A sequence of functions \(\{f_n\}, n=1,2,3,\cdots,\) defined on \(E\), and suppose the sequence of numbers \(\{f_n(x)\}\) converges for every \(x\in E\), we define the function \(f\) by \[f(x)=\lim_{n\to \infty}f_n(x)\quad (x\in E)\] We say that \(\{f_n\}\) converges to \(f\) pointwise on \(E\) and \(f\) is the limit function.
A sequence of functions \(\{f_n\}, n=1,2,3,\cdots,\) converges uniformly on \(E\) to a function \(f\) if for every \(\epsilon>0\) there is an integer \(N\) such that \(n>N\) implies \[|f_n(x)-f(x)|\le\epsilon\] for all \(x\in E\). If \(\{f_n\}\) converges uniformly on \(E\), it is possible, for each \(\epsilon>0\), to find one integer \(N\) which will do for all \(x\in E\). The series \(\sum f_n(x)\) converges uniformly on \(E\) if the sequence \(\{s_n\}\) of partial sums defined by \[\sum_{i=1}^{n}f_i(x)=s_n(x)\] converges uniformly on \(E\).
Suppose \(f_n\to f\) uniformly on a set \(E\) in a metric space. Let \(x\) be a limit point of \(E\), and suppose that \[\lim_{t\to x}f_n(t)=A_n\quad(n=1,2,3,\cdots)\] Then there exists \(N\) such that \(n\ge N\), \(m\ge N\),\(t\in E\) imply that \[|f_n(t)-f_m(t)|\le\epsilon\] Let \(t\to x\), we obtain \[|A_n-A_m|\le\epsilon\] so that \(\{A_n\}\) is a Cauchy sequence and therefore converges, say to \(A\).
Suppose \(f_n\to f\) uniformly on a set \(E\) in a metric space. Let \(x\) be a limit point of \(E\), and suppose that \[\lim_{t\to x}f_n(t)=A_n\quad(n=1,2,3,\cdots)\] Because \[|f(t)-A|\le|f(t)-f_n(t)|+|f_n(t)-A_n|+|A_n-A|\] Choose \(n\) such that \[|f(t)-f_n(t)|\le\frac{\epsilon}{3}\] for all \(t\in E\) and \[|A_n-A|\le\frac{\epsilon}{3}\] Then for this \(n\), we choose a neighborhood \(V\) of \(x\) such that \[|f_n(t)-A_n|\le\frac{\epsilon}{3}\] if \(t\in V\cap E,t\ne x\). Then \[|f(t)-A|\le\epsilon\] or \[\lim_{t\to x}f(t)=\lim_{n\to \infty}A_n\] or in other words, the conclusion is \[\lim_{t\to x}\lim_{n\to\infty}f_n(t)=\lim_{n\to \infty}\lim_{t\to x}f_n(t)\] provided \(t\in V\cap E,t\ne x\).
Suppose \(K\) is compact and (a) \(\{f_n\}\) is a sequence of continuous functions on \(K\), (b) \(\{f_n\}\) converges pointwise to a continuous function \(f\) on \(K\), (c) \(f_n(x)\ge f_{n+1}(x)\) for all \(x\in K, n=1,2,3,\cdots\), then \(f_n\to f\) uniformly on \(K\). Put \(g_n=f_n-f\). Then \(g_n\) is continuous, \(g_n\to 0\) pointwise, and \(g_n\ge g_{n+1}\). We have to prove that \(g_n\to 0\) uniformly on \(K\). Let \(\varepsilon>0\), \(K_n\) be the set of all \(x\in K\) with \(g_n(x)\ge\varepsilon\). Since \(g_n\) is continuous, then \(K_n\) is closed. Because Closed subsets of a compact set are compact \(\Biggl(\)Suppose the closed subset of compact set \(F\subset K\subset X\), \(F\) is a closed subset relative to compact set \(K\), let \(\{V_{\alpha}\}\) be an open cover of \(F\), if \(F^c\) is adjoined to \(\{V_{\alpha}\}\), we obtain an open cover \(\Omega\) of \(K\). Since \(K\) is compact, there is finite sub-collection \(\Phi\) of \(\Omega\) which covers \(K\), and hence \(F\). If \(F^c\) is a member of \(\Phi\), we may remove it from \(\Phi\) and still retain an open cover of \(F\). Then, a finite sub-collection of \(\{V_{\alpha}\}\) covers \(F\) and \(F\) is compact.\(\Biggl)\), then \(K_n\) is compact. Since \(g_n\ge g_{n+1}\) we have \(K_n\supset K_{n+1}\). Fix \(x\in K\). Since \(g_n(x)\to 0\) we see that \(x\notin K_n\) if \(n\) is sufficiently large. Thus \(x\notin \cap K_n\). In other words \(\cap K_n\) is empty. Hence \(K_n\) is empty for some \(N\). \(\Biggl(\)If \(\{K_{\alpha}\}\) is a collection of compact subsets of a metric space \(X\) such that the intersection of every finite sub-collection of \(\{K_{\alpha}\}\) is nonempty, fix a member \(K_1\) of \(\{K_{\alpha}\}\) and denote \(G_{\alpha}=K_{\alpha}^c\). Assume that no point of \(K_1\) belongs to every \(K_{\alpha}\), then the sets \(G_{\alpha}\) form an open cover of \(K_1\) and since \(K_1\) is compact, there are finitely many indices \(\alpha_1,\cdots,\alpha_n\), such that \[K_1\subset(G_{\alpha_1}\cup\cdots\cup G_{\alpha_n})\] or \[K_1\subset(K_{\alpha_1}^c\cup\cdots\cup K_{\alpha_n}^c)\] or \[K_1\subset(K_{\alpha_1}\cap\cdots\cap K_{\alpha_n})^c\] but this means \[K_1\cap(K_{\alpha_1}\cap\cdots\cap K_{\alpha_n})\] is empty, which contradicts to our hypothesis. Then there must be some point of \(K_1\) belongs to every \(K_{\alpha}\) and \(\cap K_{\alpha}\) is nonempty.\(\Biggl)\) It follows that \(0\le g_n(x)\le \varepsilon\) for all \(x\in K\) and for all \(n\ge N\). Then \(f_n\to f\) uniformly on \(K\).
If \(X\) is a metric space, \(\mathscr C(X)\) will denote the set of all complex-valued, continuous, bounded functions with domain \(X\). Thus \(\mathscr C(X)\) consists of all complex continuous functions on \(X\) if \(X\) is compact. We associate with each \(f\in\mathscr C(X)\) its supremum norm \[\lVert f \rVert=\underset{x\in X}{\text{sup}}\lvert f(x)\rvert\] Since \(f\) is bounded, \(\lVert f \rVert<\infty\). If \(h=f+g\), then \[\lvert h(x) \rvert\le\lvert f(x) \rvert+\lvert g(x) \rvert\le\lVert f \rVert+\lVert g \rVert\] for all \(x\in X\). If we difine the distance between \(f\in \mathscr C(X)\) and \(g\in \mathscr C(X)\) to be \(\lVert f-g \rVert\), then \(\mathscr C(X)\) is a metric space.
Suppose \(\{f_n\}\) is a sequence of functions, differentiable on \([a,b]\) and \(\{f_n(x_0)\}\) converges for some point \(x_0\) on \([a,b]\). If \(\{f'_n\}\) converges uniformly on \([a,b]\), choose \(N\) such that \(n>N\), \(m>N\), implies \[|f_n(x_0)-f_m(x_0)|<\frac{\varepsilon}{2}\] and \[|f'_n(t)-f'_m(t)|<\frac{\varepsilon}{2(b-a)} \quad (a\leq t\leq b)\] If we apply the mean velue theorem to the function \(f_n-f_m\), then \[|f_n(x)-f_m(x)-(f_n(t)-f_m(t))|\leq\frac{|x-t|\varepsilon}{2(b-a)}\leq\frac{\varepsilon}{2}\] for any \(x\) and \(t\) on \([a,b]\), if \(n\ge N\) and \(m\ge N\). Then \[|f_n(x)-f_m(x)|\leq |f_n(x)-f_m(x)-(f_n(x_0)-f_m(x_0))|+|f_n(x_0)-f_m(x_0)|\leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\\ (a\leq x\leq b, n\ge N,m\ge N)\] So that \(\{f_n\}\) converges uniformly on \([a,b]\).
Suppose \(\{f_n\}\) is a sequence of functions, differentiable on \([a,b]\) and \(\{f_n(x_0)\}\) converges for some point \(x_0\) on \([a,b]\). If \(f_n\to f\) converges uniformly on \([a,b]\), then \[f(x)=\lim_{n\to \infty}f_n(x)\quad (a\leq x\leq b)\]. Let fix a point \(x\) on \([a,b]\) and define \[\phi_n(t)=\frac{f_n(t)-f_n(x)}{t-x}\] \[\phi(t)=\frac{f(t)-f(x)}{t-x}\] for \(a\leq t\leq b, t\ne x\). Then \[\lim_{t\to x}\phi_n(t)=f'_n(x)\quad (n=1,2,3,\cdots)\] Because \[|\phi_n(t)-\phi_m(t)|\leq \frac{\varepsilon}{2(b-a)}\quad (n\ge N,m\ge N)\] so that \(\{\phi_n\}\) converges uniformly, for \(t\ne x\). Since \(f_n\to f\), we conclude that \[\lim_{n\to \infty}\phi_n(t)=\phi(t)\] uniformly for \(a\leq t\leq b, t\ne x\). Then \[f'(x)=\lim_{t\to x}\phi(t)=\lim_{t\to x}\lim_{n\to \infty}\phi_n(t)=\lim_{n\to \infty}\lim_{t\to x}\phi_n(t)=\lim_{n\to \infty}f'_n(x)\\ (a\leq x\leq b)\]
Let \(\{f_n\}\) is a sequence of functions defined on a set \(E\). \(\{f_n\}\) is pointwise bounded on \(E\) if the sequence \(\{f_n(x)\}\) is bounded for every \(x\in E\), that is if there exists a finite-valued function \(\phi\) defined on \(E\) such that \[|f_n(x)|<\phi(x)\quad(x\in E, n=1,2,3,\cdots)\] We say that \(\{f_n\}\) is uniformly bounded on \(E\) if there exists a number \(M\) such that \[|f_n(x)|<M\quad(x\in E, n=1,2,3,\cdots)\] Now if \(\{f_n\}\) is pointwise bounded on \(E\) and \(E_1\) is a countable subset of \(E\), it is always possible to find a subsequence \(\{f_{n_k}\}\) such that \(\{f_{n_k}\}\) converges for every \(x\in E_1\).
Let \(f\) be a mapping of a metric space \(X\) into a metric space \(Y\), we say \(f\) is uniformly continuous on \(X\) if every \(\varepsilon>0\) there exists \(\delta>0\) such that \(d_Y(f(p),f(q))<\varepsilon\) for all \(p, q\in X\) for which \(d_X(p,q)<\delta\).
A family \(\mathscr F\) of complex functions \(f\) defined on a set \(E\) in a metric space \(X\) is said to be equicontinuous on \(E\) if for every \(\varepsilon>0\) there exists a \(\delta>0\) such that \[|f(x)-f(y)|<\varepsilon\] whenever \(d(x,y)<\delta,x\in E,y\in E, f\in\mathscr F\). Every member of an equicontinuous family is uniformly continuous.
continuous + compact space=uniformly continuous Let \(f\) be a continuous mapping of a compact metric space \(X\) into a metric space \(Y\), then \(f\) is uniformly continuous on \(X\). Given \(\epsilon>0\), since \(f\) is continuous, we can associate each point \(p\in X\) a positive number \(\phi(p)\) such that \(\underset{q\in X}{d_X(p,q)}<\phi(p)\) implies that \(d_Y(f(p), f(q))<\frac{\epsilon}{2}\). Let \(J(p)\) be the set of all \(q\in X\) for which \[d_X(p,q)<\frac{1}{2}\phi(p)\], since \(p\in J(p)\), the collection of all sets \(J(p)\) is an open cover of \(X\); and since \(X\) is compact, there is a finite set of points \(p_1,\cdots,p_n\) in \(X\) that \(X\subset J(p_1)\cup\cdots\cup J(p_n)\). Let \(\delta=\frac{1}{2}\text{min}[\phi(p_1),\cdots,\phi(p_n)]\), then \(\delta>0\). Now let \(p,q\) be the points of \(X\), such that \(d_X(p,q)<\delta\), since \(X\subset J(p_1)\cup\cdots\cup J(p_n)\), there is an integer \(m, 1\le m\le n\) such that \(p\in J(p_m)\), hence \(d_X(p,p_m)<\frac{1}{2}\phi(p_m)\) and we also have \[d_X(q,p_m)\le d_X(q,p)+d_X(p,p_m)<\delta+\frac{1}{2}\phi(p_m)\le \phi(p_m)\] therefore \[d_Y(f(p),f(q))\le d_Y(f(p),f(p_m))+d_Y(f(q),f(p_m))<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\] which means \(f\) is uniformly continuous on \(X\).
converges uniformly + compact space=equicontinuous If complex-valued, continuous function \(f_n\in\mathscr C(K), n=1,2,3,\cdots,\) converges uniformly on a compact metric space \(K\), then \(\{f_n\}\) is equicontinuous on \(K\). Since \(\{f_n\}\) converges uniformly, there is an integer \(N\) such that \[\lVert f_n-f_N \rVert<\varepsilon \quad(n>N)\] and since continuous functions are uniformly continuous on compact sets, there is a \(\delta>0\) such that \[\lvert f_i(x)-f_i(y) \rvert<\varepsilon \] if \(1\leq i\leq N, d(x,y)<\delta\). If \(n>N\) and \(d(x,y)<\delta\) it follows that \[\lvert f_n(x)-f_n(y) \rvert\leq\lvert f_n(x)-f_N(x) \rvert+\lvert f_N(x)-f_N(y)\rvert+\lvert f_N(y)-f_n(y)\rvert<3\varepsilon\] Then \(\{f_n\}\) is equicontinuous on \(K\).
pointwise bounded + equicontinuous + compact space=uniformly bounded If complex-valued, continuous function \(f_n\in\mathscr C(K), n=1,2,3,\cdots,\) pointwise bounded and equicontinuous on a compact metric space \(K\), then \(\{f_n\}\) is uniformly bounded on \(K\). Since \(\{f_n\}\) is equicontinuous, let \(\varepsilon>0\) and choose \(\delta>0\), \[|f_n(x)-f_n(y)|<\varepsilon\] for all \(n\) and \(d(x,y)<\delta\). Since \(K\) is compact, there are finitely many points \(p_1,\cdots,p_r\) in \(K\) such that to every \(x\in K\) corresponds at least one \(p_i\) with \(d(x,p_i)<\delta\). Since \(\{f_n\}\) is pointwise bounded, there exist \(M_i<\infty\) such that \(|f_n(p_i)|<M_i\) for all \(n\). If \(M=\text{max}(M_1,\cdots,M_r)\), then \(|f_n(x)|<M+\varepsilon\) for every \(x\in K\), which means \(\{f_n\}\) is uniformly bounded on \(K\).
pointwise bounded + countable set= convergent subsequence If \(\{f_n\}\) is a pointwise bounded sequence of complex-valued functions on a countable set \(E\), then \(\{f_n\}\) has a subsequence of \(\{f_{n_k}\}\) such that \(\{f_{n_k}(x)\}\) converges for every \(x\in E\). Let \(\{x_i\},i=1,2,3,\cdots,\) be the points of \(E\), arranged in a sequence. Since \(\{f_n(x_1)\}\) is bounded, there exists a subsequence, which we denote as \(\{f_{1,k}\}\) such that \(\{f_{1,k}(x_1)\}\) converges as \(k\to\infty\). For the sequences \[\begin{matrix} S_1:&f_{1,1}&f_{1,2}&f_{1,3}&f_{1,4}&\cdots\\ S_2:&f_{2,1}&f_{2,2}&f_{2,3}&f_{2,4}&\cdots\\ S_3:&f_{3,1}&f_{3,2}&f_{3,3}&f_{3,4}&\cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \end{matrix}\] which have the following properties: (a) \(S_n\) is a subsequence of \(S_{n-1}\), for \(n=2,3,4,\cdots\) (b) \(\{f_{n,k}(x_n)\}\) converges, as \(k\to \infty\) (The boundedness of \(\{f_{n}(x_n)\}\)) (c) The order in which the functions appear in the same in each sequence, if one function precedes another in \(S_1\), they are in the same relation in every \(S_n\), until one or the other is deleted. We consider the sequence \[\begin{matrix} S:&f_{1,1}&f_{2,2}&f_{3,3}&f_{4,4}&\cdots\\ \end{matrix}\] the sequence \(S\) (except possibly its first \(n-1\) terms) is a subsequence of \(S_n,n=1,2,3,\cdots\), Hence (b) implies that \(\{f_{n,n}(x_i)\}\) converges as \(n\to \infty\), for every \(x_i\in E\).
pointwise bounded + equicontinuous + compact space=uniformly convergent subsequence If complex-valued function \(f_n\in\mathscr C(K), n=1,2,3,\cdots,\) pointwise bounded and equicontinuous on a compact metric space \(K\), then \(\{f_n\}\) contains a uniformly convergent subsequence. Since \(K\) is compact, let \(E\) be a countable dense subset of \(K\), then \(\{f_{n}\}\) has a subsequence \(\{f_{n_i}\}\) such that \(\{f_{n_i}(x)\}\) converges for every \(x\in E\). Let \(\varepsilon>0\) and \(\delta>0\) and let \(V(x,\delta)\) be the set of all \(y\in K\) with \(d(x,y)<\delta\). Since \(E\) is dense in \(K\) and \(K\) is compact, there are finitely many points \(x_1,\cdots,x_m\) in \(E\) such that \[K\subset V(x_1,\delta)\cup\cdots\cup V(x_m,\delta)\] Since \(\{f_{n_i}(x)\}\) converges for every \(x\in E\), there is an integer \(N\) such that \[|f_{n_i}(x_s)-f_{n_j}(x_s)|<\varepsilon\] whenever \(i\ge N,j\ge N, 1\leq s\leq m\). If \(x\in K\), then \(x\in V(x_s,\delta)\) for some \(s\), so that \[|f_{n_i}(x)-f_{n_i}(x_s)|<\varepsilon\] for every \(i\). If \(i\ge N,j\ge N\), then \[|f_{n_i}(x)-f_{n_j}(x)|\leq|f_{n_i}(x)-f_{n_i}(x_s)|+|f_{n_i}(x_s)-f_{n_j}(x_s)|+|f_{n_j}(x_s)-f_{n_j}(x)|<3\varepsilon\] which is a uniformly convergent subsequence.
continuous function uniformly converged by polynomials There exists a sequence of polynomials \(P_n\) with its limit is a continuous complex function uniformly on \([a,b]\), such that \[\lim_{n\to\infty}P_n(x)=f(x)\] We assume \([a,b]=[0,1]\) and \(f(0)=f(1)=0\), we define \(f(x)\) to be zero for \(x\) outside \([0,1]\), then \(f\) is uniformly continuous on the whole line. Choose \(c_n\) to make \[\begin{align} 1=\int_{-1}^{1}c_n(1-x^2)^ndx&=c_n\int_{-1}^{1}(1-x^2)^ndx\\ &=2c_n\int_{0}^{1}(1-x^2)^ndx\\ &\ge 2c_n\int_{0}^{1/\sqrt{n}}(1-x^2)^ndx\\ &\ge 2c_n\int_{0}^{1/\sqrt{n}}(1-nx^2)dx\\ &=\frac{4c_n}{3\sqrt{n}}\\ &>\frac{c_n}{\sqrt{n}}\\ \end{align}\] \((n=1,2,3,\cdots)\) Then \(c_n<\sqrt{n}\) for any \((0<\delta\leq|x|\leq 1)\) \(c_n(1-x^2)^n\leq \sqrt{n}(1-\delta^2)^n\) so that \(c_n(1-x^2)^n\to 0\) uniformly in \((\delta\leq|x|\leq 1)\). Now let \[P_n(x)=\int_{-1}^{1}f(x+t)c_n(1-t^2)^ndt\quad(0\leq x\leq1)\] Given \(\varepsilon>0\) we choose \(\delta>0\) such that \(|y-x|<\delta\) implies \[|f(y)-f(x)|<\frac{\varepsilon}{2}\] Let \(M=\text{sup}|f(x)|\), for \(0\leq x\leq1\), \[\begin{align} |P_n(x)-f(x)|&=\Biggl|\int_{-1}^{1}\Bigl[f(x+t)-f(x)\Bigr]c_n(1-t^2)^ndt\Biggr|\\ &\leq \int_{-1}^{1}\Bigl|f(x+t)-f(x)\Bigr|c_n(1-t^2)^ndt\\ &\leq 2M\int_{-1}^{-\delta}c_n(1-t^2)^ndt+\frac{\varepsilon}{2}\int_{-\delta}^{\delta}c_n(1-t^2)^ndt+2M\int_{\delta}^{1}c_n(1-t^2)^ndt\\ &\leq 4M\sqrt{n}(1-\delta^2)^n+\frac{\varepsilon}{2}\\ &<\varepsilon \end{align}\] for all large enough \(n\).