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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 $[0,1]$ s.t. $f(x)\leq C/x$ 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 $\frac{\ell}{\log\ell}$).

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

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 $[0,1]$ s.t. $f(x)\leq C/x$ 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 $\frac{\ell}{\log\ell}$).

• List item

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 $[0,1]$ s.t. $f(x)\leq C/x$ 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 $\frac{\ell}{\log\ell}$).

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 (just an idea; this works only for $C^1$ functions) One has $$\int_0^1 f(x)dx-f(1)=-\int_0^1 xf'(x)dx=:C.$$ Then, $f(x)=-\int_x^1f'(y)dy\leq \frac{1}{x}\int_0^1yf'(y)dy\leq C/x$.

Lemma 2. Let $f:(0,1]\to(0,+\infty)$ a continuous non-increasing function on $[0,1]$ s.t. $f(x)\leq C/x$ 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 $\frac{\ell}{\log\ell}$).

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 $[0,1]$ s.t. $f(x)\leq C/x$ 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 $\frac{\ell}{\log\ell}$).