# Failure of the Hahn-Banach in ℂₚ

The Hahn-Banach is known to be a crucial tool in functional analysis. The following are lecture notes written a while ago for my functional analysis course on the failure of Hahn-Banach in the $p$-adic completion of $\bar{\mathbb{Q}}_p$ (denoted by $\mathbb{C}_p$). For those who are well-versed in the basics of $\mathbb{Q}_p$, I would recommend skipping to around the middle of this page.

Since these are lecture notes, I present as much proof as I can without going into too much gritty detail.

Definition. The $p$-adic norm of $x\in\mathbb{Q}\setminus \{0\}$ is defined as $|x|_p=p^{-r}$ where $r$ is the unique power of $p$ dividing $x$. We define $|0|_p:=0$.

Let $x=75/2=5^2\cdot 3\cdot 2^{-1}$. Then

$|x|_5=5^2, |x|_3=3, |x|_2=2^{-1}$

Definition. A norm $|\cdot|:X\to \mathbb{R}$ is called a non-archimedean norm if

$|x+y|\leq \max\{|x|,|y|\}$

If a norm does not satisfy the above inequality, then $|\cdot|$ is a archimedean norm.

This inequality is called the strong triangle inequality. As a metric, this norm induces the ultrametric.

Proposition. The $p$-adic norm is a non-archimedean norm of the field $\mathbb{Q}$.

Proof. By definition, $|x|_p=0\iff x=0$ and $|x|_p\geq 0$ for all $x$. Since $\mathbb{Z}$ is UFD, $|xy|_p=|x|_p|y|_p$. Finally, if $x,y\in\mathbb{Q}$,

$x+y=p^r\frac{u}{v}+p^s\frac{w}{z},\quad\gcd(u,v)=1,\quad\gcd(w,z)=1$

Without loss of generality, let $s\geq r$. Then

$x+y=p^{r}\frac{uz+p^{s-r}vw}{vz}\implies |x+y|_{p}\leq|x|_p.$

where the implication follows since $\gcd(vz,p)=1$ and $\gcd(uz+p^{s-r}vw,p)=1$.
Thus in full generality,

$|x+y|_p\leq \max\{|x|_p,|y|_p\}\leq |x|_p+|y|_p.$

$\blacksquare$

Definition. The metric completion of $\mathbb{Q}$ with respect to the metric $|\cdot|_p$ is called the $p$-adic numbers, denoted by $\mathbb{Q}_p$.

Proposition. Any $x\in \mathbb{Q}_p$ has the form

$x=\sum_{i=k}^\infty a_ip^i$

where $a_k\neq 0$ and $a_i\in\{0,…,p-1\}$ for $i>k$.

Consider $\mathbb{Q}_2$ and sums of the form

$x=\sum_{i=0}^\infty a_ip^i$

where $a_i\in\{0,1\}$. Consider the infinite coordinate $\tilde{x}=(a_0,a_1,a_2,…)$. Then $\tilde{x}$ is the binary representation of $x$.

Proposition. $\zeta_n\in\mathbb{Q}_p$ if and only if $p\equiv 1\mod n$.

Proof. F.Q. Gouvêa, $p$-adic Numbers, 1997

$\blacksquare$

Corollary. $\mathbb{Q}_p$ is not algebraically closed.

Proof. For every $p$, $\zeta_{p+1}\not\in\mathbb{Q}_p$. Thus $\mathbb{Q}_p$ does not contain the root of the polynomial $x^{p+1}-1=0$.

$\blacksquare$

Definition. The algebraic closure of $\mathbb{Q}_p$ is denoted by $\overline{\mathbb{Q}}_p$.

Proposition. The norm $|\cdot|_{p}$ extends to $\overline{\mathbb{Q}}_p$.

Proof. Take $x\in\overline{\mathbb{Q}}_p$. Then $\mathbb{Q}_p(x)$ is a finite extension, of say degree $n$, since $x$ is the root of a polynomial of finite degree. Define $\psi:\overline{\mathbb{Q}}_p(x)\to\overline{\mathbb{Q}}_p(x)$ by

$w\mapsto xw$

Then $\psi$ is $\mathbb{Q}_p$-linear map, thus define

$|x|_{p}=\sqrt[n]{|\det \psi|_p}$

and check axioms.

$\blacksquare$

Proposition. $\overline{\mathbb{Q}}_p$ is not complete with respect to $|\cdot|_p$.

Sketch of Proof. Define

$a_i:=\begin{cases} \zeta_i,\quad\mbox{if } (i,p)=1,\\ 1,\quad \mbox{otherwise}. \end{cases}$

Define

$x_{n,m}:=\sum_{i=n}^m a_ip^i$

Then $\{x_{n,m}\}$ is Cauchy under the $p$-norm, but the limit is not in $\overline{\mathbb{Q}}_p$.

$\blacksquare$

Definition. The metric completion of $\overline{\mathbb{Q}}_p$ with respect to $|\cdot|_p$ is called the $p$-adic complex numbers, denoted by $\mathbb{C}_p$.

Definition. A metric space $T$ is called spherically complete if for any nested sequence of closed balls $B_1\supseteq B_2\supseteq…$ in $T$,

$\bigcap_{n\in\mathbb{N}}B_n\neq\emptyset$

Some examples of spherically complete fields are $\mathbb{R}$  and $\mathbb{C}$ by Cantor’s intersection theorem. In fact, these fields are complete with respect to nested sequences of compact sets.

Lemma. $\mathbb{Q}_p$ is spherically complete.

Lemma. $\mathbb{C}_p$ is not spherically complete.

Proof. Since $\mathbb{C}_p$ is separable (See A Course in p-adic Analysis, A. Robert.), there exists a countable dense subset $A$ which can be written as a sequence $(a_1,a_2,…)$ where $a_i\in\mathbb{C}_p$. Fix a sequence $(\epsilon_n)_{n=1}^\infty$ where $1>\epsilon_1>\epsilon_2>…>1/2$. On $\mathbb{C}_p$, define a relation $a\sim b$ iff $|a-b|_p\leq\epsilon_1$. This is an equivalence relation since

$a\sim b\sim c\implies |a-c|_p=|a-b+b-c|_p\leq \max\{|a-b|,|b-c|\}\leq \epsilon_1\implies a\sim c$

Since the other properties are trivially satisfied, we have our claim. Visually, this equivalence relation partitions $\mathbb{C}_p$ as the set of all $\epsilon_1$ balls where each ball is in fact disjoint! (For $\mathbb{R}$, note that $(-1,1)\cap (0,2)=(0,1)$, so the disjointness is indeed very surprising.)

Now observe that $|\cdot|_p$ is a norm mapping to $\mathbb{R}$, but it ONLY takes on values $p^\mathbb{Z}$ when considered under $\mathbb{Q}$. It can be shown that for $\mathbb{C}_p$, the norm takes on values $p^\mathbb{Q}$. Moreover, one may notice these values are dense in $\mathbb{R}_+^\times$. It follows that the diameter of every equivalence class is actually $\epsilon_1$.

We can now iterate. Take $B_1$ to be an equivalency class such that $a_1\notin B_1$. Then on $B_1$, define an equivalence relation $a\sim b$ iff $|a-b|_p\leq\epsilon_2$. Then take $B_2\subseteq B_1$ to be an equivalency class such that $a_2\notin B_2$. Inductively, we have a nested sequence

$B_1\supseteq B_2\supseteq…\mbox{ such that }\operatorname{diam}(B_i)=\epsilon_i\mbox{ and }a_i\not\in B_i \forall i\in\mathbb{N}$

I claim $\cap_{n\in\mathbb{N}} B_n=\emptyset$. Indeed, if not, then there exists $b\in \cap_{n\in\mathbb{N}} B_n$. Since $b$ is in each equivalency class, we have $B_i=B_{\epsilon_i}(b)$ for all $i$, so $B_{1/2}(b)\subset\cap_{n\in\mathbb{N}} B_n$ since $\epsilon_i>1/2$. By construction, no $a_i$ can be contained in $B_{1/2}(b)$, but this contradicts the density of $\overline{\mathbb{Q}}_p$ in $\mathbb{C}_p$.

$\blacksquare$

Definition. Let $X$ be any set, and let $k$ be a complete field with norm $|\cdot|_k$. Then we denote by $l^\infty(X)$ the set of all bounded functions on $X$ to $k$. That is

$l^\infty(X):=\left\{\varphi:X\to k\mid \sup_{x\in X}|\varphi(x)|_k<\infty\right\}$

Lemma. Over $\mathbb{C}_p$, $l^\infty(X)$ is spherically complete and $c_0$ is non-spherically complete.

Proof. Non-Archimedean Functional Analysis, A.C.M. Van Rooij: If $k$ has a dense norm, then $l^\infty$ is spherically complete. Since the norm of $\mathbb{C}_p$ is dense in $\mathbb{R}^+_{\geq 0}$, we are done.

$\blacksquare$

Recall

Theorem. [Hahn-Banach] Let $V$ be a vector-space over a complete archimedian field. Let $p$ be a sub-linear functional on $V$. For every linear subspace $E\subseteq V$, for all $f\in E'$ dominated by $p$, there exists a $g\in V'$ such that $f=g|_{E}$ and $||g||\leq ||p||$.

This theorem fails very harshly when the Banach space is non-archimedean; in fact, we can find examples that do not even satisfy the following properties:

Definition. A Banach space $V$ is said to satisfy the Hahn-Banach extension property (HBEP) if for every linear subspace $E\subseteq V$, for all $f\in E'$, there exists a $g\in V'$ such that $f=g|_{E}$ and $||g||=||f||$.

Theorem. HBEP does not hold for Banach spaces over $\mathbb{C}_p$.

Proof. Consider the closed subspace of $l^\infty(\mathbb{N})$

$\Delta=\left\{(a,a,…):a\in\mathbb{C}_p\right\}$

Consider the linear functional

$\varphi:\Delta\to\mathbb{C}_p\mbox{ defined by }(a,a,…)\mapsto a$

Clearly $||\varphi||=1$. Now since $\mathbb{C}_p$ is not spherically complete, there is a nested sequence of balls $\{B_{r_i}(x_i)\}$ with empty intersection in $\mathbb{C}_p$. Consider the nested sequence $B_i^*:=B_{r_i}((x_i,x_i,…))\subseteq l^\infty(\mathbb{N})$. The intersection of this family is non-empty since $l^\infty(\mathbb{N})$ is spherically complete. Thus let $x$ be in the intersection. Now suppose $\varphi$ extends to a linear map $\psi:l^\infty(\mathbb{N})\to \mathbb{C}_p$ with norm preserved. Then

$|\psi(x)-x_i|_\infty = |\psi(x)-\varphi(x_i)|_\infty \leq ||\psi||||x-x_i||_{A}=|x-x_i|_{p}\leq\max\{|x|_p,|x_i|_p\}=\max\{r,r_i\}$

so $\psi(x)\in\cap_n B_n$, a contradiction.

$\blacksquare$

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