Representing Systems of Dilations and Translations in Symmetric Function Spaces

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Let X be an arbitrary separable symmetric function space on [0, 1]. By using a combination of the frame approach and the notion of the multiplicator space \(\mathscr {M}(X)\) of X with respect to the tensor product, we investigate the problem when the sequence of dyadic dilations and translations of a function \(f\in X\) is a representing system in the space X. The main result reads that this holds whenever \(\int _0^1 f(t)\,dt\ne 0\) and \(f\in \mathscr {M}(X).\) Moreover, the condition \(f\in \mathscr {M}(X)\) turns out to be sharp in a certain sense. In particular, we prove that a decreasing nonnegative function f\(f\ne 0,\) from a Lorentz space \(\varLambda _{\varphi }\) generates an absolutely representing system of dyadic dilations and translations in \(\varLambda _{\varphi }\) if and only if \(f\in \mathscr {M}(\varLambda _{\varphi }).\)

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The work of the Sergey V. Astashkin was supported by the Ministry of Education and Science of the Russian Federation, Project 1.470.2016/1.4 and by the RFBR Grant 18-01-00414. The work of the Pavel A. Terekhln was supported by the RFBR Grant 18-01-00414. The authors are very grateful to the referee for detailed and constructive criticism that helped us improve the presentation of the paper.

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Correspondence to Sergey V. Astashkin.

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Communicated by Krzysztof Stempak.



A central role in the proof of the Filippov–Oswald theorem [8] is played by condition (2) that has allowed to Filippov and Oswald to prove the following characterization: let \(I_{k,i}=\left[ \frac{i}{2^k},\frac{i}{2^{k+1}}\right] \) for \(k=0,1,\ldots \) and \(i=0,1,\ldots ,2^k-1.\) For \(n = 2^k + i\) set \(I_n := I_{k,i}\) and \(f_n:=f_{k,i}.\) If \(f\in L_p([0,1])\) (with \(1\le p<\infty \)) satisfies condition (1), then a subsystem \(\{f_{n_l}\}\subset \{f_n\}\) is a representing system in \(L_p([0,1])\) if and only if for every \(L\in \mathbb {N}\) the set \(\cup _{l=L}^\infty I_{n_l}\) has full Lebesgue measure in the interval [0, 1].

However, we show here that condition (2) is not satisfied by Lorentz spaces on [0, 1] different from \(L_1.\) This is an immediate consequence of the following connection of (2) with the smoothness of a separable symmetric function space on [0, 1] at the function, identically equal to 1. Recall that a Banach space E is smooth at an element \(x_0\in E,\)\(\Vert x_0\Vert _E=1,\) whenever there exists a unique \(x^*\in E^*\) with \(\Vert x^*\Vert _{E^*}=x^*(x_0)=1.\)

Proposition 4

Let X be a separable symmetric function space on [0, 1]. Then, condition (2) is fulfilled for each \(f\in X\) such that \(\int _0^1 f(t)\,dt\ne 0\) if and only if X is smooth at 1.


Firstly, let condition (2) be fulfilled for each \(f\in X\) such that \(\int _0^1 f(t)\,dt\ne 0.\) Assuming that X is not smooth at 1, we find two functions \(y_1\) and \(y_2,\)\(y_1\ne y_2,\) from the dual space \(X^*=X'\) such that

$$\begin{aligned} \Vert y_1\Vert _{X^*}=\Vert y_2\Vert _{X^*}=\langle 1,y_1\rangle =\langle 1,y_2\rangle =1.\end{aligned}$$

Let \(f\in X\) be an arbitrary function such that \(a:=\langle f,y_1\rangle >0\) and \(b:=-\langle f,y_2\rangle >0.\) Obviously, we can assume that \(\int _0^1 f(t)\,dt\ne 0.\) Then, we have

$$\begin{aligned} \Vert 1-\lambda f\Vert _{X}\ge \langle 1-\lambda f,y_1\rangle =1-\lambda a\ge 1\end{aligned}$$

if \(\lambda \le 0,\) and similarly

$$\begin{aligned} \Vert 1-\lambda f\Vert _{X}\ge \langle 1-\lambda f,y_2\rangle =1+\lambda b\ge 1\end{aligned}$$

if \(\lambda >0.\) This contradicts the condition.

Conversely, suppose that X is smooth at 1 but, however, there is a function \(f\in X,\)\(\int _0^1 f(t)\,dt\ne 0,\) such that

$$\begin{aligned} \Vert 1-\lambda f\Vert _X\ge 1\quad \text{ for } \text{ all }\quad \lambda \in \mathbb {R}.\end{aligned}$$

Then, clearly, the projection \(P(a\cdot 1+b\cdot f):= a\cdot 1,\)\(a,b\in \mathbb {R},\) defined on the subspace, spanned by 1 and f,  has norm 1. Therefore, by Hahn–Banach Theorem, we have

$$\begin{aligned} 1=\inf _{\lambda \in \mathbb {R}}\Vert 1-\lambda f\Vert _X=\inf _{\lambda \in \mathbb {R}}\sup _{\Vert y\Vert _{X^*}\le 1}|\langle 1-\lambda f,y\rangle |= \sup _{\Vert y\Vert _{X^*}\le 1,\langle f,y\rangle =0}|\langle 1,y\rangle |. \end{aligned}$$

Hence, there exists a sequence \(\{y_n\}\subset X^*=X'\) such that \(\Vert y_n\Vert _{X^*}\le 1,\)\(\langle f,y_n\rangle =0,\)\(n=1,2,\ldots ,\) and \(\langle 1,y_n\rangle \rightarrow 1\) as \(n\rightarrow \infty .\) Since the closed unit ball in \(X^*\) is weakly\(^*\) compact, we can find a subsequence \(\{y_{n_k}\}\subset \{y_n\}\) such that \(y_{n_k}\rightarrow \tilde{y}\) weakly\(^*\) for some \(\tilde{y}\in X^*,\)\(\Vert \tilde{y}\Vert _{X^*}\le 1.\) This implies that \(\langle f,\tilde{y}\rangle =0\) and \(\langle 1,\tilde{y}\rangle =1.\) On the other hand, since \(\Vert x\Vert _1\le \Vert x\Vert _X\) (see Sect. 2.1), we have

$$\begin{aligned} \Vert 1\Vert _{X^*}=\langle 1,1\rangle =1.\end{aligned}$$

Therefore, taking into account that X is smooth at 1, from the preceding equations we deduce that \(\tilde{y}(t)\equiv 1\) and so \(\langle f,1\rangle =\int _0^1 f(t)\,dt=0,\) which contradicts the hypothesis.

\(\square \)

Corollary 5

Let \(\varphi \) be an increasing convex function on [0, 1],  \(\varphi (0)=0,\)\(\varphi (1)=1,\) and \(\lim _{t\rightarrow 0}\varphi (t)/t=\infty .\) Then there is a function \(f\in \varLambda _{\varphi }\) such that \(\int _0^1 f(t)\,dt\ne 0\) and for each \(\lambda \in \mathbb {R}\) we have

$$\begin{aligned} \Vert 1-\lambda f\Vert _{\varLambda _{\varphi }}\ge 1.\end{aligned}$$


Recall that isometrically \((\varLambda _{\varphi })^*=M_{\varphi },\) where \(M_{\varphi }\) is the Marcinkiewicz space with the norm

$$\begin{aligned} \Vert x\Vert _{M_{\varphi }}=\sup _{0<t\le 1}\frac{1}{\varphi (t)}\int _0^tx^*(s)\,ds \end{aligned}$$

[11, Theorem II.5.2]. One can easily check that from properties of \(\varphi \) it follows that both functions \(y_1(t)\equiv 1\) and \(y_2(t)=\varphi '(t)\) belong to \(M_{\varphi },\)\(y_1\ne y_2,\) and

$$\begin{aligned} \Vert y_1\Vert _{M_{\varphi }}=\Vert y_2\Vert _{M_{\varphi }}=\langle 1,y_1\rangle =\langle 1,y_2\rangle =1.\end{aligned}$$

This means that the space \(\varLambda _{\varphi }\) is not smooth at 1. Therefore, applying Proposition 5, we get the desired result. \(\square \)

Remark 2

A careful inspection of the proof of Lemma 2 from the paper [8] shows that, in fact, this proof is based on using the well-known Weak Greedy Algorithm. In the case of \(L_p,\)\(1\le p<\infty ,\) everything that is needed to apply it is condition (2). However, if we try to prove an analogue of the Filippov–Oswald theorem for a general separable symmetric function space X on [0, 1],  the following much more restrictive conditions are required:

(a) \(f\in \mathscr {M}(X);\)

(b) \(dist_{\mathscr {M}(X)}(1,X_{0,f})<1;\)

(c) \(\sup _{\Vert x\Vert _{\mathscr {M}(X)}\le 1}\liminf _{k\rightarrow \infty }dist_{\mathscr {M}(X)}(x,X_{k,f})<1.\)

Here, as before, \(X_{k,f}=\text {span}((f_{\alpha })_{|\alpha |=k}]),\)\(k=0,1,2,\ldots ,\) and for every Banach space Y\(L\subset Y\) and \(y_0\in Y\) we set

$$\begin{aligned} dist_{Y}(y_0,L):=\inf _{y\in L}\Vert y_0-y\Vert _Y. \end{aligned}$$

In contrast to that, according to Theorem 2, the only condition \(f\in \mathscr {M}(X)\) (together with (1)) assures that the sequence of dyadic dilations and translations of f is an absolutely representing system in the separable symmetric function space X. Thus, we see that the frame approach, used in this paper, works under less restrictive conditions and so has wider applicability than the above Weak Greedy Algorithm, used in [8] (cf. [18]).

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Astashkin, S.V., Terekhin, P.A. Representing Systems of Dilations and Translations in Symmetric Function Spaces. J Fourier Anal Appl 26, 13 (2020).

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  • Sequence of dilations and translations
  • Symmetric function space
  • Representing system
  • Tensor product
  • Frame
  • Lorentz space

Mathematics Subject Classification

  • Primary 46E30
  • Secondary 46B70
  • 42C15
  • 46B15