Full Text
UDC 513.88:513.83
MATHEMATICS
M. I. KADETS
TOPOLOGICAL EQUIVALENCE
OF ALL SEPARABLE BANACH SPACES
(Presented by Academician L. V. Kantorovich, XI 1, 1965)
In this note the following is proved.
Theorem. All separable Banach spaces are homeomorphic.
Thus an affirmative answer is obtained to the question posed by M. Fréchet \((^1)\) in 1928 and S. Banach \((^2)\) in 1932.
C. Bessaga and A. Pełczyński \((^3)\) showed that if a separable Banach space \(E\) contains a subspace \(X\) homeomorphic to \(l_2\), then it itself is also homeomorphic to \(l_2\). It is well known that every infinite-dimensional Banach space contains an infinite-dimensional subspace with a basis. Hence, by the Bessaga–Pełczyński theorem, it is enough to establish a homeomorphism of all Banach spaces with a basis.
Proposition 1. Let \(X\) be a Banach space with basis \(\{e_k\}_1^\infty\); denote the conjugate basis system by \(\{f_k\}_1^\infty\) \((f_k \in X^*)\). Suppose, further, that the norm of the space is subject to the following condition: if
\[ \|x_\nu\|=\|x\|=1 \quad (\nu=1,2,\ldots);\qquad \lim_{\nu\to\infty} f_n(x_\nu)=f_n(x) \quad (n=1,2,\ldots), \]
then
\[ \lim_{\nu\to\infty}\|x_\nu-x\|=0. \]
Assume also that on the unit ball \(U\) of the space \(X\) there is defined a functional \(F(x)\) having the following properties:
-
The functional \(F(x)\) is continuous.
-
\(F(x)>0\) for \(\|x\|<1\); \(F(x)=0\) for \(\|x\|=1\); \(F(0)=1\).
-
If
\[ \lim_{n\to\infty} F(S_n)=0 \qquad \left(S_n=\sum_{k=1}^{n} a_k e_k\right), \tag{1} \]
then the series \(\displaystyle \sum_{k=1}^{\infty} a_k e_k\) converges.
- For fixed \(n\) and \(a_k\) \((k<n)\), the function
\[ \psi(a)=F(S_{n-1}+ae_n) \]
is strictly increasing for \(a<0\) and strictly decreasing for \(a>0\).
Then the space \(X\) is homeomorphic to the space \(l_2\).
We shall precede the proof of Proposition 1 by two lemmas. To each normalized element
\[ x=\sum_{k=1}^{\infty} f_k(x)e_k \]
we assign a pair of numerical sequences:
\[ \gamma_0(x)=1;\qquad \gamma_n(x)=F(S_nx) \quad \left(S_nx=\sum_{k=1}^{n} f_k(x)e_k,\ n=1,2,\ldots\right), \]
\[ \theta_n(x)=\operatorname{sign} f_n(x) \quad (n=1,2,\ldots). \]
We note that
\[ 1\geq \gamma_1(x)\geq \gamma_2(x)\geq \cdots,\qquad \lim_{n\to\infty}\gamma_n(x)=0. \]
Lemma 1. A normalized sequence \(x_\nu\) converges to an element \(x\) if and only if
\[ \lim_{\nu\to\infty}\theta_n(x_\nu)\,[\gamma_{n-1}(x_\nu)-\gamma_n(x_\nu)] = \theta_n(x)\,[\gamma_{n-1}(x)-\gamma_n(x)] \qquad (n=1,2,\ldots). \]
Lemma 2. Whatever the pair of numerical sequences \(\{\gamma_k\}_1^\infty\) and \(\{\theta_k\}_1^\infty\), subject to the conditions
\[ 1\geq \gamma_1\geq \gamma_2\geq \cdots,\qquad \lim_{n\to\infty}\gamma_n=0, \]
\[ \theta_n=\pm 1,\quad \text{if } \gamma_n\ne \gamma_{n-1};\qquad \theta_n=0,\quad \text{if } \gamma_n=\gamma_{n-1}, \]
there exists a unique normalized element \(x\) such that
\[ \gamma_n(x)=\gamma_n,\qquad \theta_n(x)=\theta_n \qquad (n=1,2,\ldots). \]
Proof of Proposition 1. Let \(X\) and \(Y\) be spaces satisfying all the requirements of Proposition 1. Suppose that in each of them the functionals \(\gamma_n(x),\theta_n(x)\) and, respectively, \(\gamma_n(y),\theta_n(y)\) have been introduced. To each normalized element \(x\in X\) we assign that normalized element \(y=Tx\in Y\) for which
\[
\theta_n(y)\gamma_n(y)=\theta_n(x)\gamma_n(x)\qquad (n=1,2,\ldots).
\]
By Lemma 2 this correspondence is one-to-one. By Lemma 1 it is bicontinuous. The homeomorphism thus obtained extends to the whole space \(X\):
\[
Tx=\|x\|T(x/\|x\|),\qquad T(\theta)=\theta.
\]
It remains to show that in the space \(l_2\) there exists a functional \(F(x)\) with the required properties. It turns out that in this case it is sufficient to put \(F(x)=1-\|x\|\) \((x\in l_2)\).
We now show that every Banach space with a basis can, by means of an equivalent renorming, be made to satisfy the conditions of Proposition 1.
Proposition 2. In every separable Banach space \(X\) with basis \(\{e_k\}_1^\infty\) one can introduce a norm \((\|\ \|)\), equivalent to the original one \((\|\ \|_0)\), satisfying the following requirements:
a) The space \((X,\|\ \|)\) is locally uniformly convex \((UR_L\) in the notation of M. Day, \((^4)\), p. 188);
b) For any normalized elements \(x_\nu\) and \(x\), the condition
\[ \lim_{\nu\to\infty} f_n(x_\nu)=f_n(x)\qquad (n=1,2,\ldots) \]
implies strong convergence:
\[
\lim_{\nu\to\infty}\|x_\nu-x\|=0.
\]
c) With respect to the new norm the basis \(\{e_k\}\) is orthogonal, i.e.
\[ \left\|\sum_{k=1}^{n-1}a_ke_k\right\| < \left\|\sum_{k=1}^{n}a_ke_k\right\|, \qquad \text{if } a_n\ne 0 \quad (n=1,2,\ldots). \]
A norm satisfying the indicated requirements was constructed in \((^{5,6})\) (meaning the norm defined by formula (3) of the note \((^6)\)).
Proposition 3. In a Banach space whose norm satisfies the conditions a), b), c) of Proposition 2, there exists a functional \(F(x)\) possessing properties 1–4 formulated in Proposition 1.
Here we shall need two more lemmas. Consider the functional
\[ \varepsilon(x,\delta)=\frac12\sup_{z\in G(x,\delta)}\|x-z\| \qquad (\|x\|=1,\ 0\leq \delta\leq 1), \]
where
\[ G(x,\delta)=\{z:\|z\|\leq 1;\ \min_{0\leq \lambda\leq 1}\|\lambda x+(1-\lambda)z\|\geq 1-\delta\}. \]
If the space is locally uniformly convex, then
\[ \lim_{\delta\to 0}\varepsilon(x,\delta)=0. \tag{2} \]
Lemma 3. For any Banach space the functional \(\varepsilon(x,\delta)\) satisfies the inequalities:
\[
\delta \leqslant \varepsilon(x,\delta) \leqslant 1,\qquad
\varepsilon(x,\delta)\leqslant \tfrac12\|x-y\|+\varepsilon(y,\delta+\|x-y\|)
\]
\[
\varepsilon(x,\delta+h)-\varepsilon(x,\delta)\leqslant 3h/\delta
\qquad (0\leqslant \delta\leqslant \delta+h\leqslant 1),
\tag{3}
\]
whence, in particular, it follows that \(\varepsilon(x,\delta)\) is uniformly continuous on the set \(S\times[\delta_0,1]\) for every \(\delta_0>0\) (\(S\) is the unit sphere of the space). If the space is locally uniformly convex, then \(\varepsilon(x,\delta)\) is continuous on \(S\times[0,1]\).
Let us consider the functional
\[
\Phi(x)=\varepsilon\left(\frac{x}{\|x\|},\,1-\|x\|\right)\qquad (0<\|x\|\leqslant 1);
\]
for \(x=\theta\) put \(\Phi(\theta)=1\).
Lemma 4. The functional \(\Phi(x)\) is continuous for \(\|x\|\leqslant 1\) and uniformly continuous for \(\|x\|\leqslant 1-\delta_0\) for every \(\delta_0>0\).
For each element \(x\) (\(\|x\|\leqslant 1\)) define the set
\[
L(x)=\bigcup_n L_n(x),
\]
where \(L_n(x)\) is the segment joining \(S_{n-1}x\) and \(S_nx\) \((n=1,2,\ldots)\); \(S_0x=\theta\). Finally, define the functional
\[
F(x)=\left(1-\tfrac12\|x\|\right)\inf_{z\in L(x)}\Phi(z)
\qquad (0\leqslant \|x\|\leqslant 1).
\tag{4}
\]
Proof of Proposition 3. We have to verify that the functional \(F(x)\) has properties 1–4 of Proposition 1. Properties 1 and 2 follow almost immediately from the definition of \(F(x)\) and Lemma 4. Let us prove property 3. Suppose (1) is fulfilled. For each \(S_n\) consider a point \(\sigma_n\in L(S_n)\) at which the lower bound in expression (4) is attained:
\[
\sigma_n=S_{m-1}+\alpha e_m
\qquad (m=m(n),\ 0<|\alpha|\leqslant |a_m|,\ \alpha a_m>0).
\tag{5}
\]
According to (2), (3), and (5), the diameter of the set \(G(\sigma_n/\|\sigma_n\|;\,1-\|\sigma_n\|)\) tends to zero as \(n\) increases. Consider also the decreasing sequence of closed sets
\[
Q(\sigma_n)=\{z:\ \|z\|\leqslant 1;\quad f_k(z)=a_k\ (k=1,2,\ldots,m)\}
\qquad (n=1,2,\ldots).
\]
Using the orthogonality of the basis \(\{e_k\}\), one can establish that
\[
G(\sigma_n/\|\sigma_n\|;\,1-\|\sigma_n\|)\supset Q(\sigma_n).
\]
Thus the diameter of the set \(Q(\sigma_n)\) tends to zero as \(n\) increases. The unique element \(x\) lying in the intersection of all \(Q(\sigma_n)\) will also be the limit of the sequence
\[
S_n:\quad x=\sum_{k=1}^{\infty} a_k e_k .
\]
Finally, let us prove property 4. Consider the function
\[
\psi(\alpha)=F(S_{n-1}+\alpha e_n).
\]
Let \(|\alpha_1|<|\alpha_2|\), \(\alpha_1\alpha_2\geqslant 0\). Then
\[
\|S_{n-1}+\alpha_1 e_n\|<\|S_{n-1}+\alpha_2 e_n\|
\tag{6}
\]
by virtue of the orthogonality of the basis, and
\[
L(S_{n-1}+\alpha_1 e_n)\subset L(S_{n-1}+\alpha_2 e_n)
\tag{7}
\]
by the definition of the set \(L(x)\). From (4), (6), and (7) we obtain that \(\psi(\alpha_1)>\psi(\alpha_2)\).
The theorem formulated at the beginning of the note is a direct consequence of Propositions 1–3 and the theorem of Bessaga–Pełczyński.
I express my deep gratitude to V. I. Gurarii and A. Pełczyński for a number of valuable suggestions.
Kharkov Institute
of Municipal Engineers
Received
25 X 1965
REFERENCES
\(^{1}\) M. Fréchet, Les espaces abstraits, Paris, 1928.
\(^{2}\) S. Banach, Course of Functional Analysis, Kiev, 1948.
\(^{3}\) C. Bessaga, A. Pełczyński, Bull. Acad. Polon., 8, 757 (1960).
\(^{4}\) M. M. Day, Normed Linear Spaces, Moscow, 1961.
\(^{5}\) M. I. Kadets, Izv. vyssh. uchebn. zaved., Matematika, No. 6, 51 (1959).
\(^{6}\) M. I. Kadets, Izv. vyssh. uchebn. zaved., Matematika, No. 6, 186 (1961).