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$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}\newcommand{\H}{\mathcal{H}}$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.

Let $A$ be a commutative Banach ring. The above suggests to write elements $\abs{\blank}_x$ 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 considered (for $A = C(X)$) the maximal power-multiplicative seminorm $\rho(f) := \sup_{x \in \M(A)} \abs{f(x)}$. Note that because $\M(A)$ is compact Hausdorff, one can actually replace $\sup$ by $\max$. One can show for a general Banach ring $A$ that $\rho(x)$ can be defined intrinsically through $(A,\lVert{\blank}\rVert)$ via the Gelfand formula for the spectral radius: $\rho(f) = \lim_{k \to \infty} \sqrt[k]{\lVert{f^k}\rVert}$.

$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}\newcommand{\H}{\mathcal{H}}$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.

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.

$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}\newcommand{\H}{\mathcal{H}}$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.

Let $A$ be a commutative Banach ring. The above suggests to write elements $\abs{\blank}_x$ 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 considered (for $A = C(X)$) the maximal power-multiplicative seminorm $\rho(f) := \sup_{x \in \M(A)} \abs{f(x)}$. Note that because $\M(A)$ is compact Hausdorff, one can actually replace $\sup$ by $\max$. One can show for a general Banach ring $A$ that $\rho(x)$ can be defined intrinsically through $(A,\lVert{\blank}\rVert)$ via the Gelfand formula for the spectral radius: $\rho(f) = \lim_{k \to \infty} \sqrt[k]{\lVert{f^k}\rVert}$.

$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}\newcommand{\H}{\mathcal{H}}$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.

Let $A$ be a commutative Banach ring. The above suggests to write elements $\abs{\blank}_x$ 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 considered (for $A = C(X)$) the maximal power-multiplicative seminorm $\rho(f) := \sup_{x \in \M(A)} \abs{f(x)}$. Note that because $\M(A)$ is compact Hausdorff, one can actually replace $\sup$ by $\max$. One can show for a general Banach ring $A$ that $\rho(x)$ can be defined intrinsically through $(A,\lVert{\blank}\rVert)$ via the Gelfand formula for the spectral radius: $\rho(f) = \lim_{k \to \infty} \sqrt[k]{\lVert{f^k}\rVert}$.

$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}$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.

$\newcommand{\M}{\mathcal{M}}\newcommand{\abs}[1]{\lvert #1 \rvert}\newcommand{\blank}{{-}}\newcommand{\from}{\colon}\newcommand{\IRpos}{\mathbb{R}_{\ge 0}}\newcommand{\H}{\mathcal{H}}$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.

Let $A$ be a commutative Banach ring. The above suggests to write elements $\abs{\blank}_x$ 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 considered (for $A = C(X)$) the maximal power-multiplicative seminorm $\rho(f) := \sup_{x \in \M(A)} \abs{f(x)}$. Note that because $\M(A)$ is compact Hausdorff, one can actually replace $\sup$ by $\max$. One can show for a general Banach ring $A$ that $\rho(x)$ can be defined intrinsically through $(A,\lVert{\blank}\rVert)$ via the Gelfand formula for the spectral radius: $\rho(f) = \lim_{k \to \infty} \sqrt[k]{\lVert{f^k}\rVert}$.