I teach topology and differential calculus this semester. This is the occasion to play with many nice mathematical concepts, including one of my favorite Banach spaces : $\ell^p$ spaces. These spaces are at the same time simple, important, and subtle. This tiny post collects some basic properties of these spaces, just for pleasure. I will improve this tiny post from time to time.

In what follows, $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$.

$\ell^p$ spaces. For all $p\in[1,\infty)$, $\ell^p=\ell^p(\mathbb{N},\mathbb{K})$ is the set of sequences $(x_n)_{n\in\mathbb{N}}$ in $\mathbb{K}$ such that $$\|x\|_p:=\sum_n|x_n|^p<\infty.$$ We also define $\ell^\infty=\ell^\infty(\mathbb{N},\mathbb{K})$ the set of sequences $(x_n)_{n\in\mathbb{N}}$ in $\mathbb{K}$ such that $$\|x\|_\infty:=\sup_n|x_n|<\infty.$$ These spaces naturally generalize $\mathbb{R}^n$ to $\mathbb{R}^\infty$. Note by the way that $[0,1]^\infty\subset\ell^p$, $p\in[1,\infty]$, but the topology induced by $\ell^p$ on $[0,1]^\infty$ is not the product topology, it is stronger, in particular for $p=\infty$ it corresponds to uniform convergence rather than convergence of each coordinate.

We always have $\ell^{p_1}\subsetneq\ell^{p_2}$ for all $1\leq p_1<p_2\leq\infty$.

Three spaces are of special importance: $\ell^1$, $\ell^2$, and $\ell^\infty$.

Functional point of view. A sequence $(x_n)_{n\in\mathbb{N}}$ in $\mathbb{K}$ is a function $\mathbb{N}\to\mathbb{K}$, $x(n):=x_n$. If we equip $\mathbb{N}$ with the discrete topology, the discrete $\sigma$-field, and the counting measure $\mathrm{d}n$, then $$\ell^p(\mathbb{N},\mathbb{K})=L^p(\mathbb{N},\mathrm{d}n,\mathbb{K})\quad\text{and}\quad \|x\|_p^p=\int |x(n)|^p\mathrm{d}n,$$ while $\ell^\infty=L^\infty(\mathbb{N},\mathrm{d}n,\mathbb{K})=\mathcal{C}_b(\mathbb{N},\mathbb{K})$. But let us study these spaces from scratch.

For all $n\in\mathbb{N}$, we set $e_n:=\mathbf{1}_n$, in such a way that for all $x\in\ell^p$, $x=\sum_nx_ne_n$.

Completeness. The spaces $\ell^p$, $p\in[1,\infty]$, are Banach spaces : complete normed vector spaces. The vector space nature and the norm axioms are not difficult to check. To establish the completeness of $\ell^p$ when $p\in[1,\infty)$, we consider a Cauchy sequence $(x^{(m)})$ in $\ell^p$. For all $n$, $(x^{(m)}_n)$ is a Cauchy sequence in $\mathbb{R}^n$ which is complete, hence we get a sequence $x^*=(x^*_n)$ such that $x^{(m)}_n\to x^*_n$ as $m\to\infty$, for all $n$. Next, since $(x^{(m)})$ is bounded in $\ell^p$, for all $N$, $$\sum_{n=0}^N|x^*_n|^p=\lim_{m\to\infty}\sum_{n=0}^N|x^{(m)}_k|^p\leq\sup_m\|x^{(m)}\|_p^p<\infty$$ thus $x^*\in\ell^p$ by taking the limit $N\to\infty$. Next, $$\sum_{n=0}^N|x^{(m)}_n-x^*_n|^p=\lim_{k\to\infty}\sum_{n=0}^K|x^{(m)}_n-x^{(k)}_n|^p\leq\varlimsup_{k\to\infty}\|x^{(m)}-x^{(k)}\|_p^p,$$ which implies that $x^{(m)}\to x^*$ in $\ell^p$ since $(x^{(m)})$ is Cauchy in $\ell^p$. This also works for $\ell^\infty$.

For $p=2$, we get a Hilbert space $\ell^2$ with dot product $x\cdot y:=\sum_nx_n\overline{y_n}$, while when $p\neq2$, the norm $\left\|\cdot\right\|_p$ does not satisfy the parallelogram identity and is thus not Hilbertian.

Not locally compact. The argument can be adapted to any infinite dimensional normed vector space. Namely, for $\ell^p$, $p\in[1,\infty]$, for any $r>0$, $x_n:=re_n=r\mathbf{1}_n$ satisfies $\|x_n\|_p=r$ and $\|x_n-x_m\|=r2^{1/p}$ if $n\neq m$, also for any $\varepsilon>0$, if $\overline{B}(0,r)$ was coverable by a finite number of balls of radius $\varepsilon>0$, then one of these ball would contain at least two distinct points $x_n$ and $x_m$, and this would give $r2^{1/p}=\|x_n-x_m\|\leq2\varepsilon$, which is impossible when $\varepsilon<r2^{1/p-1}$.

Non separability of $\ell^\infty$. For $I\subset\mathbb{N}$, set $e_I:=\sum_{i\in I}e_i$. Then for $I\cap J$, we have $\|e_I-e_J\|_\infty=1$, thus $B(e_I,\tfrac{1}{2})\cap B(e_J,\tfrac{1}{2})=\varnothing$, thus $\ell^\infty$ can be covered by an uncountable family of disjoint non-empty balls, since the set of non-empty subsets of $\mathbb{N}$ is uncountable. This argument does not work for $\ell^p$ with $p\in[1,\infty)$, since in this case $e_I\in\ell^p$ imposes that $I$ is finite, and the set of non-empty finite subsets of $\mathbb{N}$ is countable.

Separability of $\ell^p$, $p\in[1,\infty)$. If suffices to consider $A:=\cup_nA_n$ where $$A_n:=\Bigr\{\sum_{i=0}^nq_ie_i:q\in\mathbb{Q}_{\mathbb{K}}^{n+1}\Bigr\}.$$ Hence for all $x\in\ell^p$ and all $\varepsilon>0$, denoting $\pi_n(x):=\sum_{i=0}^{n-1}x_ie_i$, we have $\|x-\pi_n(x)\|_p\leq\varepsilon$ for a large enough $n$, and then $\|\pi_n(x)-y_n\|_p\leq\varepsilon$ for some $y\in A_n$ thanks to the density of $\mathbb{Q}_{\mathbb{K}}$ in $\mathbb{K}$. Note that the approximation $\|x-\pi_n(x)\|_p\leq\varepsilon$ of $x$ by the finitely supported sequence $\pi_n(x)$ does not work in $\ell^\infty$. Actually the closure in $\ell^\infty$ of the set of finitely supported sequences is the set of sequences which tend to $0$ at $\infty$, denoted $\ell^\infty_0$, which is strictly smaller than $\ell^\infty$, and which is a separable Banach subspace of $\ell^\infty$.

Hölder inequality. For all $p\in[1,\infty]$, $q:=p/(p-1)\in[1,\infty]$, all $x\in\ell^p$, $y\in\ell^q$, $$\sum_n|x_ny_n|\leq\|x\|_p\|y\|_q.$$ It follows from the Hölder inequality on $\mathbb{R}^n$ for $\pi_n(x)=\sum_{i=0}^{n-1}x_ie_i$ and $\pi_n(y)=\sum_{i=0}^{n-1}y_ie_i$. The equality is achieved when $|x_n|^p$ and $|y_n|^q$ are proportional (possibly asymptotically).

Dual of $\ell^p$, $p\in[0,\infty)$. Recall that the dual of a normed vector space $X$ is the normed vector space $X’=L(X,\mathbb{K})$ of continuous linear forms, namely continuous linear mappings $X\to\mathbb{K}$.

For all $p\in[1,\infty)$, denoting $q:=1/(1-1/p)=p/(p-1)\in(1,\infty]$ the Hölder conjugate of $p$, the map $\Phi:\ell^q\mapsto(\ell^p)’$ defined for all $y\in\ell^q$ and $x\in\ell^p$ by $$\Phi(y)(x):=\sum_nx_ny_n$$ is a bijective linear isometry. In particular $$(\ell^p)’\equiv\ell^q.$$ In particular $(\ell^2)’\equiv\ell^2$, $(\ell^1)’\equiv\ell^\infty$, and $((\ell^p)’)’\equiv\ell^p$ (reflexivity) for $p\in(1,\infty)$.

Proof. The map $\Phi$ is well defined thanks to the Hölder inequality, and $\|\Phi(x)\|_{(\ell^p)’}\leq\|x\|_q$. Let us establish the equality, hence it is injective or into, and then that it is surjective or onto.

Let us consider the case $p=1$ ($q=\infty$). Let $y\in\ell^\infty$. There exists $(n_k)$ such that $|y_{n_k}|\to\|y\|_\infty$ when $k\to\infty$. For all $k$, $x_k:=\tfrac{|y_{n_k}|}{y_{n_k}}e_{n_k}\in\ell^1$ satifies $\|x_k\|_1=1$ and $\Phi(y)(x_k)=|y_{n_k}|\to\|y\|_\infty$ when $k\to\infty$, hence $\|\Phi\|=\|y\|_\infty$, and $\Phi$ is an isometry and thus an injection. For the surjectivity, if h$\varphi\in(\ell^1)’$ then for all $x\in\ell^1$, $\varphi(x)=\sum_ny_nx_n$ with $y_n:=\varphi(e_n)$, but $|y_n|\leq\|\varphi\|\|e_n\|_1=\|\varphi\|<\infty$ hence $y\in\ell^\infty$ and $\varphi=\Phi(y)$.

Let us consider the case $p>1$ ($q<\infty$). The equality case in the Hölder inequality $x_n:=\tfrac{y_n}{|y_n|}|y_n|^{q-1}$ satisfies $\|x\|_p^p=\|y\|_q^q$ since $p(q-1)=q$, hence $\Phi(y)(x)=\|y\|_q^q=\|y\|_q\|x\|_p$, hence $\|\Phi(y)\|_{(\ell^p)’}=\|y\|_q$, and $\Phi$ is an isometry hence it is injective. Let us show that $\Phi$ is surjective. Let $\varphi\in(\ell^p)’$, then for all $x\in\ell^p$, $\varphi(x)=\sum_nx_ny_n$ where $y_n:=\varphi(e_n)$ with $e_n:=\mathbf{1}_n$. Suppose by contradiction that $y\not\in\ell^q$. Denoting $\pi_N(z):=(z_0,z_1,\ldots,z_{N-1},0,0,\ldots)$, and $x_n:=\frac{|y_n|}{y_n}|y_n|^{q-1}$, since $y\not\in\ell^q$,
$$\frac{\Phi(y)(\pi_N(x))}{\|\pi_N(x)\|_p} =\|\pi_N(y)\|_q\underset{n\to\infty}{\longrightarrow}\infty, \quad\text{which contradicts \Phi(y)\in(\ell^p)’}.$$

Dual of $\ell^\infty$. The dual of $\ell^\infty=(\ell^1)’$ is strictly larger than $\ell^1$. More precisely, the map $\Phi:\ell^1\mapsto(\ell^\infty)’$ defined by $\Phi(y)(x):=\sum_nx_ny_n$ is a linear isometry which is injective but not surjective. In other words, $\ell^1\subsetneq(\ell^\infty)’=((\ell^1)’)’$, in other words $\ell^1$ is not reflexive.

Proof. The isometry (and thus the injectivity) comes from the Hölder inequality
$|\Phi(y)(x)|\leq\|x\|_\infty\|y\|_1$ and its equality case $x_n=\tfrac{|y_n|}{y_n}$, which gives  $\|\Phi(y)\|=\|y\|_1$. It remains to establish that $\Phi$ is not surjective. Consider the following subspace of $\ell^\infty$:
$S:=\Bigr\{(x_n):x_*:=\lim_{n\to\infty}x_n\text{ exists}\Bigr\}.$
The linear functional $x\mapsto x_*$ is bounded has unit norm on $S$. Thanks to the Hahn–Banach (for the non separable Banach space $\ell^\infty$) we can extend it into $L:\ell^\infty\to\mathbb{K}$ in such a way that $|Lx|\leq\|x\|_\infty=\sup_n|x_n|$ and $Lx=\lim_{n\to\infty}x_n$ of $(x_n)$ converges. We have thus constructed a “limit” to each bounded sequence, which respects linearity, and which coincides with the usual limit for converging sequences. In particular this proves that $(\ell^\infty)’\supsetneq\ell^1$. Indeed, if we had $L(x)=\sum_nx_ny_n$ for some $(y_n)\in\ell^1$, then, if we define, for a fixed $m$ and $\ell\neq0$, $x_n:=0$ if $n<m$ and $x_n:=\ell$ if $n\geq m$, then we would get a contradiction : $$\ell=L(x)=\ell\sum_{n\geq m}y_n\xrightarrow[m\to\infty]{}0.$$

The space $S\subset\ell^\infty$ above can be seen as the space of continuous functions on the Alexandrov compactification $\overline{\mathbb{N}}=\mathbb{N}\cup\{\infty\}$ of $\mathbb{N}$. The topology on this space is metrizable, and by a Riesz theorem, every linear form on $S$ can be seen as a measure on $\overline{\mathbb{N}}$. The linear form $L$ above corresponds then clearly to the Dirac mass at $\infty$. This functional respects the additive structure but not necessarily the multiplicative one, in other words we do not have necessarily $L(xy)=L(x)L(y)$, and similarly we do not have necessarily $L(f(x))=f(L(x))$ for all $f$.

Representation.

• Every separable Banach space is isomorphic to a quotient $\ell^1/S$ where $S$ is a closed subspace of $\ell^1$.
• Every separable Hilbert space of infinite dimension is isomorphic to $\ell^2$ (via a Hilbert basis).