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Suppose we are given a C*-algebra $\mathcal{X}$ that has no unit. Consider the unitization $$\mathcal{X}_1 \equiv \{x + \lambda \mid x \in \mathcal{X}, \lambda \in \mathbb{C}\}$$ of $\mathcal{X}$, endowed with the natural operations. Define a norm on $\mathcal{X}_1$ by $\|x+\lambda\| \equiv \sup \{ \|xa + \lambda a\| \colon a \in \mathcal{X}, \|a\| \leq 1\}$. How can one prove that $\mathcal{X}_1$ is complete with respect to this new norm?

Take a Cauchy sequence in $\mathcal{X}_1$, say $\{x_n + \lambda_n\}_n$. Then $\forall \epsilon>0 \exists N(\epsilon)$ such that for all $n,m \geq N(\epsilon)$ and for all $a \in \mathcal{X}, \|a\| \leq 1$ we have $\|(x_n-x_m)a + (\lambda_n-\lambda_m)a\| < \epsilon$. From that we know $\{x_n a + \lambda_n a\}_n$ is Cauchy in $\mathcal{X}$ for all $a \in \mathcal{X}, \|a\| \leq 1$. Hence for each $a \in \mathcal{X}, \|a\| \leq 1$ there exists $y(a) \in \mathcal{X}$ with $x_n a + \lambda_n a \to y(a)$ by completeness of $\mathcal{X}$. At this point I am stuck, since I don't quite see how to get an $\epsilon$-bound for $\|x_n + \lambda_n - y\|$.

Can anyone give hints on how to proceed?

h3fr43nd
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    Your bigger problem is that you didn't define a single element $y$, but a family of elements $y(a)$ for $|a|\le 1$. Can we maybe prove $|\lambda|,,|x|\le |x+\lambda|$? – Berci Mar 22 '20 at 21:19
  • That makes sense, yet I don't quite see how to prove that inequality! I think my question is flawed though, shouldn't I at least assume, that $\mathcal{X}$ is a C*-algebra? – h3fr43nd Mar 23 '20 at 08:14
  • But, can we even prove that the new norm extends the original one, $|x+0|=|x|$? – Berci Mar 23 '20 at 09:39
  • Yeah but that only holds for C-algebras, right? Since, clearly $|xa| \leq |x|$ for all $|a| \leq 1$, and therefore $|x+0| \leq |x|$. On the other hand, if $\mathcal{X}$ were a C algebra, we could define $a=\frac{x^}{|x|}$ and then by the identity $|x^x|=|x|^2$ in C* algebras we deduce $|xa|=|x|$, so $|x+0|=|x|$. Maybe I am missing something, but I think without the C* property this won't work in general – h3fr43nd Mar 23 '20 at 09:51
  • It's from Conway's functional analysis book (page 233, Prop 1.9) and in there the completeness of this unitization of a C*-algebra is left as an exercise to the reader. I thought for the completeness part it is enough to have a Banach algebra, but that's false I think. – h3fr43nd Mar 23 '20 at 10:05
  • I added an answer. Have a look. – J. De Ro Mar 23 '20 at 10:06
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    I don't think you really need the $C$-structure for this question, but you definitely need the $$-structure. The $C$-structure becomes necessary when you want to prove that $\mathcal{\chi}_1$ is itself a $C$-algebra, which is true (I verifies this a couple of days ago). – J. De Ro Mar 23 '20 at 10:08
  • Nice thanks for your help! – h3fr43nd Mar 23 '20 at 10:10
  • I made a mistake. My answer does use the $C^*$-structure. – J. De Ro Mar 23 '20 at 10:13
  • Do you see why the question asks that $\mathcal{X}$ cannot have a unit? I.e. what goes wrong when $\mathcal{X}$ has a unit? – J. De Ro Mar 23 '20 at 10:22
  • I think the problem is that, if you consider the element $x = -1_\mathcal{X} + 1 \in \mathcal{X}_1$, then we would have $|x| = 0$, yet $x \neq 0$. – h3fr43nd Mar 23 '20 at 10:54
  • So, did you check that $\Vert x \Vert = 0 \implies x = 0$ for your norm? This is exactly where things go wrong if we have a unit. – J. De Ro Mar 23 '20 at 11:27
  • If we assume our C*-algebra has no unit, then this defines a norm. If we have a unit however, then as I said before, we have no norm. Or am I mistaken? – h3fr43nd Mar 23 '20 at 11:28
  • Yes, but did you explicitely check that $\Vert x \Vert = 0 \implies x = 0$ in the case we have no unit? This is part of the question to verify. – J. De Ro Mar 23 '20 at 11:30
  • Ah i see, thanks for pointing that out. Suppose $|x+\lambda| = 0$. In the case, where either $x, \lambda = 0$, the statement is obvious (since we have an isometry). So assume $x, \lambda \neq 0$. If $|x+\lambda| = 0$, then for all $|a| \leq 1$ we have $x a + \lambda a = (x+\lambda)a = 0$. Hence $-\lambda^{-1} x a = a$ for all $a \in \mathcal{X}$ (by scaling), so we have a left unit in $\mathcal{X}$, namely $-\lambda^{-1} x$. Since $\mathcal{X}$ is a C*-algebra, this also yields a unit, contradiction. Right? – h3fr43nd Mar 23 '20 at 11:55
  • Yes, that's definitely the idea! In a *-algebra, if you have a left unit you also have a right unit and thus a unit. Best of luck further! (If you have a related question about Banach algebra's and related stuff, you can ask me in this comment section to have a look at your question!). – J. De Ro Mar 23 '20 at 12:19
  • Thanks a lot! I will do that :) – h3fr43nd Mar 23 '20 at 13:59
  • @ε-δ Maybe you could have a look: https://math.stackexchange.com/questions/3597846/one-point-compactification-of-maximal-ideal-space – h3fr43nd Mar 27 '20 at 17:46

1 Answers1

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Note that $\mathcal{X}_1 =\mathcal{X}\oplus \mathbb{C}$ as $\mathbb{C}$-vector spaces.

We also have an isometric inclusion:

$$i: \mathcal{X} \to \mathcal{X}_1: a \mapsto a+0$$

Indeed, for $a \neq 0$

$$\Vert L_a\Vert = \sup_{\Vert x \Vert \leq 1} \Vert a x \Vert \leq \Vert a\Vert = \frac{\Vert aa^*\Vert}{\Vert a \Vert} = \frac{\Vert L_a a^* \Vert}{\Vert a \Vert} \leq \Vert L_a \Vert \frac{\Vert a^*\Vert}{\Vert a \Vert }= \Vert L_a \Vert$$

and thus $\Vert i(a) \Vert = \Vert L_a \Vert = \Vert a \Vert$. Note that the $C^*$-structure was used to justify the second equality above.

Thus, we see that $\mathcal{X_1}/i(\mathcal{X}) \cong \mathbb{C}$ is a finite-dimensional vector space.

Now, use the following result to conclude:

If $V$ is a normed space and $W$ is a linear subspace of $V$ that is complete (for the norm induced by $V$) and the quotient space $V/W$ with the natural quotient norm is finitedimensional, then $V$ is a Banach space.

J. De Ro
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  • By "2 out of 3" I meant, that the statement you use for concluding essentially tells you that if $W$ and $V/W$ are complete, then so is $V$. Similarily, if $V$ and $W$ are complete, then so is $V/W$. Therefore 2 out of 3. – h3fr43nd Mar 23 '20 at 10:26
  • Ah yes. That makes sense. – J. De Ro Mar 23 '20 at 11:26