# What standard Banach space is isomorphic to the completion of this different normed structure on $\ell^1$?

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.

• 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 $\ell^2$. Mar 30, 2023 at 12:46
• 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. Mar 30, 2023 at 13:37
• 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. Mar 30, 2023 at 21:03
• @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. Mar 31, 2023 at 4:39
• 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 $\|x\| = \sup_{q \in [2,\infty)} \|x\|_q$. Apr 6, 2023 at 14:06

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 $$\frac{\ell}{\log\ell}$$).

• Since $x_j\to 0$, you could then also use $\Psi(t)=-|t|/\log |t|$. Apr 10, 2023 at 17:28
• @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. Apr 10, 2023 at 20:38
• 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. Apr 11, 2023 at 10:49
• 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. Apr 11, 2023 at 20:53
• Thank you very much Lorenzo for your answer Apr 12, 2023 at 12:15

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.

• Thank you very much for your answer Apr 6, 2023 at 18:54
• I revised this with more bulletpoints and fewer words, putting the results first and the justifications afterwards. Please correct anything I have garbled!
– user44143
Apr 6, 2023 at 20:27

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.

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.

• Thank you very much for your answer Apr 12, 2023 at 12:14