Recent content by HeinzBor

I realized one can show boundedness of $$\rho$$ by showing that $$||\rho(x)||_{\infty} = max \{ ||L_{x}||_{\infty}, \lambda \},$$ since both $$|\lambda| \ and \ ||L_{x}||_{\infty}$$ are finite

So According to the definition which I found $$||\rho(x)||_{\infty} $$ should be the maximum value of $$1: L_{x} + |0|$$ vs $$2: |0| + |\lambda|$$. So $$||\rho(x)||_{\infty} = |\lambda| \ or ||\rho(x)||_{\infty} = |L_{x}| $$ depending on whether lambda or L_x is the largest in terms of...

In a previous exercise I have shown that for a $$C^{*} algebra \ \mathcal{A}$$ which may or may not have a unit the map $$L_{x} : \mathcal{A} \rightarrow \mathcal{A}, \ L_{x}(y)=xy$$ is bounded. I.e. $$||L_{x}||_{\infty} \leq ||x||_{1}$$, $$x=(a, \lambda) \in \mathcal{\hat{A}} = \mathcal{A}...

Yes starting with the vector space structure axiom, I wasn't sure I could use it in that way in this setting.. Thanks again! I will try to go through that it has been too long ago since I took an algebra course!

For the last one I end up here. $$(b, \beta)^{*} (a, \alpha)^{*} = (b^{*}, \overline{\beta})(a^{*}, \overline{\alpha}) = (b^{*} a^{*} + \overline{\beta}a^{*} + \overline{\alpha}b^{*}, \overline{\beta} \overline{\alpha}) = (a^{*} B^{*} +...

big time , fixed now

Alright now it is clear! I was stuck because I didn't know that I was allowed to set $$(\mu a + vb)^{*}) = (\overline{\mu} a^{*} + \overline{v} b^{*})$$. (which should just be applying the involution) But from your last step it is just one time applications of vector space structure, pull out...

Let $$\mathcal{A}$$ be a $C*$-algebra which may or may not have a unit with norm $$||.||$$, and put $$\mathcal{\overline{A}} = \mathcal{A} \oplus \mathbb{C}$$ as a vector space with mupltiplication: $$(a, \lambda) (b, \mu) = (ab + \lambda b + \mu a , \lambda \mu)$$, $$(a, \lambda)^{*} =...

Okay so for linearity: $$ ((f + \lambda g) \circ h)(x) = f((h(x)) + \lambda g (h(x))) = (f \circ h + \lambda g \circ h)(x) $$ Multiplicativity: $$ (f \circ g) ( g \circ h) (x) = f(g \circ h)(x) = ((fg) \circ h) (x). $$ To show the mapping...

Ahhh yes ! of course.. $$X \rightarrow V$$ can be given the structure of a vector space over $$\mathbb{F}$$, if for any $$f,g : X \rightarrow V$$, $$x \in X$$, $$\lambda \in \mathbb{F}$$ $$(f+g)(x) = f(x) + g(x)$$ and $$(\lambda f)(x) = \lambda f(x)$$ and composition is associative..

Yea I assume I can work with the axioms of a vector space since the functions belong to C(X) and C(Y). However, I haven't had an algebra course for so long, so I am a bit confused on how to work with those definitions, when I have the composition instead of the usual product showing up. Am I...

If ##X## and ##Y## are homeomorphic compact Hausdorff spaces, then ##C(X)## and ##C(Y)## are ##star##-isomorphic unital ##C^{*}##-algebras. So I got the following map to work with (AND RECALL THAT ##C(X)## and ##C(Y)## are vector spaces). $$C(h) : C(Y) \rightarrow C(X) \ : \ f \mapsto f \circ...

ahh yes of course I see it now thanks a lot!

I think it should be fixed now

Thanks a lot for your detailed explanation. Okay so what I thought is the following, following your hint: $$0 \leq ||T \oplus S (h \oplus k)||^{2} = ||T(h)||^{2} + ||S(k)||^{2}\\ \leq || T ||^{2} || h ||^{2} + || S ||^{2} || k ||^{2} \ (using \ boundedness \ of \ T,S)$$ $$\leq max \{...