(adsbygoogle = window.adsbygoogle || []).push({}); Edit:I originally wrote that ##\mathcal A## is a Banach algebra. The assumption that goes into the theorem is stronger. It's a C*-algebra. I am however still mainly interested in the claim that ##\mathcal A_1##, as defined below, is a Banach sub-algebra of ##\mathcal B(\mathcal A)##.

Let ##\mathcal A## be a C*-algebra without identity. For each ##x\in\mathcal A##, define ##L_x:\mathcal A\to A## by ##L_xy=xy## for all ##y\in\mathcal A##. It's trivial to show that ##L_x\in\mathcal B(\mathcal A)## for all ##x\in\mathcal A##. The map ##x\mapsto L_x## with domain ##\mathcal A## will be denoted by L. It's easy to show that L is an isometric algebra homomorphism into ##\mathcal B(\mathcal A)##. Define ##\mathcal A_1=\{L_x+\lambda|x\in\mathcal A,\, \lambda\in\mathbb C\}##. The notation ##\mathcal A_1=L(\mathcal A)+\mathbb C## makes this definition easier to remember. ##\mathcal A_1## is closed under addition, scalar multiplication and multiplication. Supposedly* (this is what I want to prove), it's also complete. Since it's a subset of ##\mathcal B(\mathcal A)##, every Cauchy sequence is convergent. It's just not obvious that the limit of a convergent sequence in ##\mathcal A_1## is in ##\mathcal A_1##. So I want to prove that it is.

*) The claim is made by Conway in "A course in operator theory", in the proof of theorem 1.5, on page 3. http://books.google.com/books?id=gt...way operator&hl=sv&pg=PA3#v=onepage&q&f=false. His notation is slightly different from mine. (In particular, he writes λ where I write L)

Some of my thoughts: Let ##(T_n)_{n=1}^\infty## be an arbitrary convergent sequence in ##\mathcal A_1##. Let ##(x_n)## and ##(\lambda_n)## be sequences in ##\mathcal A## and ##\mathbb C## respectively, such that ##T_n=L_{x_n}+\lambda_n##. It would be nice if we could use that ##(T_n)## is Cauchy, to show that these two are Cauchy, and therefore convergent. Then we can define ##x=\lim_nx_n## and ##\lambda=\lim_n\lambda_n##, and perhaps show that ##L_{x_n}+\lambda_n\to L_x+\lambda##. But I don't see a way to proceed from

$$\varepsilon>\|T_n-T_m\|=\|(L_{x_n}+\lambda_n)-(L_{x_m}+\lambda_m)\| =\|L_{x_n-x_m}+(\lambda_n-\lambda_m)\|.$$ When we're dealing with Hilbert spaces, the usual way to continue a calculation like this would be to square both sides of the inequality and then use the pythagorean theorem (assuming that the terms are orthogonal), but I don't see what to do here.

It looks like (I haven't worked through that part of the proof yet) that we can prove that the C*-identity holds, without proving completeness first, when the involution is defined by ##(L_x+\lambda)^* =L_{x^*}+\bar\lambda##. I was thinking that maybe we can use that somehow, but when I tried, it just made the calculation longer, and I ran into the same issue as above.

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# Banach sub-algebra

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