THE METRIC SPACE OF CONTINUOUS CURVES OF A METRIC SPACE

L. E. Portnov

(Presented by Prof. P. P. Kufarev)

By \(L_R\) we shall mean the set of all continuous curves of the metric space \(R\) [1].

Theorem 1. Let \(\varphi_1(t), \varphi_2(t)\in M\), where \(M=\{\varphi(t)\}\) is the set of all possible continuous nondecreasing functions \(\varphi(t)\) such that \(\varphi(0)=0,\varphi(1)=1\). There exist \(\varphi_3(t), \varphi_4(t)\in M\) such that

\[ \varphi_1[\varphi_3(t)] = \varphi_2[\varphi_4(t)]. \]

Proof. Let \(\varepsilon_n=\dfrac{1}{2^n}\), \(n=1,2,\ldots\). Partition the interval \([0,1]\) by the points

\[ 0=t_0<t_1<t_2<\ldots<t_{2^n}=1 \]

so that

\[ (k-1)\varepsilon_n\leq \varphi_1(t)\leq k\varepsilon_n \quad \text{for } t\in [t_{k-1},t_k], \]

and by the points

\[ 0=t'_0<t'_1<\ldots<t'_{2^n}=1 \]

so that

\[ (k-1)\varepsilon_n\leq \varphi_2(t)\leq k\varepsilon_n \quad \text{for } t\in [t'_{k-1},t'_k],\quad k=1,2\ldots 2^n. \]

Put

\[ \varphi_3^n(t)=\sum_{k=1}^{2^n}\psi_k(t), \]

where

\[ \psi_k(t)= \begin{cases} \dfrac{t_{k-1}(t'_k-t)+t_k(t-t'_{k-1})}{t'_k-t'_{k-1}} \end{cases} \]

for \(t\in [t'_{k-1},t'_k]\), and for \(t\in [0,1]\setminus [t'_{k-1},t'_k]\).

It is not difficult to see that

\[ \varphi_3^n(t)\in M \quad \text{and} \]

\[ |\varphi_2[\varphi_3^n(t)]-\varphi_1(t)|\leq \varepsilon_n, \]

i.e.,

\[ \lim_{n\to\infty}\varphi_2[\varphi_3^n(t)]=\varphi_1(t). \]

By theorem [2], for some subsequence \(\{\varphi_3^m(t)\}\subset\{\varphi_3^n(t)\}\) one can indicate a uniformly convergent sequence of functions \(\{\varphi_m(t)\}\) such that the sequence of functions \(\varphi_3^m[\varphi_m(t)]\) will also converge uniformly on \([0,1]\).

Denote:
\[ \lim_{m\to\infty}\psi_m(t)=\varphi_3(t), \qquad \lim_{m\to\infty}\varphi_3^{m}[\psi_m(t)]=\varphi_4(t). \]

It is not difficult to see that
\[ \varphi_3(t),\quad \varphi_4(t)\in M \quad \text{and} \]
\[ \varphi_1[\varphi_3(t)]=\varphi_2[\varphi_4(t)]. \]

Corollary. Let \(f_0(t)\) be a continuous mapping of the interval \([0,1]\) into a metric space \(R\). Then the family of continuous mappings
\[ \{f[\varphi(t)]\}, \tag{1} \]
where \(\varphi(t)\in M\), has the equivalence property.

Proof. Let \(f_1(t), f_2(t)\in(1)\). This means that
\[ f_1(t)=f_0[\varphi_1(t)],\qquad f_2(t)=f_0[\varphi_2(t)]. \]
To prove our assertion it is enough to find such \(\psi_1(t)\) and \(\psi_2(t)\) that
\[ \varphi_1[\psi_1(t)]=\varphi_2[\psi_2(t)], \]
the possibility of which follows from the preceding theorem.

Theorem 2. Let \(\{f(t)\}\), \(\{g(t)\}\) be two equivalence families of continuous mappings of the interval \((0,1)\) into a metric space \(R\) [1]. Let
\[ f_0(t)\in\{f(t)\},\qquad g_0(t)\in\{g(t)\}. \]
For any
\[ f_1(t)\in\{f(t)\},\qquad g_1(t)\in\{g(t)\}, \]
there exist \(\varphi_1(t), \varphi_2(t)\in M\) such that, if
\[ y_1(t)=\rho[f_1(t),g_1(t)], \]
\[ y_0(t)=\rho\{f_0[\varphi_1(t)],\ g_0[\varphi_2(t)]\}, \]
then
\[ E_{y_1}=E_{y_0}, \]
where \(E_{y_1}\) is the set of values of the function \(y_1(t)\), and \(E_{y_0}\) is the set of values of the function \(y_0(t)\).

Proof. Since \(f_1(t)\) and \(f_0(t)\) belong to one equivalence family, there exist \(\delta_1(t), \delta_2(t)\in M\) such that
\[ f_1[\delta_1(t)]=f_0[\delta_2(t)]. \]
Similarly, for \(g_1(t), g_0(t)\) one can indicate \(\gamma_1(t), \gamma_2(t)\in M\) such that
\[ g_1[\gamma_1(t)]=g_2[\gamma_2(t)]. \]
By Theorem (1), for \(\delta_1(t), \gamma_1(t)\) one can indicate \(\alpha_1(t), \beta_1(t)\) such that
\[ \delta_1[\alpha_1(t)]=\gamma_1[\beta_1(t)]. \]
Consider
\[ y_2(t)=\rho\{f_1\{\delta_1[\alpha_1(t)]\},\ g_1\{\gamma_1[\beta_1(t)]\}\} \]
and
\[ y_3(t)=\rho\{f_0\{\delta_2[\alpha_1(t)]\},\ g_0\{\gamma_2[\beta_1(t)]\}\}. \]
It is clear that
\[ E_{y_2}=E_{y_3}, \]
since
\[ f_1[\delta_1(t)]=f_0[\delta_2(t)],\qquad g_1[\gamma_1(t)]=g_0[\gamma_2(t)] \]
and
\[ E_{y_2}=E_{y_1}. \]
Therefore, putting
\[ \varphi_1(t)=\delta_2[\alpha_1(t)],\qquad \varphi_2(t)=\gamma_2[\beta_1(t)], \]
we shall have
\[ E_{y_0}=E_{y_3}=E_{y_1}. \]

Theorem 3. Let \(\{f(t)\}\) and \(\{g(t)\}\) be two equivalence families of continuous mappings of the interval \([0,1]\) into a metric space \(R\). Then the set of values of the function
\[ F[f(t),g(t)]=\sup \rho\{f(t),g(t)\}, \]
where \(f(t)\in\{f(t)\}\), \(g(t)\in\{g(t)\}\), is closed.

Proof. It is enough to show that if
\[ F[f_n(t),g_n(t)]=\sup_{t\in[0,1]}\rho[f_n(t),g_n(t)]=a_n \]
and
\[ \lim_{n\to\infty}a_n=a, \]
\[ f_n(t)\in\{f(t)\},\qquad g_n(t)\in\{g(t)\}, \]

then there exist

\[ \bar f(t)\in \{f(t)\}\qquad \text{and}\qquad \bar g(t)\in \{g(t)\} \]

such that \(F[\bar f(t),\bar g(t)]=a\).

Let \(f_0(t)\in\{f(t)\}\), \(g_0(t)\in\{g(t)\}\); then, by theorem [2], we can find sequences \(\{\varphi_n(t)\}\), \(\{\psi_n(t)\}\), with \(\varphi_n(t)\), \(\psi_n(t)\) such that

\[ F[f_n(t),g_n(t)] = F\{f_0[\varphi_n(t)],g_0[\psi_n(t)]\} = a_n . \]

The sequences \(\{\varphi_n(t)\}\), \(\{\psi_n(t)\}\), taking theorem [2] into account, may be assumed to converge uniformly.

Let

\[ \lim_{n\to\infty}\varphi_n(t)=\varphi(t) \qquad\text{and}\qquad \lim_{n\to\infty}\psi_n(t)=\psi(t), \]

then

\[ F\{f_0[\varphi(t)],g_0[\psi(t)]\}=a. \]

Denoting

\[ \bar f(t)=f_0[\varphi(t)],\qquad \bar g(t)=g_0[\psi(t)], \]

we have

\[ \sup \rho[\bar f(t),\bar g(t)]=a. \]

In particular, \(a\) may be equal to

\[ \inf_{\{f(t)\},\{g(t)\}} F[f(t),g(t)]. \]

It is easy to show that, with the aid of the function (1), the space can be metrized.

Theorem 4. For the space \(L_R\) to be complete, it is necessary and sufficient that the space \(R\) be complete.

Proof. Necessity. The elements of the space \(R\) may be regarded as elements of the space \(L_R\), considering continuous mappings \(f(t)\) that assign to all of \([0,1]\) one element of \(R\). Hence the necessity of the completeness of \(R\) for the completeness of \(L_R\) is clear. Sufficiency. Let \(R\) be complete, and let \(\{L_n\}\) be a fundamental sequence of elements of \(L_R\), and

\[ \{f_1(t)\},\quad \{f_2(t)\},\ldots,\{f_n(t)\} \tag{3} \]

the corresponding sequence of equivalence classes. Take \(f_1^\circ(t)\in\{f_1(t)\},\ldots,f_n^\circ(t)\in\{f_n(t)\}\ldots\) and consider the classes of mappings

\[ \{f_1^\circ[\varphi(t)]\},\ldots,\{f_n^\circ[\varphi(t)]\},\ldots \tag{4} \]

\[ \rho(L_n,L_{n+1}) = \inf \rho[f_n(t),f_{n+1}(t)] = (\text{by theorems }2,3) = \]

\[ = \rho\{f_n^\circ[\varphi_n(t)],f_{n+1}^\circ[\varphi_{n+1}(t)]\}. \]

Hence we can write the system of equalities

\[ \rho(L_1,L_2)=\rho\{f_1^\circ[f_1(t)],f_2^\circ[\varphi_2(t)]\}, \]

\[ \rho(L_2,L_3)=\rho\{f_2^\circ[\varphi'_2(t)],\varphi_3^\circ[\varphi_3(t)]\}, \]

\[ \cdots \tag{5} \]

\[ \rho(L_n,L_{n+1})=\rho\{f_n^\circ[\varphi'_n(t)],f_{n+1}(t)\}. \]

\[ \cdots \]

By theorem 1, for \(\varphi_n[\beta_{n-2}(t)]\) and \(\varphi'_n(t)\) there are \(\alpha_{n-1}(t)\) and \(\beta_{n-1}(t)\) such that

\[ \varphi_n\{\beta_{n-2}[\alpha_{n-1}(t)]\}=\varphi'_n[\beta_{n-1}(t)]. \]

We agree to denote \(\alpha[\beta(t)]=(\alpha\beta)\). Put
\(\alpha'_n(t)=(\alpha_n\hat\delta_n)\), \(\beta'_n(t)=(\beta_n\hat\delta_n)\), where the function
\[ \hat\delta_n(t)=\sum_{k=1}^{2^n}\psi_k(t) \]
is constructed, as in Theorem 1, so that
\[ \left|(\alpha_1\hat\delta_1\alpha_2\hat\delta_2\ldots\alpha_{n-1}\hat\delta_{n-1}) - (\alpha_1\hat\delta_1\alpha_2\hat\delta_2\ldots\alpha_n\hat\delta_n)\right| < \frac{1}{2^n}. \]

It is clear that
\[ \varphi_n\{\beta_{n-2}[\alpha'_{\,n-1}(t)]\} = \varphi'_n[\beta'_{\,n-1}(t)]. \]

Substituting all \(\alpha'_n(t)\) and \(\beta'_n(t)\) in (4), we obtain
\[ \begin{aligned} \rho(L_1,L_2)&=\rho\{f^0_1[\gamma_0(t)]\, f^0_2[\gamma_1(t)]\},\\ \rho(L_2,L_3)&=\rho\{f^0_2[\gamma_1(t)],\, f^0_3[\gamma_2(t)]\},\\ &\cdots\\ \rho(L_n,L_{n+1})&=\rho\{f^0_n[\gamma_{n-1}(t)],\, f^0_{n+1}[\gamma_n(t)]\}, \end{aligned} \tag{6} \]
where
\[ \gamma_0(t)=\varphi_1\left[\lim_{n\to\infty}(\alpha_1\hat\delta_1\alpha_2\hat\delta_2\ldots\alpha_n\hat\delta_n)\right], \]
\[ \gamma_1(t)=\varphi_2\left[\lim_{n\to\infty}(\alpha_1\hat\delta_1\alpha_2\hat\delta_2\ldots\alpha_n\hat\delta_n)\right], \]
\[ \gamma_2(t)=\varphi_3\beta'_1\left[\lim_{n\to\infty}(\alpha_2\hat\delta_2\ldots\alpha_n\hat\delta_n)\right], \]
\[ \cdots \]
\[ \gamma_{n-1}(t)=\varphi_n\beta'_{\,n-2}\left[\lim_{m\to\infty}(\alpha_{n-1}\hat\delta_{n-1}\alpha_n\hat\delta_n\ldots\alpha_{n+m}\hat\delta_{n+m})\right]. \]
\[ \cdots \]

Since
\[ \lim_{n\to\infty}(\alpha_1\hat\delta_1\alpha_2\hat\delta_2\ldots\alpha_n\hat\delta_n)\in M, \]
and
\[ \lim_{n\to\infty}(\alpha_k\hat\delta_k\ldots\alpha_n\hat\delta_n)\in M \]
for any \(k\), it follows that \(\gamma_0(t), \gamma_n(t)\in M\) \((n=1,2,\ldots)\), and then the fundamental sequence
\[ \{f^0_n[\gamma_{n-1}(t)]\} \]
of continuous mappings of the interval \([0,1]\) into \(R\) converges, by completeness of \(R\), to a continuous mapping \(f^0(t)\) of \([0,1]\) into \(R\). The fundamentality of (7) follows from (6) and the fundamentality of the sequence \(\{L_n\}\). The equivalence class of continuous mappings of the interval \([0,1]\) into \(R\), to which \(f_0(t)\) belongs, determines in \(L_R\) a curve \(L_0\) such that for every \(\varepsilon>0\) and (2) there is an \(N=N(\varepsilon)\) such that for all \(n\ge N\)
\[ \rho(L_0,L_n)<\varepsilon. \]
And this means that \(L_R\) is complete, since \(\{L_n\}\) is an arbitrary fundamental sequence from \(L_R\).

Theorem 5. The space \(L_{C_{(0,1)}}\) is separable.

Proof. We shall consider all possible mappings of the interval \([0,1]\) into the space of continuous functions \(C_{(0,1)}\) of the following form
\[ P_n(t)=\sum_{k=1}^{2^n} f_k(t) \]
\[ f_k(t)= \begin{cases} \dfrac{P_{k-1}(x)(S_k-t)+P_k(x)(t-S_{k-1})}{S_k-S_{k-1}}, & \text{for } t\in [S_{k-1},S_k],\\[1.2em] 0, & \text{for } t\in [0,1]\setminus [S_{k-1},S_k], \end{cases} \tag{8} \]
and where \(S_0=0\), \(S_k=\dfrac{k}{2^n}\), \(k=1,2,\ldots,2^n\), \(p_0(x), p_1(x),\ldots,p^n_2(x)\) are polyno-

with rational coefficients.

\(P_n(t)\) is a continuous mapping of the interval \([0,1]\) into \(C_{(0,1)}\). It is not difficult to see that the set of all mappings of the form (8) is countable. It is not difficult to see that the set of mappings of the form (8) is dense in the set of all continuous mappings of the interval \((0,1)\) into \(C_{(0,1)}\). Let \(g(t)\) be a continuous mapping of the interval \([0,1]\) into \(C_{(0,1)}\), and let \(\varepsilon>0\) be arbitrary. By virtue of the compactness of \([0,1]\) and the continuity of \(g(t)\) on \([0,1]\), one can specify such an \(n\) that, for \(|t_1-t_2|\leq \dfrac{1}{2^n}\), we shall have

\[ \rho[g(t_1),g(t_2)]<\frac{\varepsilon}{2}. \tag{9} \]

Since the set of polynomials with rational coefficients is dense in \(C_{(0,1)}\), for \(g(S_k)=g\left(\dfrac{k}{2^n}\right)\) there exists a polynomial \(P_k(x)\) such that

\[ \rho\left[P_k(x),\ g\left(\frac{k}{2^n}\right)\right]<\frac{\varepsilon}{2}, \tag{10} \]

where \(k=0,1,2,\ldots,2^n\).

Let us show that \(\rho[g(t),P_n(t)]<\varepsilon\), where

\[ P_n(t)=\sum_{k=1}^{2^n} f_k(t) \]

and

\[ f_k(t)= \begin{cases} \dfrac{P_{k-1}(x)(S_k-t)+P_k(x)(t-S_{k-1})}{S_k-S_{k-1}}, & \text{for } t\in [S_{k-1},S_k],\\[1.2em] 0, & \text{for } t\notin [S_{k-1},S_k], \end{cases} \]

and \(P_k(x)\) satisfy inequality (10). Let \(t_0\in[0,1]\); then \(t_0\in [S_{k-2},S_k]\), and let \(g(t_0)=\varphi(x)\).

Denote

\[ \frac{S_k-t_0}{S_k-S_{k-1}}=\alpha,\qquad \frac{t_0-S_{k-1}}{S_k-S_{k-1}}=\beta. \]

It is easy to see that \(\alpha+\beta=1\),

\[ \rho[g(t_0),P_n(t_0)] = \max_{x\in[0,1]}\left|\varphi(x)-\alpha P_{k-1}(x)+\beta P_k(x)\right|, \tag{11} \]

\[ |\varphi(x)-(\alpha P_{k-1}(x))+\beta P_k(x)| = |\alpha\varphi(x)-\alpha P_{k-1}(x)+\beta(x) -\beta P_k(x)| \leq \alpha\max_{x\in[0,1]}|\varphi(x)-P_{k-1}(x)| + \beta\max_{x\in[0,1]}|\varphi(x)-P_k(x)|. \tag{12} \]

Since \(|t_0-S_{k-1}|\leq \dfrac{1}{2^n}\) and \(|t_0-S_k|\leq \dfrac{1}{2^n}\), taking (9) and (10) into account, we have

\[ \max_{x\in[0,1]}|\varphi(x)-P_{k-1}(x)|<\varepsilon, \]

\[ \max_{x\in[0,1]}|\varphi(x)-P_k(x)|<\varepsilon. \tag{13} \]

From (11), (12), and (13) we have

\[ \rho[g(t_0),P_n(t)]<\varepsilon. \]

since \(t_0\in[0,1]\) is arbitrary, it follows that

\[ \sup_{t\in[0,1]}\rho[g(t),P^n(t)]>\varepsilon . \]

Thus, the set of mappings of the form (8) is dense in the set of all continuous mappings of the interval \([0,1]\) into \(C_{(0,1)}\).

Consider the set of equivalence classes of continuous mappings of the interval \([0,1]\) in the space \(C_{(0,1)}\)

\[ \{P\}. \tag{14} \]

We shall say that the equivalence class \(\{f(t)\}\in(14)\) if there exists a mapping of the form (8), \(P_n\), such that

\[ P_n(t)\in\{f(t)\}. \]

Since one mapping cannot belong to different equivalence classes, the set (14) is at most countable. Number the classes (14) as

\[ \{f_1(t)\},\quad \{f_2(t)\},\ldots,\{f_n(t)\},\ldots, \]

and to all mappings of the form (8) that belong to \(\{f_k(t)\}\) assign the number \(k\), denoting them by \(P_k(t)\). The set of curves determined by the classes (8) will be denoted by

\[ \{L_k\}. \tag{15} \]

We shall show that (15) is dense in \(L_{C_{(0,1)}}\).

Let \(L\in L_{C_{(0,1)}}\), let \(\{g(t)\}\) be the equivalence class defining \(L\), and let \(\varepsilon>0\) be arbitrary. Take \(g_0(t)\in\{g(t)\}\). Since the set of mappings of the form (8) is dense in the set of all continuous mappings of the interval \((0,1)\) into \(C_{0,1}\), for the chosen \(g_0(t)\) and \(\varepsilon>0\) there is a mapping of type (8), \(P_k(t)\), such that

\[ \sup \rho[g_0(t),P_k(t)]<\varepsilon . \]

Let \(\{f_k(t)\}\) be the equivalence class containing \(P_k(t)\); then

\[ \rho(L,L_k)=\inf_{\{g(t)\},\,\{f_k(t)\}}\{\sup \rho[g(t),f_k(t)]\}\leq \]

\[ \leq \sup_{t\in[0,1]}\rho[g_0(t),P_k(t)]<\varepsilon . \]

Thus, (15) is a countable and dense subset of \(L_{C_{(0,1)}}\).

Corollary. For \(L_R\) to be separable it is necessary and sufficient that \(R\) be separable.

Proof. Necessity is obvious, since \(R\) itself may be regarded as a subset of \(L_R\), by considering all possible continuous mappings that assign to the entire interval \((0,1)\) one element of \(R\).

Sufficiency. Let \(R\) be separable. By the Banach–Mazur theorem, in \(C_{(0,1)}\) there exists a subset isometric to \(R\), i.e., there exists a homeomorphic mapping \(\Phi\) of the space \(R\) into \(C_{(0,1)}\) such that \(\Phi(R)\subset C_{(0,1)}\) is isometric to \(R\). It is not hard to see that \(\Phi(R)\) is separable, since \(L\Phi(R)\subset L_{C_{(0,1)}}\), and by the preceding theorem \(L_{C_{(0,1)}}\) is separable.

Let \(\{L_k\}\) be a countable dense subset of \(L\Phi(R)\); then, as is easy to see,

\[ \{\Phi^{-1}(L_k)\} \]

is a countable dense subset of \(L_R\).

REFERENCES

1. A. N. Kolmogorov, S. V. Fomin. Elements of the Theory of Functions and Functional Analysis. Moscow State University, § 20, 1954.

2. L. E. Portnov. On sequences of monotone functions. Izvestiya TPI, vol. 131, 1965.