Norms as Points in $C(X)$

Question

$\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$.

For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is,

1. $\abs{f}_x \geq 0$.

2. $\abs{cf}_x = \abs c \cdot \abs f_x$.

3. $\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$.

4. $\abs{f + g}_x \leq \abs f_x + \abs g _x$.

Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are:

1. How might we put a topology on $Y$ using only the information of $C(X)$?

2. Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?

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So in this view, multiplicative seminorms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global".

Answer 1

$\newcommand\Abs[1]{\lVert{#1}\rVert}\newcommand\abs[1]{\lvert#1\rvert}$If one also enforces non-triviality and continuity, then there is a one-to-one correspondence between multiplicative seminorms and points. Namely:

Proposition. Let $X$ be compact Hausdorff, and let $\Abs{ }: C(X \to {\bf R}) \to {\bf R}$ be a continuous map that obeys the homogeneity property $\Abs{cf} = \abs c\Abs f$, the multiplicative property $\Abs{fg}=\Abs f\Abs g$, and the triangle inequality $\Abs{f+g} \leq \Abs f + \Abs g$, and such that $\Abs{ }$ is not identically zero. Then there exists a unique $x_0 \in X$ such that $\Abs f = \abs{f(x_0)}$.

Proof. Since $\Abs f \neq 0$ for at least one $f$, and $f \cdot 1 = f$, we conclude from multiplicativity that $\Abs1 = 1$.

Call a point $x \in X$ typical if there is a function $f \in C(X \to {\bf R})$ that vanishes in a neighbourhood of $x$ with $\Abs f \neq 0$. If every point was typical, then by compactness one could find finitely many functions $f_1, \dots,f_k \in C(X \to {\bf R})$, all of nonzero norm, with each $f_i$ vanishing on an open set $U_i$ with the $U_i$ covering $X$. But then $\Abs0 = \Abs{f_1 \dotsb f_k} = \Abs{f_1} \dotsb \Abs{f_k} \neq 0$, which is absurd. Thus there must be a point $x_0$ which is not typical; thus $\Abs f=0$ whenever $f$ vanishes near $x_0$, and hence by the triangle inequality $\Abs f=\Abs g$ whenever $f$, $g$ agree near $x_0$.

For any $f \in C(X \to {\bf R})$ and any $\varepsilon>0$, one can use Urysohn's lemma to find a function $f_\varepsilon$ that lies with $\varepsilon$ of $f(x_0)$ and agrees with $f$ on a neighbourhood of $x_0$. Then $\Abs f = \Abs{f_\varepsilon}$; taking limits as $\varepsilon \to 0$, we conclude that $\Abs f = \Abs{f(x_0)} = \abs{f(x_0)}$, so we have located an $x_0$ with the required properties, which is unique since $C(X \to {\bf R})$ separates points. $\Box$

Without the continuity axiom there appear to be additional, weirder seminorms arising from other distributions supported at points than Dirac masses. For instance, for $X = [-1,1]$, the map $\Abs f \mathrel{:=} \abs{f(0)} \exp( \alpha \frac{f'(0)}{f(0)} )$, for any fixed real $\alpha$, gives a densely defined map on $C([-1,1] \to {\bf R})$ that obeys the homogeneity, multiplicativity, and triangle inequality (the latter is a calculation using the convexity of $t \mapsto \exp(\alpha t)$), but is discontinuous. It may be possible to extend this map to the rest of $C([-1,1] \to {\bf R})$ by a Zorn's lemma argument, though I have not checked this. [EDIT: actually, the triangle inequality only seems to work for functions with $f(0)>0$. Not sure whether this example can be fixed to cover the opposite case $f(0)<0$. I'll leave it up here in case anyone else has an idea.]

Answer 2

A slight improvement to Terry Tao's answer: It is not necessary to assume a priori that the multiplicative seminorm $\|\cdot\|$ is continuous with respect to the sup-norm on $C(X)=C(X\to \mathbb R)$ to conclude $\|f\|=|f(x_0)|$ for some $x_0\in X$.

This follows from a theorem of Mazur [Sur les anneaux linéaires, C. R. Acad. Sci. Paris, 2 0 7 (1938), 1025-1027] which says that the only real algebras with a multiplicative norm are (isometrically isomorphic to) the real numbers, the complex numbers, or the quaternions.

We will apply this to the the quotient algebra $C(X)/I$ with the ideal $I=\{f\in C(X): \|f\|=0\}$. Denoting by $\tilde f$ the equivalence class of $f\in C(X)$ one easily sees that $\tilde f \mapsto \|f\|$ is a well-defined multiplicative norm on $C(X)/I$ to which Mazur's theorem applies. Since this quotient algebra is commutative it cannot be the quaternions and we show that it neither can be ismorphic to $\mathbb C$. Indeed, assume that $\pi: C(X)\to \mathbb C$ is an algebra morphism onto $\mathbb C$ with kernel $I$. If $g\in C(X)$ satisfies $\pi(g)=i$ we get $\pi(g^2+1)=0$ so that $g^2+1 \in I$ which is absurd because $g^2+1$ is invertible in $C(X)$ and thus does not belong to the (proper) ideal $I$.

We conclude that $C(X)/I$ is isomorphic to $\mathbb R$ so that $I$ is a maximal ideal in $C(X)$ and hence of the form $I=\{f\in C(X): f(x_0)=0\}$ for some $x_0\in X$. Finally, given $f\in C(X)$ write $f=f-f(x_0)+f(x_0)$ (where the number $f(x_0)$ is identified with the constant function $f(x_0)\mathrm 1)$ to get $$\|f\|=\|f(x_0)\mathrm 1\|=|f(x_0)| \|\mathrm 1\|=|f(x_0)|.$$

Answer 3

$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}\newcommand{\H}{\mathcal{H}}$For a commutative Banach ring $A$, e.g. a Banach Algebra over the field $\mathbb{R}$, let me write $\M(A)$ for the set of all multiplicative (for me, this includes $\abs{1} = 1$) seminorms $\abs{\blank} \from A \to \IRpos$, which are bounded by the given norm on $A$. The natural topology on this set is the coarsest topology such that for every $f \in A$, the function $\M(A) \to \IRpos$, $\abs{-} \mapsto \abs{f}$ is continuous. One can show in this generality that the space $\M(A)$ is a compact Hausdorff space (this is Theorem 1.2.1 in the monograph by Berkovich) and it is commonly referred to as the Berkovich spectrum of $A$. It was systemetically studied (though mainly in the case of non-archimedean Banach algebras) in the book Spectral Theory and Analytic Geometry over Non-Archimedean Fields by V. G. Berkovich.

Terry Tao has shown above that if $X$ is a compact Hausdorff Space, then the natural map $X \to \M(C(X))$ is a bijection. Equipping the right hand side with the above topology, it is continuous by construction and hence a homeomorphism since both sides are compact Hausdorff.

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Let $A$ be a commutative Banach ring. The above suggests to write bounded multiplicative seminorms $\abs{\blank}_x$ as points $x \in \M(A)$ and to pretend that $\abs{f}_x$ is the absolute value of the “function” $f$ at the point $x$. This can even be made precise in a Grothendieck-like manner: Given $x \in \M(A)$, the set $\mathfrak{p}_x :=\{f \in A \,|\, \abs{f}_x = 0\}$ is a closed prime ideal of $A$. Hence we can consider the quotient Banach ring $A / \mathfrak{p}_x$. The multiplicative norm extends in a well-defined way to the fraction field $\mathrm{Frac}(A / \mathfrak{p}_x)$ and we let $\H(x)$ be its completion. Then we write $f(x)$ for the image of $f$ under the natural map $A \to A / \mathfrak{p}_x \to \mathrm{Frac}(A / \mathfrak{p}_x) \to \H(x)$. By construction, $\abs{f(x)}$ is exactly $\abs{f}_x$. Note that if we started with a real Banach Algebra, then $\H(x)$ can only possibly be $\mathbb{R}$ or $\mathbb{C}$ by the Gelfand-Mazur theorem.

If $A = C(X)$, and we write $x$ both for a point $x \in X$ and for its image under $X \cong \M(C(X))$, then $\H(x) = \mathbb{R}$ and $\abs{f(x)}$ agrees with its original meaning.

In your question you also noted for $A = C(X)$ that the norm of $C(X)$ can be recovered from the norms $\abs{\blank}_x$ via $\lVert f \rVert = \rho(f) := \sup_{x \in X} \abs{f(x)}$. Note that because $\M(A)$ is compact Hausdorff, one can actually replace $\sup$ by $\max$. For a general Banach ring $A$ the seminorm $\rho(f) := \sup_{x \in \M(A)}\abs{f(x)}$ can still be expressed in terms of the original Banach norm: It can be computed via the Gelfand formula for the spectral radius: $\rho(f) = \lim_{k \to \infty}\sqrt[k]{\lVert{f^k}\rVert}$. One has $\rho = \lVert \blank \rVert$ if and only if $\lVert\blank\rVert$ was power-multiplicative.