fix \(V \subseteq X\) open, \(V^c\) is closed in \(X\) thus compact, hence \(f[V^c]\) is compact subset of \(Y\) and consequently \(f[V^c]\) is closed. Since \(f:X \to Y\) is one-to-one and onto, hence \[f[V]=\left(f[V^c]\right)^c\] and, consequently, \(f[V]\) is open as desired.$$\tag*{$\blacksquare$}$$
Continuity and connectivity
Continuity and connectivity
Theorem. If \(f:X \to Y\) is continuous mapping of a metric space \(X\) into a metric space \(Y\) and if \(E\) is a connected subset of \(X\) then \(f[E]\) is connected in \(Y\).
Proof. Assume for a contradiction that \(f[E]=A \cup B\), where \(A\) and \(B\) are nonempty separated sets in \(Y\). Put \[G=E \cap f^{-1}[A] \quad \text{ and }\quad H=E \cap f^{-1}[B].\]
Then \(E=G \cup H\) and neither \(G\) nor \(H\) is empty.
Since \(A \subseteq {\rm cl\;}(A)\) we have \(G \subseteq f^{-1}[{\rm cl\;}(A)]\) and the latter set is closed since \(f\) is continuous hence \({\rm cl\;}(G) \subseteq f^{-1}[{\rm cl\;}(A)]\).
Hence \[f[{\rm cl\;}(G)] \subseteq f[f^{-1}[{\rm cl\;}(A)]] \subseteq {\rm cl\;}(A).\]
Since \(f[H] \subseteq B\) and \({\rm cl\;}(A) \cap B = \varnothing\) we conclude that \[f[H \cap {\rm cl\;}(G)] \subseteq f[{\rm cl\;}(G)] \cap f[H] \subseteq {\rm cl\;}(A) \cap B=\varnothing,\] so \(H \cap {\rm cl\;}(G)=\varnothing\).
The same argument shows that \({\rm cl\;}(H) \cap G=\varnothing\).
Thus \(G\) and \(H\) are separated sets, which is a contradiction since \(E\) is connected.$$\tag*{$\blacksquare$}$$
Darboux property
Darboux property (intermediate value theorem). Let \(f\) be a continuous function on the interval \([a,b]\). If \(f(a)<f(b)\) and if \(c\) is a number such that \(f(a)<c<f(b)\), then there is a point \(x \in (a,b)\) such that \[f(x)=c.\] A similar result holds if \(f(a)>f(b)\).
Proof. \([a,b]\) is connected so \(f\big[[a,b]\big]\) is connected in \(\mathbb{R}\) as well by the previous theorem. Thus if \(f(a)<c<f(b)\), then \(c \in f\big[[a,b]\big]\), so there is \(x \in [a,b]\) so that \(f(x)=c.\)$$\tag*{$\blacksquare$}$$
Remark. The theorem stated above is sometimes called Darboux property or the intermediate value theorem.
Example
Exercise. Prove that the equation \[x^3-x^2+2x+3=0\] has a solution \(x_0\) such that \(-1 \leq x_0 \leq 0\).
Solution. Consider a continuous function \[f(x)=x^3-x^2+2x+3.\] We calculate \[f(-1)=-1, \quad \text{ and } \quad f(0)=3.\] It follows by the Darboux property that there is \(c \in [-1,0]\) such that \(f(c)=0\). Thus \(c\) is a solution of our equation as desired.$$\tag*{$\blacksquare$}$$
\(f(x)=x^3-x^2+2x+3\), \(x_0 \approx -0.8437\)
Example
Exercise. Prove that the equation \[x^3=20+\sqrt{x}\] has solution \(x_0\).
Solution. Consider a continuous function \[f(x)=x^3-\sqrt{x}-20.\] We calculate \[f(1)=-20<0, \quad \text{ and } \quad f(4)=42>0.\]
It follows by the Darboux property that there is \(c \in [1,4]\) such that \(f(c)=0\). Thus \(c\) is a solution of our equation as desired.$$\tag*{$\blacksquare$}$$
\(f(x)=x^3-\sqrt{x}-20\), \(x_0 \approx 2,7879\)
