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A colleague asked me the following question:

"What can one do with the following norm on $\ell^1$: $|x|=\int_1^2 |x|_pdp$ where $| \;\; |_p$ is the standard norm on $\ell_p$?"

This interesting norm is introduced in Continuous mixed p-norm adaptive algorithm for system identification, Hadi Zayyani 2014, IEEE Signal Processing letters, doi:10.1109/LSP.2014.2325495

I observed that this is not a complete norm since it is dominated by $\ell^1$ norm but $\ell^1$ norm is not dominated by this norm. So I am curious about the structure of the completion of this norm.

What is a standard Banach space which is isomorphic to the completion of this norm on $\ell^1$? Is it a reflexive Banach space?

Note: Since 2 month ago I try to find a possible feature of "continuous field of Banach spaces" but I do not know how to apply this theory to this particular norm. Any suggestion or help is very appreciated.

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    $\begingroup$ Can we show that it is the space of sequences $x$ such that $|x|=\int_1^2 |x|_p dp$ is finite? Is that complete in this norm? This space seems to be larger than $\bigcup_{p<2} \ell^p$ but smaller than $\ell^2$. $\endgroup$ Mar 30, 2023 at 12:46
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    $\begingroup$ For $x$ in $l_1$ the function $|x|_p$ $1\leq p \leq 2$ is decreasing hence the norm is finite and roughly speaking is the average of $| x| _p$. The basis is symmetric and boundedly complete. Hence if the space does not contain $l_1$ it is reflexive. To show that this happens it is enough to prove that the norm of the averages of the normalized block sequences tend to zero when their size increases to infinity. $\endgroup$
    – S Argyros
    Mar 30, 2023 at 13:37
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    $\begingroup$ I'd like to write down a simple observation. The norm $\|x\|:= \sum_{n=0}^{\infty} 2^{-n} |x|_{p_n}$ where $p_n = 1+2^{-n}$ is an equivalent norm. So this space is isomorphic to $(\bigoplus \ell^{p_n})_{\ell^1}$ that is not reflexive. $\endgroup$
    – Onur Oktay
    Mar 30, 2023 at 21:03
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    $\begingroup$ @OnurOktay Ermm... Not to the sum, but to the subspace of that sum cut by the diagonal $x^{1}=x^{2}=\dots$, which seems to change the game entirely. But yeah, an integral of a decreasing function is, indeed, equivalent to the sum you suggested. $\endgroup$
    – fedja
    Mar 31, 2023 at 4:39
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    $\begingroup$ If yes, then it would be natural to check next if the dual space is given by $\bigcup_{q \in [2,\infty)} \ell^q$ with the norm $\|x\| = \sup_{q \in [2,\infty)} \|x\|_q$. $\endgroup$ Apr 6, 2023 at 14:06

3 Answers 3

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In what follows I will show that the closure of $\ell^1$ under the norm $|x|=\int_1^2|x|_pdp$ is nothing but the Orlicz space $L_\Phi$, where $\Phi$ is the function $$\Phi(t)=\frac{t^2-|t|}{\ln|t|}$$ (extended by continuity to $\Phi(0)=0$). More precisely, I will prove the following claim.

Claim. Let $x\in\bigcap_{p>1}\ell^p(\mathbb N)$. Then, $$ \int_1^2|x|_pdp<\infty \iff \int_1^2|x|^p_pdp<\infty \iff \sum_{j\in\mathbb N} \Phi(x_j)<\infty. $$

Remarks.

  1. The function $\Phi$ has a nice graph (it is convex, the derivative is $\Phi'(t)=\frac{t}{(\ln t)^2}$).
  2. The function $\Phi$ can be modified to behave essentially arbitrarily at infinity, since functions with the above conditions are bounded. One can take any Young function that behaves like $-|t|/\ln|t|$ as $t\to 0$ for the definition of the Orlicz space.
  3. The hypothesis $x\in\bigcap_{p>1}\ell^p(\mathbb N)$ is just for context, one simply needs $x\in \mathbb R^{\mathbb N}$ (with the convention that the $\ell^p-$norm of a sequence not belonging to $\ell^p$ is infinite).
  4. The proof is immediate for the zero sequence, so I will assume that $x\not\equiv 0$.

We now begin to prove the claim.


Lemma 1. Let $f:(0,1]\to(0,+\infty)$ a continuous non-increasing function on $(0,1]$. Then, $$ \int_0^1 f(x)dx<\infty \implies \sup_{x\in(0,1]} xf(x)<+\infty. $$

Proof. One has $$\int_0^1 f(y)dy\geq \int_0^x f(y)dy \geq xf(x)dx.$$


Lemma 2. Let $f:(0,1]\to(0,+\infty)$ a continuous non-increasing function on $(0,1]$ such that $f(x)\leq C/x$ for some $C>0$. Then, $$ \int_0^1 f(x)dx<\infty\iff \int_0^1 |f(x)|^{x+1}dx<\infty.$$

Proof. It comes from the fact that $$ \lim_{x\to 0^+} \frac{|f(x)|^{x+1}}{f(x)}=\lim_{x\to 0^+} |f(x)|^x=1,$$ where the last limit follows by monotonicity as $\lim_{x\to 0^+} 1/x^x=1$ and $\lim_{x\to 0^+} c^x=1$, $c>0$.


Proposition. The first equivalence of the claim holds.

Proof. The direction '$\implies$' follows combining the two Lemmas. The other direction follows by monotonicity.


The remaining part of the claim follows by Fubini's theorem: $$ \int_1^2|x|^p_pdp=\int_1^2\sum_{j\in\mathbb N}|x_j|^pdp=\sum_{j\in\mathbb N}\int_1^2|x_j|^pdp= $$ $$ =\sum_{j\in\mathbb N}\frac{|x_j|^2-|x_j|}{\ln|x_j|}. $$


Final remarks. The claim (together with the observations from other answers) tells that the completion of $\ell^1$ under this norm is the space of the sequences $x$ such that $$ \sum_{j\in\mathbb N}\Phi(x_j)<\infty. $$ Since this condition is invariant under multiplying $x$ by a non-negative scalar, the space is precisely the Orlicz space $L_\Phi$. It follows from general facts that the norm $|\cdot|$ is equivalent to $$ \|x\|_\Phi:=\inf\left\{k\in(0,\infty)\;\Big|\;\sum_{j\in\mathbb N}\Phi\left(\frac{x_j}{k}\right)\leq 1\right\}. $$ I am not sure whether there is a standard way of denoting the space $L_\Phi$ (maybe $\frac{\ell}{\log\ell}$).

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    $\begingroup$ Since $x_j\to 0$, you could then also use $\Psi(t)=-|t|/\log |t|$. $\endgroup$ Apr 10, 2023 at 17:28
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    $\begingroup$ @ChristianRemling thank you for the remark. I left it like that because $t/\ln t$ diverges at $t=1$, and also to make $\Phi$ convex (so that it plays well with the definition of Orlicz space), but it's right that the behaviour of phi as $t\to\infty$ is irrelevant concerning the characterization as long as $\Phi$ stays away from zero. $\endgroup$ Apr 10, 2023 at 20:38
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    $\begingroup$ Nice result! I wonder if the alternative description of the norm could give a positive answer to the following: Consider a sequence $(x_n)_n $ in $l_1$ with each $x_n$ a finite linear combination of the basis, $|x_n| =1$ and $lim_n |x_n|_1 = \infty$. Then the norm of $x_n$ is asymptotically concentrated around one( in the same manner as happens for the normalized averages of the basis).If the answers is positive then it seems to me that the space is $l_1$ saturated. $\endgroup$
    – S Argyros
    Apr 11, 2023 at 10:49
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    $\begingroup$ That this space is $\ell_1$ saturated follows from general results; e.g., the papers of Lindenstrauss-Tzafriri on Orlicz sequence spaces. Then it is also complementably $\ell_1$ saturated because the space has an unconditional basis. $\endgroup$ Apr 11, 2023 at 20:53
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    $\begingroup$ Thank you very much Lorenzo for your answer $\endgroup$ Apr 12, 2023 at 12:15
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Let $X$ be the completed space with your norm. We collect several facts about $X$:

  • $x\in X$ if and only if $P_Nx \in X$ for all $N$ and $|P_N x|$ is uniformly bounded, where $P_Nx$ is the finite sequence that equals $x$ in the first $N$ slots and has $0$'s afterwards.

  • $\ell^1$ is dense in $X$ using the $|\cdot|$ norm.

  • Finite sequences are also dense in $X$, since the set of finite sequences are dense in $\ell^1$ with respect to the $|\cdot|_1$ norm, which dominates $|\cdot|$.

  • $X$ embeds into $\ell^2$, because the norm dominates the $\ell^2$ norm on sequences, because $|x|_p \leq |x|_q$ for $p \geq q$.

  • $X\subseteq \ell^p$ for all $p > 1$; if we had $x\in \ell^2 \setminus \ell^p$ for any $p > 1$, then $|P_N x|_q$ for all $q \in [1,p]$ would not be uniformly bounded,

  • Jochen's formulation holds, i.e. $x\in X$ if and only if $x\in \cap_{1 < p \leq 2} \ell^p$ and $|x| < \infty$.

  • $X$ embeds into every $\ell^p$ continuously, by the monotonicity of the $\ell^p$ norms and $$ (p-1)|x|_p = \int_1^p |x|_p ~dq \leq \int_1^p |x|_q ~dq \leq |x| $$

We also collect facts about $X^*$:

  • $\ell^q \subset X^*$ for all finite $q$.

  • $X^* \subseteq \ell^\infty$, by the embedding of $\ell^1$ into $X$

  • $\cup_{q < \infty} \ell^q \subsetneq X^*$, since we can find $\sum y_k$ in $X^*-\cup_{q < \infty} \ell^q*$:

    • Let $y_k \in \ell^{2^k + 1}\setminus \ell^{2^{k-1} + 1}$ for $k \in \mathbb{N}_0$, be such that $|y_k|_{2^k + 1} \leq 2^{-k}$.
    • $\sum_k y_k$ converges to some sequence in $\ell^\infty$.
    • $\sum y_k \in X^*$ because $|\langle x, \sum y_k\rangle| \leq \sum |x|_{1 + 2^{-k}} |y_k|_{1 + 2^k} \leq \sum 2^{-k} |x|_{1 + 2^{-k}} \approx |x|$.

I am hopeful that this leads to a characterization of $X^*$, but am not sure how to prove it, nor what would be a nicer norm for it.

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    $\begingroup$ Thank you very much for your answer $\endgroup$ Apr 6, 2023 at 18:54
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    $\begingroup$ I revised this with more bulletpoints and fewer words, putting the results first and the justifications afterwards. Please correct anything I have garbled! $\endgroup$
    – user44143
    Apr 6, 2023 at 20:27
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We will show that the space contains isomorphically the space $l_1$ therefore the space is not reflexive. We start with the following that I posed as a question in a previous comment.

Fact 1: For every $0<\delta < 1 $ $lim_n \frac{\int_1^{1+\delta} n^{1/p} dp } {\int_1^2 n^{1/p} dp } = 1$.

Proof: Since the function $n^{1/p}$, $1\leq p \leq 2$ is decreasing we have that $\int_1^{1+\delta} n^{1/p} dp >\delta n^{1/1+\delta}$ and
$n^ {1/1+2\delta} > \int_{1+2\delta}^2 n^{1/p} dp$
Now for $n\in N$ we have that $ \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2 n^{1/p} dp } < \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^{1+\delta} n^{1/p} dp } < \frac{n^{1/1+2\delta }} {\delta n^{1/1+ \delta }} = \frac{1} { \delta} \frac {1}{n^{\delta/ (1+\delta)(1+2\delta)}}$.
Hence for every $0<\delta <1/2$

$lim_n \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2 n^{1/p} dp } =0$ which finishes the proof of Fact 1.

Fact 2: We start with the following classical result.
If $(f_n)_n$ is a normalized sequence in $L^1 [1,2]$ which is not uniformly integrable (i.e. there exists $\epsilon>0$ such that for every $\delta>0$ there exists a Borel set $A$ with $\lambda (A)<\delta$ and $\int_A |f_n| > \epsilon$ for infinite $n\in N$ ) then $(f_n)_n$ has a subsequnce equivalent to $l_1$ basis.
This result is due to Kadec and Pelczynski ( see J. Diestel: Sequences and Series in Banach Spaces (Graduate Texts in Mathematics, 92) p. 93).
Next in the space we denote $(e_i)_i\in N$ the basis of $l_1$ which is a symmetric basis for the space.For $n\in N $ we set $z_n = \sum_ {i=1} ^{n} e_i$ and $x_n = \frac{1} {\int_1^2 n^{1/p} dp } z_n$ which has norm 1.
Consider the function $f_n(p) = |x_n|_p $ $1\leq p \leq 2 $ and Fact 1 yields that the sequence $(f_n)$ it is not uniformly integrable.Therefore $(f_n)$ has a subsequence equivalent to $l_1$ basis which implies that $(x_n)$ satisfies the same property in the norm of the space. We will adapt Kadec - Pelczynski's proof in the setting of $(x_n)$.
Step 1 : There exists a decreasing sequence $(\delta_k )$ and a subsequence $(x_{n_k})$ such that for all $k$ we have that $\int_{1+\delta_{k+1}}^{1+\delta_k} |{x_{n_k}}|_p dp > \frac{1} { 4}$.
The proof uses induction and the following :
From Fact 1 for $ 0< \delta < 1$ there exists $n\in N$ such that $\int_1^{1+\delta}| {x_n}|_p dp > \frac{1} {2}$. For this $n$ there exists $\delta_1 < \delta $ such that $\int_1^{1+\delta_1}|{x_n}|_p dp < \frac{1} {4}$. Hence $\int_{1+\delta_1}^{1+\delta} |{x_n}|_p dp > \frac{1} {4}$.
Step 2: We set $A_k = [\delta_{k+1}, \delta_{k }]$. There exists an infinite $I \subset N $ such that for every $k\in I$ setting $ B_k = \cup \{ A_j : j\in I, j\neq k \} $ we have that $\int_{B_k} |{x_{n_k}|_p}dp < \frac {1} {8}$.
This is a classical result due to H. P. Rosenthal and an elegant and short proof was given by J. Kupka ( see Page 82 in the aforementioned reference). We assume that $ I = N $

Claim: The sequence $(x_{n_k} )_{k\in N}$ is equivalent to $l_1$ basis.

Indeed
$\int_1^2 |\sum_{j=1}^k \lambda_{j} x_{n_j} |_p dp > \int_{\cup_{i\in N} A_i} |\sum_{j=1}^k \lambda_{j} x_{n_j} |_pdp \geq \sum_{j=1} ^{k} (\int_{A_j} |\lambda_{j} x_{n_j}|_pdp - \int_{B_j} |\lambda_{j} x_{n_j}|_p dp)\geq \frac {1}{8} \sum_{j=1}^{k} |\lambda_j|$.

I have two questions related to this result.

Question 1 : Does the space contain a complemented subspace isomorphic to $l_1$ ?

Edit: The answer to Question 1 is affirmative hence the dual of the space contains isomorphically the space $l_\infty $.

The functionals $f_A^x $ where A is a Borel subset of [1,2] and $x=\sum_{i=1}^n \lambda_i e_i (\lambda_i \geq 0). $

For $x$ as above and $ p\in (1,2] $ we set $f_p^x = \frac{1} {(\sum_{i=1}^n \lambda_i^p)^1/q } \sum_{i=1}^n \lambda_i^{1/q-1} e_i$ where $ 1/p + 1/q = 1 $.
The functional $f^x_p$ is the unique normalized element of $l_q$ that norms $x$ as an element of $l_p $.
Observe that for a given $x$ as above and $z\in l_1$ the function $f_p ^x (z) $ with variable $ p \in (1,2]$ is continuous hence for a Borel $A \subset (1,2]$ the integral
$f_A ^x(z)= \int_A f_p^x (z) dp$
is well defined for all $z \in l_1$ and $f_A ^x$ is linear.

Properties of $f_A ^x$.

We denote by $|.|$ the norm of the space and by $|.|_*$ the norm of its dual.

For all $x$ , $A$ $| f_A ^x |_* \leq 1$ moreover if $ |x|\leq 1$ and $\int_A |x|_p dp \geq c > 0 $ then $| f_A ^x |_* \geq c$.

For $(A_k)_{k=1}^m $ disjoint Borel sets , $(x_k)_{k=1}^m$ in $l_1$ and $(\alpha_k)_{k=1}^m$ reals we have that.

$ | \sum_{k=1}^{m} \alpha_{k}f_{A_k}^{x_k} |_* \leq max \{|\alpha_k| : k=1,...,m \}$
In particular every sequence $(f_{A_k}^{x_k} )_k$ with $(A_k))_k$ disjoint Borel sets is weakly null since every n-average of them has norm less or equal to $1/n$.
Furthermore if for every $k \in N$ $|f_{A_k}^{x_k}|_*\geq c >0 $ then $(f_{A_k}^{x_k} )_k$ has a subsequence which is Schauder basic which yields that this subsequence is equivalent to $c_o$ basis.

The dual of the space contains isomorphically $c_0$.

We set $A_k= (\delta_{k+1}, \delta_k)$ and $x_k = x_{n_k}$ as they appeared in Step 1 above. Then the sequence $(f_{A_k}^{x_k} )_k$ satisfies $ |f_{A_k}^{x_k}|_* \geq \frac{1} {4}$ and $(A_k)_k$ are disjoint. Hence it has a subsequence equivalent to $c_o$ basis.

The space has a complemented subspace isomorphic to $l_1$.
This is an immediate consequence of the previous result. A classical Theorem states that if $c_0$ is isomorphic to a subspace of $X^*$ then $l^1$ is isomorphic to a complemented subspace of $X$.

Question 2: Is the space $l_1$ saturated?

Edit 1 The answer to this question is also affirmative. In particular the following holds:

Fact 3: Every closed infinite dimensional subspace $Z$ of the space has a further subspace $Y$ which is complemented in the space and isomorphic to $l_1$.

To prove this we first observe that it is enough to consider block subspaces namely subspaces generated by a normalized block sequence $(x_n )_n$. For such a sequence we will show the following:

Fact 4: For every normalized block sequence $( x_n)_n$ there exists a sequence $(F_k )_k $ with $F_k \subset N$, $\#F_k = n_k$ such setting $z_k = \sum_{n\in F_k}x_n$ the following holds.

For every $q > 1 $ $ lim_k \frac {\int_1^{q} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} = 1 $.

With this result and repeating the proof of Fact 2 above for the sequence $(z_k)_k $ we conclude that it has a subsequence equivalent to $l_1$ basis. moreover the subspace generated by a further subsequence is complemented in the space. This follows from the answer to Question 1 above with a small modification at the last part.

Proof of Fact 4: We set $q_k = 1 + \frac{1}{2^k} $. Observe that for all $k \in N$
$sup \{ |x_n |_{q_k} : n\in N \} \leq 2^k $.

We assume that $ lim_n |x_n |_{q_k} = C_k $ for all $k\in N $ (otherwise we pass to a subsequence).
Observe that $(C_k)_k$ is increasing.

Claim: For every $k\in N$ there exists $F_k \subset N$ such that setting $z_k = \sum_ {n\in F_k} x_n $ we have that

$ \frac {\int_1^{q_k} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} > 1- \frac{1} {k} $

Proof of the Claim: Since $ |x_n|_{q_k} \rightarrow C_k$ and $|x_n|_{q_{k+1}} \rightarrow C_{k+1}$ for every $l\in N$ we may select $F_l \subset N$ such that $\#F_l =l$ and

$ \frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^{-k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$
where $z_l = \sum _{n\in F_{l}} x_n$.

As in Fact 1 we have that:

$ \frac {\int_{q_k}^2 |z_l|_p dp } {\int_1^{2} |z_l|_p dp} < \frac {\int_{q_k}^2 |z_k|_p dp } {\int_1^{q_{k+1}} |z_k|_p dp} < \frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } \leq (\frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^ {-k+1}C_{k+1} l^{1/q_{k+1}}}) + \frac { C_{k} l^{1/{q_k}}} { 2^{-k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$.

We choose $l_k$ such that $\frac {\int_{q_k}^2 |z_{l_k}|_p dp } {\int_1^{2} |z_{l_k}|_p dp} <\frac {1} { k}$ and we set $ F_k = F_{l_k} $ and $ z_k = z_{l_k} $. Clearly $F_k , z_k $ satisfy the conclusion of the claim.

Conclusion: The sequence $(z_k )_k $ satisfies:

For every $q > 1 $ $ lim_k \frac {\int_1^{q} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} = 1 $.Hence as in Fact 2 there is a subsequence $(z_{k_m})_m$ equivalent to $l_1$ basis. This subsequence has a further subsequence with the property that the space that generates is complemented in the whole space.

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    $\begingroup$ Thank you very much for your answer $\endgroup$ Apr 12, 2023 at 12:14

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