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Andromeda
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Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $B$\operatorname{End}_{\mathbb{C}}(K)$ automatically contractive?

Let $A$ be a unital $C^*$-algebra and let $K$ be an inner product space (not necessarily complete!). Let $\pi: A \to B(K)$$\pi: A \to \operatorname{End}_{\mathbb{C}}(K)$ be a unital algebra homomorphism such that $$\langle \pi(a)\xi, \eta\rangle = \langle \xi, \pi(a^*)\eta\rangle$$ for all $a \in A$ (i.e. the adjoint of $\pi(a)$ exists and equals $\pi(a^*)$). Is it true that $\|\pi(a)\|\le \|a\|$ for all $a \in A$? If $K$ is a Hilbert space, this result is well-known. However, since $K$ is no longer complete, $B(K)$ is not Banach and in particular not a $C^*$-algebra. Does the result remain true?

I'm mainly interested in knowing the answer for the $C^*$-algebra $A= \ell^\infty\prod_{i \in I} M_{n_i}(\mathbb{C})$.

Thanks for your help!

Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $B(K)$ automatically contractive?

Let $A$ be a unital $C^*$-algebra and let $K$ be an inner product space (not necessarily complete!). Let $\pi: A \to B(K)$ be a unital algebra homomorphism such that $$\langle \pi(a)\xi, \eta\rangle = \langle \xi, \pi(a^*)\eta\rangle$$ for all $a \in A$ (i.e. the adjoint of $\pi(a)$ exists and equals $\pi(a^*)$). Is it true that $\|\pi(a)\|\le \|a\|$ for all $a \in A$? If $K$ is a Hilbert space, this result is well-known. However, since $K$ is no longer complete, $B(K)$ is not Banach and in particular not a $C^*$-algebra. Does the result remain true?

I'm mainly interested in knowing the answer for the $C^*$-algebra $A= \ell^\infty\prod_{i \in I} M_{n_i}(\mathbb{C})$.

Thanks for your help!

Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $\operatorname{End}_{\mathbb{C}}(K)$ automatically contractive?

Let $A$ be a unital $C^*$-algebra and let $K$ be an inner product space (not necessarily complete!). Let $\pi: A \to \operatorname{End}_{\mathbb{C}}(K)$ be a unital algebra homomorphism such that $$\langle \pi(a)\xi, \eta\rangle = \langle \xi, \pi(a^*)\eta\rangle$$ for all $a \in A$ (i.e. the adjoint of $\pi(a)$ exists and equals $\pi(a^*)$). Is it true that $\|\pi(a)\|\le \|a\|$ for all $a \in A$? If $K$ is a Hilbert space, this result is well-known. However, since $K$ is no longer complete, $B(K)$ is not Banach and in particular not a $C^*$-algebra. Does the result remain true?

I'm mainly interested in knowing the answer for the $C^*$-algebra $A= \ell^\infty\prod_{i \in I} M_{n_i}(\mathbb{C})$.

Thanks for your help!

Source Link
Andromeda
  • 175
  • 2
  • 17

Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $B(K)$ automatically contractive?

Let $A$ be a unital $C^*$-algebra and let $K$ be an inner product space (not necessarily complete!). Let $\pi: A \to B(K)$ be a unital algebra homomorphism such that $$\langle \pi(a)\xi, \eta\rangle = \langle \xi, \pi(a^*)\eta\rangle$$ for all $a \in A$ (i.e. the adjoint of $\pi(a)$ exists and equals $\pi(a^*)$). Is it true that $\|\pi(a)\|\le \|a\|$ for all $a \in A$? If $K$ is a Hilbert space, this result is well-known. However, since $K$ is no longer complete, $B(K)$ is not Banach and in particular not a $C^*$-algebra. Does the result remain true?

I'm mainly interested in knowing the answer for the $C^*$-algebra $A= \ell^\infty\prod_{i \in I} M_{n_i}(\mathbb{C})$.

Thanks for your help!