Abstract Generated abstract
The paper studies whether all unconditional Schauder bases in certain countably Hilbert spaces are quasi-equivalent, extending earlier results from nuclear spaces and finite centers of Hilbert scales. It considers the infinite center of a completely continuous Hilbert scale, represented by weighted Hilbert spaces with weights tending to infinity, and uses interpolation properties of nested Hilbert spaces together with estimates for renormalized and rearranged bases. The main theorem proves that in the space \(E_\infty(a_k)\) every unconditional basis is quasi-equivalent to the principal unit-vector basis, by constructing an appropriate permutation and normalization and showing that the induced linear operator is an isomorphism.
Full Text
UDC 513.88
MATHEMATICS
V. P. ZAKHARYUTA
ON THE QUASI-EQUIVALENCE OF BASES IN AN INFINITE CENTER OF A HILBERT SCALE
(Presented by Academician L. V. Kantorovich, March 28, 1969)
1°. Let \(\{x_k\}\) be an unconditional Schauder basis in a linear topological space \(E\). It is obvious that the system
\[ y_k=\lambda_k T(x_{n_k}) \]
is an unconditional basis in \(E\) for any permutation of the natural sequence \(\{n_k\}\), any sequence of numbers \(\lambda_k\ne 0\), and any isomorphism \(T:E\to E\). The bases \(\{x_k\}\) and \(\{y_k\}\) are called quasi-equivalent \((^{1,2})\).
A natural question arises: are all unconditional bases in \(E\) quasi-equivalent? This problem was solved positively for an entire class of nuclear countably normed spaces \((^{1-4})\). The assumption of nuclearity of the space was removed in \((^{5})\) for the case of so-called finite centers of Hilbert scales**. In the present paper the quasi-equivalence of all unconditional bases in infinite centers of completely continuous Hilbert scales is established. The proof of quasi-equivalence in infinite centers differs qualitatively from the arguments of \((^{5})\). This is due to the fact that Lemma 3 of \((^{5})\) has no analogue for infinite centers.
2°. Every positive self-adjoint bounded operator \(A\) in a Hilbert space \(H\) generates a Hilbert scale \(\{H_\alpha\}\) of Hilbert spaces \(H_\alpha\) with scalar products
\[ (x,y)_{H_\alpha}=(A^{-\alpha}x,A^{-\alpha}y), \]
where \((x,y)\) is the scalar product in \(H=H_0\) \((^{1,6})\). We shall call the scale \(\{H_\alpha\}\) completely continuous if the operator \(A\) is completely continuous. In the latter case one may assume that \(H_\alpha=l_2(a_k^\alpha)\), where \(a_k=1/\lambda_k\), \(\lambda_k\) are the eigenvalues of the operator \(A\), arranged in decreasing order, i.e. \(a_k\uparrow\infty\).
The countably Hilbert space
\[ E_\alpha(a_k)\equiv \lim_{\lambda<\alpha}\operatorname{pr} l_2(a_k^\lambda), \]
following \((^{1})\), will be called the center of the scale \(\{H_\lambda\}\)—finite if \(\alpha<\infty\), and infinite if \(\alpha=\infty\). For \(x=(\xi_1,\xi_2,\ldots,\xi_k,\ldots)\in H_\lambda\) we shall denote
\[ \|x\|_\lambda=\|x\|_{H_\lambda}=\left(\sum |\xi_k|^2 a_k^{2\lambda}\right)^{1/2}. \]
If \(H_0\supset H_1\) are two Hilbert spaces with a (completely) continuous embedding, then there exists a (completely continuous) Hilbert scale \(\{G_\alpha\}\) such that \(G_0=H_0,\ G_1=H_1\). We shall say that \(\{G_\alpha\}\) is the scale generated by \(H_0\) and \(H_1\), and use the notation
\[ G_\alpha=(H_0)^{1-\alpha}(H_1)^\alpha. \]
We shall repeatedly use the following simple consequence of the known interpolation theorem \((^{7})\).
* This question was first posed by M. G. Haplanov for the space of analytic functions.
** Here and below the terminology of paper \((^{5})\) is used.
Lemma (cf. \((^9)\)). Let \(H_1 \supset H_2 \supset H_3 \supset H_4\) be a quadruple of Hilbert spaces with (complete) continuous embeddings. Then for any \(\alpha: 0 \leq \alpha \leq 1\) the following (complete) continuous embedding holds:
\[
(H_1)^{1-\alpha}(H_3)^\alpha \supset (H_2)^{1-\alpha}(H_4)^\alpha .
\]
\(3^\circ\). We shall prove the main result.
Theorem. In the space \(E=E_\infty(a_k)\), \(a_k \uparrow \infty\), all unconditional bases are quasiequivalent.
Let \(\{x_k\}\) be an arbitrary unconditional basis in \(E\). It is required to show that \(\{x_k\}\) is quasiequivalent to the principal basis \(\{e_k\}\), consisting of the unit vectors. In other words, one has to find a permutation of the natural series \(\{n_k\}\) and a numerical sequence \(\sigma_k>0\) such that the bases \(\{y_k=\sigma_k x_{n_k}\}\) and \(\{e_k\}\) are equivalent, i.e. the linear operator \(T\) defined by the formula
\[
T\left(\sum \xi_k e_k\right)=\sum \xi_k y_k,
\tag{1}
\]
is an isomorphism of \(E\) onto itself.
Denote by \(\{x_k'\}\) the system biorthogonal to the basis \(\{x_k\}\). By Lemma 1 \((^5)\), the system of Hilbert norms
\[
|x|_{\lambda}^{(0)}=\left(\sum |x_k'(x)|^2 \,\|x_k\|_\lambda^2\right)^{1/2}, \qquad \lambda<\infty,
\]
is equivalent to the original system of norms \(\{\|x\|_\lambda,\ \lambda<\infty\}\). Denote by \(H_\lambda^{(0)}\) the Hilbert space obtained from \(E\) by completion with respect to the norm \(\|x\|_\lambda^{(0)}\). From the equivalence of the systems of norms it follows that there exist numbers \(\mu_1<\nu_1<\nu_2<\mu_2,\ \lambda_1<\lambda_2\) such that the embeddings
\[
H_{\mu_1}\supset H_{\lambda_1}^{(0)}\supset H_{\nu_1}\supset H_{\nu_2}\supset H_{\lambda_2}^{(0)}\supset H_{\mu_2}
\]
hold with continuous embeddings.
Take a permutation of the natural series \(\{n_k\}\) and a numerical sequence \(\sigma_k>0\) such that the system \(y_k=\sigma_k x_{n_k}\) satisfies the conditions
\[
\|y_k\|_{\lambda_1}^{(0)}=1,\qquad \|y_k\|_{\lambda_2}^{(0)}=b_k\uparrow\infty .
\]
Below it will be shown that this permutation and renormalization are the desired ones, i.e. the operator (1) is an isomorphism of \(E\) onto itself.
Applying Lemma 2 \((^5)\) first to the quadruple of spaces
\[
H_{\mu_1}\supset H_{\lambda_1}^{(0)}\supset H_{\lambda_2}^{(0)}\supset H_{\mu_2}
\]
and the bases \(\{a_k^{-\mu_1}e_k\}\), \(\{y_k\}\), and then to the quadruple of spaces
\[
H_{\lambda_1}^{(0)}\supset H_{\nu_1}\supset H_{\nu_2}\supset H_{\lambda_2}^{(0)}
\]
and the bases \(\{y_k\}\), \(\{a_k^{-\nu_1}e_k\}\), we obtain the inequalities:
\[
a_k^{\nu_2-\nu_1}\prec b_k=\|y_k\|_{\lambda_2}^{(0)}\prec a_k^{\mu_2-\mu_1}.
\tag{2}
\]
Here and in what follows the notation \(a_k\prec b_k\) means \(a_k=O(b_k)\).
It follows from the lemma that for any \(\sigma: 0\leq \sigma\leq 1\) the continuous embeddings
\[
H_{\mu_1(1-\sigma)+\nu_2\sigma}\supset G_\sigma\supset H_{\nu_1(1-\sigma)+\mu_2\sigma},\qquad
G_\sigma=(H_{\lambda_1}^{(0)})^{1-\sigma}(H_{\lambda_2}^{(0)})^\sigma
\]
hold.
Hence, if one takes into account that \(\|y_k\|_{G_\sigma}=b_k^\sigma\), the inequalities \((\nu_1<\nu<\nu_2)\)
\[
b_k^{\sigma_2}\prec \|y_k\|_\nu\prec b_k^{\sigma_1},
\tag{3}
\]
follow, where
\[
\sigma_1=\sigma_1(\nu)=(\nu-\nu_1)/(\mu_2-\nu_1), \qquad
\sigma_2=\sigma_2(\nu)=(\nu-\mu_1)/(\nu_2-\mu_1).
\]
Comparing inequalities (2) and (3), we obtain:
\[
a_k^{\gamma_2}\prec \|y_k\|_\nu\prec a_k^{\gamma_1},
\tag{4}
\]
where
\[
\gamma_1=\sigma_1(\mu_2-\mu_1), \qquad \gamma_2=\sigma_2(\nu_2-\nu_1).
\]
Now take an arbitrary \(\lambda>\lambda_2^{(0)}\) and \(\tau=\tau(\lambda)\ge \nu_2\), \(\chi=\chi(\lambda)\) such that the continuous embeddings hold:
\[ H_\tau \supset H_\lambda^{(0)} \supset H_\chi \]
and \(\tau(\lambda)\uparrow\infty\) as \(\lambda\uparrow\infty\).
Let \(\{y_{n_k}\}\) be a rearrangement of the basis \(\{y_k\}\) in decreasing order of the norms in \(H_\lambda^{(0)}\), i.e.,
\[ \|y_{n_k}\|_\lambda^{(0)}=s_k\uparrow\infty . \]
Application of Lemma 2 \((^5)\), analogously to the preceding one, gives the inequalities:
\[ a_k^{\tau-\nu_1}\prec \|y_{n_k}\|_\lambda^{(0)}=s_k\prec a_k^{\chi-\mu_1}. \tag{5} \]
As before, after applying the lemma of the present paper we obtain the estimates \((\nu_1<\nu<\tau)\):
\[ s_k^{\widetilde{\sigma}_2}\prec \|y_{n_k}\|_\nu \prec s_k^{\widetilde{\sigma}_1}, \tag{6} \]
where \(\widetilde{\sigma}_1=\widetilde{\sigma}_1(\nu,\lambda)=(\nu-\nu_1)/(\chi-\nu_1)\),
\(\widetilde{\sigma}_2=\widetilde{\sigma}_2(\nu,\lambda)=(\nu-\mu_1)/(\tau-\mu_1)\).
The following inequalities are the result of comparing inequalities (5) and (6):
\[ a_k^{\widetilde{\gamma}_2}\prec \|y_{n_k}\|_\nu \prec a_k^{\widetilde{\gamma}_1}, \tag{7} \]
where \(\widetilde{\gamma}_1=\widetilde{\sigma}_1(\chi-\mu_1)\), \(\widetilde{\gamma}_2=\widetilde{\sigma}_2(\tau-\nu_1)\).
Comparing inequalities (4) and (7), we obtain:
\[ a_{n_k}^{\gamma_2/\widetilde{\gamma}_1}\prec a_k\prec a_{n_k}^{\gamma_1/\widetilde{\gamma}_2}. \]
Together with inequality (5) this gives
\[ a_{n_k}^{\rho_2}\prec \|y_{n_k}\|_\lambda^{(0)}\prec a_{n_k}^{\rho_1}, \tag{8} \]
where
\[ \rho_1=\rho_1(\lambda)=\frac{\gamma_1}{\widetilde{\gamma}_2}(\chi-\mu_1) =\frac{(\nu-\nu_1)(\mu_2-\mu_1)(\tau-\mu_1)} {(\mu_2-\nu_1)(\nu-\mu_1)(\tau-\nu_1)}(\chi-\mu_1), \]
\[ \rho_2=\rho_2(\lambda)=\frac{\gamma_2}{\widetilde{\gamma}_1}(\tau-\nu_1) =\frac{(\nu-\mu_1)(\nu_2-\nu_1)(\chi-\nu_1)} {(\nu_2-\mu_1)(\nu-\nu_1)(\chi-\mu_1)}(\tau-\nu_1). \]
Obviously, as \(\lambda\uparrow\infty\),
\[ \rho_2(\lambda)\uparrow\infty . \tag{9} \]
Since \(\{n_k\}\) is a rearrangement of the natural sequence, formula (8) may be rewritten in the form
\[ C_2 a_k^{\rho_2}\le \|y_k\|_\lambda^{(0)}\le C_1 a_k^{\rho_1},\qquad C_1=C_1(\lambda)<\infty,\quad C_2=C_2(\lambda)>0 . \tag{10} \]
It is now easy to establish that the operator \(T\), defined by formula (1), is an isomorphism.
I take this opportunity to thank B. S. Mityagin, who drew my attention to the fact that a number of essential points in the proof of Theorem 1 in paper \((^5)\) had been used earlier (for another purpose) in \((^2)\) (see the proof of Theorem 11).
Rostov State University
Received
27 II 1969
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