On Sequences of Monotone Functions
Abstract Generated abstract
This paper studies sequences in the class of continuous nondecreasing functions on the unit interval normalized by the endpoint values 0 and 1. After proving auxiliary results on pointwise limits at continuity points and on uniform convergence under composition, it establishes a subsequence selection theorem: from any such sequence one can choose a subsequence and a uniformly convergent sequence of monotone reparametrizations so that the composed functions converge uniformly. The proof analyzes the discontinuities of a monotone pointwise limit, constructs intervals and linear modifications around the jumps, and uses these constructions to replace jump behavior by continuous transitions under composition.
Full Text
On Sequences of Monotone Functions
L. E. Portnov
(Presented by Prof. P. P. Kufarev)
Denote by \(M=\{\varphi(t)\}\) the set of all continuous nondecreasing functions \(\varphi(t)\), defined on \([0,1]\), such that \(\varphi(0)=0\), \(\varphi(1)=1\).
Lemma 1. Suppose that the sequence of functions
\[ \{\varphi_n(t)\}\subset M \tag{1} \]
converges and
\[ \lim_{n\to\infty}\varphi_n(t)=\varphi(t). \]
If \(t_0\) is a point of continuity of the function \(\varphi(t)\), then for any sequence \(\{t_n\}\), where \(t_n\in[0,1]\) and \(\lim_{n\to\infty} t_n=t_0\), the sequence \(\{\varphi_n(t_n)\}\) (2) converges and \(\lim_{n\to\infty}\varphi_n(t_n)=\varphi(t_0)\).
We shall regard the sequence \(\{t_n\}\) as strictly monotone, which can be done without loss of generality.
Since \(\varphi_n(t)\in M\) and the sequence \(\{t_n\}\) is monotone, the inequality
\[ \begin{aligned} |\varphi_{n+m}(t_{n+m})-\varphi_{n+m}(t_0)| &\le |\varphi_{n+m}(t_n)-\varphi_{n+m}(t_0)| \\ &\le |\varphi_{n+m}(t_n)-\varphi(t_n)|+ |\varphi_{n+m}(t_0)-\varphi(t_n)| \\ &\le |\varphi_n(t_0)-\varphi(t_n)|+ |\varphi_{n+m}(t_0)-\varphi_n(t_0)|+ \tag{3}\\ &\quad +|\varphi_{n+m}(t_n)-\varphi(t_n)| \\ &\le |\varphi_n(t_0)-\varphi(t_0)|+ |\varphi(t_n)-\varphi(t_0)|+ \\ &\quad +|\varphi_{n+m}(t_0)-\varphi_n(t_0)|+ |\varphi_{n+m}(t_n)-\varphi(t_n)| \end{aligned} \]
is valid.
Take \(\varepsilon>0\). It is easy to see that
\[ |\varphi_n(t_0)-\varphi(t_0)|<\frac{\varepsilon}{5} \qquad \text{for } n>N_1(\varepsilon,t_0), \tag{4} \]
\[ |\varphi_n(t_n)-\varphi(t_0)|<\frac{\varepsilon}{5} \qquad \text{for } n>N_2(\varepsilon,t_0), \]
\[ |\varphi_{n+m}(t_0)-\varphi_n(t_0)|<\frac{\varepsilon}{5} \qquad \text{for } n>N_3(\varepsilon,t_0), \]
\[ |\varphi_{n+m}(t_n)-\varphi(t_n)|<\frac{\varepsilon}{5} \qquad \text{for } n>N_4(\varepsilon,t_n). \]
Hence
\[ |\varphi_{n+m}(t_{n+m})-\varphi_{n+m}(t_0)|<\frac45\varepsilon \]
for
\[ n+m>N_5=\max_{i=1,2,3}N_i+N_4. \tag{5} \]
Consider the inequality
\[ |\varphi_n(t_n)-\varphi(t_0)|\leq |\varphi(t_0)-\varphi_n(t_0)|+|\varphi_n(t_n)-\varphi_n(t_0)|. \]
From (4) and (5) we have
\[ |\varphi(t_0)-\varphi_n(t_0)|<\frac{\varepsilon}{5} \qquad \text{for } n>N_3, \]
\[ |\varphi_n(t_n)-\varphi_n(t_0)|<\frac45\varepsilon \qquad \text{for } n>N_5, \]
and
\[ |\varphi_n(t_n)-\varphi(t_0)|<\varepsilon \qquad \text{for } n>N_5. \]
Lemma 2. Let the sequences of functions \(\{\varphi_n(t)\}\subset M\), \(\{\psi_n(t)\}\subset M\) be uniformly convergent.
Then the sequence of functions \(\{\varphi_n[\psi_n(t)]\}\) is also uniformly convergent.
The proof follows easily from the inequality
\[ |\varphi[\psi(t)]-\varphi_n[\psi_n(t)]|\leq |\varphi[\psi(t)]-\varphi[\psi_n(t)]|+ |\varphi[\psi_n(t)]-\varphi_n[\psi_n(t)]|. \]
Theorem. Let \(\{\varphi_n(t)\}\subset M\), \(n=1,2,\ldots\) (6). For some sequence \(\{\varphi_m(t)\}\subset\{\varphi_n(t)\}\), one can specify a uniformly convergent sequence of functions
\[ \{\psi_m(t)\}\subset M \tag{7} \]
such that the sequence of functions
\[ \{\varphi_m[\psi_m(t)]\} \tag{8} \]
will converge uniformly on \([0,1]\).
It suffices to prove the theorem for the case of a pointwise convergent sequence \(\{\varphi_n(t)\}\) to a discontinuous function \(\varphi(t)\). We shall carry out the proof under the assumption that the number of discontinuity points of \(\varphi(t)\) is infinite. In other cases the proof is simplified.
Proof. Since the function \(\varphi(t)\) is monotone, the points of the interval \([0,1]\) at which it is discontinuous can be numbered.
Let \(\varphi(t)\) have discontinuities at the points
\[ t_1,t_2,\ldots,t_i,\ldots \tag{9} \]
Put \(\varphi(t_i-0)=\alpha_1(i)\), \(\varphi(t_i+0)=\alpha_2(i)\), \(i=1,2,\ldots\).
For each point \(t_i\) construct sequences
\[ x_1(i),x_2(i),\ldots,x_n(i),\ldots \tag{10} \]
\[ y_1(i),y_2(i),\ldots,y_n(i),\ldots \tag{11} \]
satisfying the conditions:
\[ \text{1) }\lim_{n\to\infty}\varphi_n[x_n(i)]=\alpha_1(i),\qquad \lim_{n\to\infty}\varphi_n[y_n(i)]=\alpha_2(i), \]
\[ \text{2) }\lim_{n\to\infty}x_n(i)=\lim_{n\to\infty}y_n(i)=t_i, \]
\[ \text{3) for } i\ne j\quad (x_n^{(i)},y_n^{(i)})\cap (x_n^{(j)},y_n^{(j)})=0. \]
We shall show that such sequences (10) and (11) can be constructed. Introduce for consideration the sequences
\[ x_1^0(i),x_2^0(i),\ldots,x_n^0(i),\ldots \tag{12} \]
\[ y_1^0(i),\,y_2^0(i),\ldots,y_n^0(i),\ldots \tag{13} \]
where \(x_n^0(i)\) is such that \(\varphi_n[x_n^0(i)]=a_1(i)\) and, for \(t>x_n^0(i)\), \(\varphi_n(t)>a_1(i)\); \(y_n^0(i)\) is such that \(\varphi_n[y_n^0(i)]=a_2(i)\) and, for \(t<y_n^0(i)\), \(\varphi_n(t)<a_2(i)\).
Let \(d_i=a_2(i)-a_1(i)\). Since \(\varphi(t_i-0)=a_1(i)\), for each natural number \(k\) one can specify a point \(p_k\) such that
\[ |\varphi(p_k)-a_1(i)|<\frac{d_i}{2^{k+1}},\qquad |p_k-t_i|<\frac{1}{2^{k+1}}. \tag{14} \]
We obtain a sequence of points \(\{p_k\}\).
It is clear that \(\lim\limits_{k\to\infty}p_k=t_i\).
For each point \(p_k\in\{p_k\}\) and the sequence (6), one can specify an index \(m_k\) such that, for \(n>m_k\),
\[ |\varphi_n(p_k)-\varphi(p_k)|<\frac{d_i}{2^{k+1}}. \tag{15} \]
Thus, to the sequence \(\{p_k\}\) we put in correspondence a sequence of numbers \(\{m_k\}\). For the chosen \(m_k\), we shall choose \(m_{k+1}\) so that \(m_{k+1}>m_k\).
Now construct the sequence \(\{x_n(i)\}\)
\[ x_1(i)=x_1^0(i),\quad x_2(i)=x_2^0(i),\ldots,x_{m_1}^{(i)}=x_{m_1}^0 \]
\[ \left. \begin{aligned} x_{m_1+2}^{(i)}=x_{m_1+2}^{(i)}=\cdots=x_{m_2}^{(i)}&=p_1,\\ x_{m_2+1}^{(i)}=x_{m_2+2}^{(i)}=\cdots=x_{m_3}^{(i)}&=p_2,\\ &\cdots\\ x_{m_k+1}^{(i)}=x_{m_k+2}^{(i)}=\cdots=x_{m_{k+1}}^{(i)}&=p_k,\\ &\cdots \end{aligned} \right\} \]
We define it in this way for all \(n>m_k\) such that \(\varphi_n(p_k)-\varphi(p_k)>0\); if, however, \(\varphi_n(t)\) is such that \(\varphi_n(p_k)-\varphi(p_k)\le 0\), then we put \(x_n^{(i)}=x_n^0(i)\).
Similarly we construct the sequence \(\{y_n(i)\}\). The sequences \(\{x_n^{(i)}\}\), \(\{y_n^{(i)}\}\) constructed by us may be regarded, respectively, as the sequences (10) and (11).
1) \(\lim\limits_{n\to\infty}\varphi_n[x_n(i)]=a_1(i)\), which follows from (10), since \(\lim\limits_{k\to\infty}\varphi(p_k)=a_1(i)\).
Similarly, \(\lim\limits_{n\to\infty}\varphi_n[y_n(i)]=a_2(i)\).
2) \(\lim\limits_{n\to\infty}x_n(i)=\lim\limits_{n\to\infty}y_n(i)=t_i\),
3) \((x_n^{(i)},y_n^{(i)})\cap(x_n(j),y_n(j))=0\) for \(i\ne j\); this follows from the fact that \((x_n^0(i),y_n^0(i))\cap(x_n^0(j),y_n^0(j))=0\) for \(i\ne j\), and our sequences \(\{x_n(i)\}\), \(\{y_n(i)\}\) are constructed in such a way that \((x_n(i),y_n(i))\subset(x_n^0(i),y_n^0(i))\), since from (9) and (10) it is clear that \(\varphi_n(x_n(i),y_n(i))\subset(a_1(i),a_2(i))\).
Construct a sequence of functions
\[ \{\tilde{\varphi}_n(t)\}\subset M, \tag{16} \]
where
\[ \tilde{\varphi}_n(t)=\varphi_n(t)\quad \text{for } t\notin U(x_n(i),y_n(i)), \text{ and is linear on } [x_n(i),y_n(i)]. \]
Construct sequences of functions
\[ \{\delta_n^k(t)\}\quad k=1,2\ldots \tag{17} \]
for \(k=k_0,\ n=1,2,\ldots\) in the following manner:
\[ \delta_n^1(t)= \begin{cases} x_n(1), & \text{for } t=x_1(1),\\ y_n(1), & \text{for } t=y_1(1),\\ \text{linear on } [0,x_1(1)],\ [x_1(1),y_1(1)],\ [y_1(1),1]. \end{cases} \]
It is clear that \(\delta'(t)=\lim\limits_{n\to\infty}\delta'_n(t)\) is a continuous function on \([0,1]\), and on \([0,x_1(1)]\), \([y_1(1),1]\) the function \(\delta'(t)\) is strictly monotone; we may assume that \(0\ne t_1\ne 1\) and, for \(t\in[x_1(1),y_1(1)]\), \(\delta'(t)=t_1\).
Let \(\delta'(t)=t_2\) for \(t=\tau_2\), \(\tau_2\in\Delta\), where \(\Delta\) is an interval of strict monotonicity of the function \(\delta'(t)\).
Set
\[ \delta^2(t)= \begin{cases} \delta'(t) & \text{for } t\in \overline{(R_1,S_1)},\\ t_2 & \text{for } t\in \overline{\delta}_{\tau_2}=[a_1,b_1],\\ \text{where } \tau_2\in[a_1,b_1]\subset(R_1,S_1)\text{ and the segment }[a_1,b_1]\text{ is so small}\\ \text{that }|\delta^1(t)-\delta^2(t)|<\dfrac14. \end{cases} \]
Construct a sequence of functions
\[ \{\delta_n^2(t)\}\subset M, \]
as follows:
\[ \delta_n^2(t)= \begin{cases} \delta'_n(t) & \text{for } t\in \overline{(R_1,S_1)},\\ x_n(2) & \text{for } t=a_1,\\ y_n(2) & \text{for } t=b_1,\\ \text{linear on }[R_1,a_1],\ [a_1,b_1],\ [b_1,S_1]. \end{cases} \]
It is clear that
\[ \lim_{n\to\infty}\delta_n^2(t)=\delta^2(t). \]
Suppose that \(\delta^3(t)\) and \(\{\delta_n^3(t)\},\ldots,\delta^k(t)\) and \(\{\delta_n^k(t)\}\) have already been constructed, and suppose that \(\delta^k(t)=t_{k+1}\) for \(t=\tau_{k+1}\), \(\tau_{k+2}\in\Delta\), where \(\Delta=[R_k,S_k]\) is an interval of strict monotonicity of the function \(\delta^k(t)\).
Set
\[ \delta^{k+1}(t)= \begin{cases} \delta^k(t) & \text{for } t\in \overline{[R_k,S_k]},\\ t_{k+1} & \text{for } t\in \overline{\delta}_{\tau_{k+1}}=[a_k,b_k], \end{cases} \]
\[ \delta^{k+1}(t)= \begin{cases} \delta^k(t) & \text{for } t\in \overline{[R_k,S_k]},\\ t_{k+1} & \text{for } t\in \overline{\delta}_{\tau_{k+1}}=[a_k,b_k],\\ \text{where } \tau_{k+1}\in[a_k,b_k]\subset(R_k,S_k)\text{ and the segment }[a_k,b_k]\text{ is}\\ \text{so small that }|\delta^k(t)-\delta^{k+1}(t)|<\dfrac{1}{2^{k+1}}. \end{cases} \tag{18} \]
We also construct a sequence of functions
\[ \{\delta_n^{k+1}(t)\}\subset M, \]
where
\[ \delta_n^{k+1}(t)= \begin{cases} \delta_n^k(t) & \text{for } t\in \overline{(R_k,S_k)},\\ x_n(k+1) & \text{for } t=a_k,\\ y_n(k+1) & \text{for } t=b_k,\\ \text{linear on }[R_k,a_k],\ [a_k,b_k],\ [b_k,S_k]. \end{cases} \]
It is clear that
\[ \lim_{n\to\infty}\delta_n^{k+1}(t)=\delta^{k+1}(t). \]
Thus, we have uniformly convergent sequences of functions
\[ \delta_1'(t),\ldots,\delta_n'(t),\ldots\ \lim \delta_n'=\delta'(t), \]
\[ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot \]
\[ \delta_1^k(t),\ldots,\delta_n^k(t),\ldots\quad \lim \delta_n^k(t)=\delta^k(t). \tag{19} \]
By virtue of (18), \(\delta(t)=\lim\limits_{k\to\infty}\delta^k(t)\) is a continuous function.
Consider the functions \(\varphi_n(t)\in(11)\), \(\gamma_n(t)\in(16)\)
\[ \varphi_n[x_n(i)]=\gamma_n[x_n(i)],\qquad \varphi_n[y_n(i)]=\gamma_n[y_n(i)] \quad\text{(by virtue of (16)).} \tag{20} \]
Put
\[ \varepsilon_n(i)= \frac{\gamma_n[y_n(i)]-\gamma_n[x_n(i)]}{2^n}. \]
Partition the interval \([x_n(i),y_n(i)]\) by the points
\(x_n(i)=l_0(i)<l_1(i)<\ldots<l_{2^n}(i)=y_n(i)\), so that
\(\varphi_n[x_n(i)]+(p-1)\varepsilon_n(i)\leq \varphi_n(t)\leq
\varphi_n[x_n(i)]+p\varepsilon_n(i)\) for
\(t\in[l_{p-1}^{(i)},l_p(i)]\), and by the points
\(x_n(i)=S_0(i)<S_1(i)<\ldots<S_{2^n}(i)=y_n(i)\), so that
\(\gamma_n[x_n(i)]+(p-1)\varepsilon_n(i)\leq \gamma_n(t)\leq
\gamma_n[x_n(i)]+p\varepsilon_n(i)\), \(t\in[S_{p-1}(i),S_p(i)]\),
\(p=1,2\ldots 2^n\).
Put
\[ \eta_n^k(t)=\alpha_n^k(t)+\sum_{i=1}^{k}\sum_{p=1}^{2^n}\beta_p^i(t), \qquad k=1,2,\ldots \]
for \(k=k_0\), \(n=1,2\ldots\),
\[ \beta_p^i(t)= \begin{cases} \displaystyle \frac{l_p^{(i)}(t-S_{p-1}^{(i)})+l_{p-1}^{(i)}(S_p^{(i)}-t)} {S_p^{(i)}-S_{p-1}^{(i)}} & \text{for } t\in[S_{p-1}(i),S_p(i)],\\[1.2em] 0 & \text{for } t\in[0,1]\setminus [S_{p-1}(i),S_p(i)], \end{cases} \]
\[ \alpha_n^k(t)= \begin{cases} t & \text{for } t\in[0,1]\setminus\displaystyle\bigcup_{i=1}^{k}[x_n(i),y_n(i)],\\[0.8em] 0 & \text{for } t\in\displaystyle\bigcup_{i=1}^{k}[x_n(i),y_n(i)]. \end{cases} \]
It is easy to see that the functions of the sequence \(\{\eta_n^k(t)\}\) (20) are such that
\(\eta_n^k(t)\in M\), \(\eta_n^k[x_n(i)]=x_n(i)\),
\(\eta_n^k[y_n(i)]=y_n(i)\), and, for \(t\in[x_n(i),y_n(i)]\),
\[ |\varphi_n[\eta_n^k(t)]-\gamma_n(t)|\leq \max_i \varepsilon_n(i)\leq \frac{1}{2^n}. \tag{21} \]
Since \(\lim\limits_{n\to\infty}x_n(i)=\lim\limits_{n\to\infty}y_n(i)=t_i\), it follows that
\(\gamma_1^k(t)=\lim\limits_{n\to\infty}\eta_n^k(t)=t\), i.e., each sequence of functions
\(\{\eta_n^k(t)\}\), \(k=1,2,\ldots\), converges uniformly on \([0,1]\) ([1]).
Thus, we have sequences of uniformly convergent functions
\[ \left. \begin{array}{c} \gamma_1'(t),\ldots,\gamma_n'(t),\ldots\quad \lim\limits_{n\to\infty}\gamma_n'(t)=t,\\ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\\ \gamma_1^k(t),\ldots,\gamma_n^k(t),\ldots\quad \lim\limits_{n\to\infty}\gamma_n^k(t)=t,\\ \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot\quad \cdot \end{array} \right\} \tag{22} \]
Let \(\{k_m\}\), \(m=1,2,\ldots\), be a sequence of numbers of the natural series such that, for \(i>k_m\),
\[ d_i=\alpha_2(i)-\alpha_1(i)<\frac{1}{2^m}. \tag{23} \]
Such a sequence exists, since \(\sum_i d_i < 1\). Next, let \(\{n_m\}\) be a sequence of natural numbers such that
\[ \left|\eta_{n_m}^{\kappa_m}(t)-t\right|<\frac{1}{2^m}, \qquad \left|\delta_{n_m}^{\kappa_m}(t)-\hat{\delta}^{\kappa_m}(t)\right|<\frac{1}{2^m}, \]
and we construct the sequence \(\{n_m\}\) so that \(n_m>m\). Such a sequence of numbers \(\{n_m\}\) can be chosen, since the sequences (19) and (22) converge uniformly. Hence, taking into account (16), (20), (21), (23), and the fact that \(n_m>m\), we have
\[ \left|\varphi_{n_m}\left[\eta_{n_m}^{\kappa_m}(t)\right]-\gamma_{n_m}(t)\right|<\frac{1}{2^m}, \tag{24} \]
\[ \lim_{m\to\infty}\gamma_{n_m}^{\kappa_m}(t)=t, \qquad \lim_{m\to\infty}\delta_{n_m}^{\kappa_m}(t)=\lim_{m\to\infty}\hat{\delta}^{\kappa_m}(t)=\delta(t). \]
For simplicity, denote
\[ \{\varphi_{n_m}(t)\}=\{\varphi_m(t)\}, \qquad \{\gamma_{n_m}(t)\}=\{\gamma_m(t)\}, \]
\[ \{\eta_{n_m}^{\kappa_m}(t)\}=\{\eta_m(t)\}, \qquad \{\delta_{n_m}^{\kappa_m}(t)\}=\{\delta_m(t)\}. \]
Inequality (24) will be written as follows:
\[ \left|\varphi_m[\eta_m(t)]-\gamma_m(t)\right|<\frac{1}{2^m}. \tag{24′} \]
Let \(\psi_m(t)=\eta_m[\delta_m(t)]\). By Lemma (2), the sequence of functions \(\{\psi_m(t)\}\) (25) converges uniformly.
Now, to prove the theorem it remains to show that the sequence of functions \(\{\gamma_m[\delta_m(t)]\}\) (26) converges uniformly.
Indeed, let \(\varepsilon>0\) be arbitrary.
\[ \begin{aligned} &\left|\varphi_m[\psi_m(t)]-\varphi_{m+n}[\psi_{m+n}(t)]\right| =\left|\varphi_m\{\eta_m[\delta_m(t)]\}-\varphi_{m+n}\{\eta_{m+n}[\delta_{m+n}(t)]\}\right| \leq \\ &\leq \left|\varphi_m\{\eta_m[\delta_m(t)]\}-\gamma_m[\delta_m(t)]\right| +\left|\varphi_{m+n}\{\eta_{m+n}[\delta_{m+n}(t)]\}-\gamma_{m+n}[\delta_{m+n}(t)]\right| \\ &\qquad +\left|\gamma_m[\delta_m(t)]-\gamma_{m+n}[\delta_{m+n}(t)]\right|. \end{aligned} \tag{27} \]
From the obtained inequality (27), taking into account (24′), it follows that the sequence of functions \(\{\varphi_m[\psi_m(t)]\}\) (28) converges uniformly on the interval \([0,1]\), if the sequence of functions (26) converges uniformly.
We shall show that the sequence of functions (21) converges uniformly. For this it is enough, taking [1] into account, to show that \(\xi(t)=\lim_{m\to\infty}\gamma_m[\delta_m(t)]\) is continuous on \([0,1]\).
For \(t\in[a_i,b_i]\),
\[ \gamma_m[\delta_m(t)] = \frac{\gamma_m[y_m(i)]-\gamma_m[x_m(t)]}{b_i-a_i}(t-b_i) +\gamma_m[y_m^{(i)}]. \]
But since \(\lim_{m\to\infty}\gamma_m[x_m(i)]=\alpha_1(i)\), \(\lim_{m\to\infty}\gamma_m[y_m(i)]=\alpha_2(i)\), it follows that
\[ \xi(t)=\frac{\alpha_2(i)-\alpha_1(i)}{b_i-a_i}(t-b_i)+\alpha_2(i), \qquad t\in[a_i,b_i]. \]
Let
\[ E=[0,1]\setminus\bigcup_i t_i, \]
where \(t_i\) is a point of discontinuity of the function \(\varphi(t)\).
On \(E\) the function \(\varphi(t)\) is continuous. We shall show that \(\xi(t)=\varphi[\delta(t)]\) for \(t\in\delta^{-1}(E)\),
then, since \(\delta(t)\) is continuous on \([0,1]\), and \(\varphi(t)\) is continuous on \(E\), the function \(\xi(t)=\varphi[\delta(t)]\) is continuous on
\(\delta^{-1}(E)=[0,1]\setminus\bigcup_i [a_i,b_i]\).
Let \(t_0\in\delta^{-1}(E)\). Then \(\delta(t_0)\in E\), i.e., \(\delta(t_0)\) is a point of continuity of the function \(\varphi(t)\).
Since \(\lim\limits_{m\to\infty}\delta_m(t_0)=\delta(t_0)\), it follows by Lemma 1 that
\[ \lim_{m\to\infty}\gamma_m[\delta_m(t_0)] = \lim_{m\to\infty}\varphi_m[\delta_m(t_0)] = \varphi[\delta(t_0)]. \]
For \(t\in([0,1]\setminus\bigcup_i [a_i,b_i])\), \(\gamma_m[\delta_m(t)]=\varphi_m[\delta_m(t)]\) by virtue of (13) and (16).
It is now not hard to see that
\(\xi(t)=\lim\limits_{m\to\infty}\gamma_m[\delta_m(t)]\) is a function continuous at all points of \([0,1]\), since
\[ \xi(t)= \begin{cases} \varphi[\delta(t)], & \text{for } t\in\delta^{-1}(E)=[0,1]\setminus\displaystyle\bigcup_i [a_i,b_i],\\[6pt] \dfrac{\alpha_2(i)-\alpha_1(i)}{b_i-a_i}(t-b_i)+\alpha_2(i), & \text{for } t\in [a_i,b_i], \end{cases} \]
and if \(t_k\in\delta^{-1}(E)\), then, as \(t_k\to a_i\),
\(\lim\limits_{k\to\infty}\varphi[\delta(t_k)]=\alpha_1(i)\), since \(\delta(t_k)\to(t_i-0)\), while as \(t_k\to b_i\),
\(\lim\limits_{k\to\infty}\varphi[\delta(t_k)]=\alpha_2(i)\), since \(\delta(t_k)\to(t_i+0)\).
If, however, \(\{t_k\}\in[0,1]\setminus\delta^{-1}(E)\), then as \(t_k\to a_i\),
\(\lim\limits_{k\to\infty}\xi(t_k)=\alpha_1(i)\), and as \(t_k\to b_i\),
\(\lim\limits_{k\to\infty}\xi(t_k)=\alpha_2(i)\), i.e., \(\xi(t)\) is continuous at the points \(a_i,b_i\).
REFERENCES
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