5
$\begingroup$

Let $A$ be a $C^*$-algebra and $\pi:A\rightarrow B(H)$ a faithful $*$-representation, so $M=\pi(A)''$ is a von Neumann algebra and $A\rightarrow M$ is an inclusion. von Neumann's Bicommutant Theorem tells us that $A=\pi(A)$ is weak$^*$-dense in $M$, and the Kaplansky Density Theorem says that further, the unit ball of $A$ is weak$^*$-dense in that of $M$.

Suppose now I have $a\in A$ and $k\geq 0$ fixed, and there is $x\in M$ with $$ \|a-x\|\leq k, \quad \|x\|\leq 1. $$

Is $x$ is the weak$^*$-closure of the set $\{ b\in A : \|a-b\|\leq k, \|b\|\leq 1 \}$?

If $a=0$ this is just Kaplansky density.

Let's weaken this, and just ask: is there $b\in A$ with $\|a-b\|\leq k$ and $\|b\|\leq 1$? This follows from just the triangle-inequality, because it is easy to see that $$ \inf\{k\geq 0 : \exists b, \|a-b\|\leq k, \|b\|\leq 1\} = \max(0, \|a\|-1). $$ So if $\|a-x\|\leq k$ and $\|x\|\leq 1$ then $\|a\|\leq k+1$ and so $\|a\|-1\leq k$.

So, let's make this new problem harder. Let $a_0\in A^+$ (so $a_0$ is positive: this is motivated by other considerations) and ask the following:

Suppose there is $x\in M$ with $\|a-a_0x\|\leq k$ and $\|x\|\leq 1$. Is there $b\in A$ with $\|a-a_0b\|\leq k$ and $\|b\|\leq 1$?

One could also consider more general maps $T:A\rightarrow A$ which extend to $M\rightarrow M$; here $T(b) = a_0b$.

$\endgroup$
4
$\begingroup$

The answer to the first question is YES (and probably the second as well ).

First of all, we may replace the original representation $\pi(A)\subset B(H)$ with a universal representation $\pi\oplus\sigma$, by replacing $x \in \pi(A)''$ with $x\oplus \frac{1}{k+1}\sigma(a) \in A^{**}$. Then, Kaplansky's density theorem is upgraded to the following (which is an easy consequence of the Hahn--Banach separation theorem).

Lemma 1: Let $z \in A^{**}$, $w \in A$, and a net $(z_i)_i$ in $A$. If $\| z - w \| \le 1$ and $z_i \to z$ weak*, then $$\lim_j\mathrm{dist}(w,\mathrm{conv}\{ z_i : i \geq j\})\le1$$

The advantage of using convex combination is that it can be iterated without destroying the previously obtained approximation estimate. From Lemma 1, one immediately obtains

Lemma 2: Let $x\in A^{**}$ and $a\in A$ be such that $\|x\|\le1$ and $\| x - a \|\le k$. Then for any $\epsilon_1>0$, the element $x$ is weak*-approximated by $y_1\in A$ such that $\|y_1\|\le 1+\epsilon_1$ and $\| y_1 - a \|\le k+\epsilon_1$.

We are done once we show the approximant $y_1$ in Lemma 2 is norm-close to an element that satisfies the exact norm inequalities:

Lemma 3: Let $y_1 \in A$ and $a \in A$ be such that $\|y_1\|\le1+\epsilon_1$, $\|y_1-a\|\le k+\epsilon_1$, and $\|a\|\le k+1$. Then, there is $y\in A$ such that $\|y\|\le1$, $\|y-a\|\le k$, and $\|y-y_1\|\approx_{\epsilon_1}0$. Here $\approx_{\epsilon_1}$ means that the difference is at most $h(\epsilon_1)$ for some explicit continuous function $h\geq0$ such that $h(0)=0$.

Now Lemma 3 is proved by iterating the following approximate version and finding a suitable convergence sequence $(y_n)_n$:

Lemma 4: Let $y_1 \in A$ and $a \in A$ be such that $\|y_1\|\le1+\epsilon_1$, $\|y_1-a\|\le k+\epsilon_1$, and $\|a\|\le k+1$. Then, for any $\epsilon_2>0$, there is $y_2\in A$ such that $\|y_2\|\le1+\epsilon_2$, $\|y_2-a\|\le k+\epsilon_2$, and $\|y_2-y_1\|\approx_{\epsilon_1}0$.

Proof of Lemma 4: By Lemma 1, it suffices to find $y_2$ in $A^{**}$ (as opposed to in $A$). Put $\alpha=\beta=(2\epsilon_1)^{1/2}\approx_{\epsilon_1}0$. Let $a=v|a|$ be the polar decomposition,
$p:=1_{[k+1-\alpha,k+1]}(|a|)$, and $q:=vpv^*$. Since $ap\approx_{\epsilon_1}(k+1)vp$, $\|y_1p\|\approx_{\epsilon_1}1$, $\| y_1p - ap \|\approx_{\epsilon_1}k$, and $vp$ is a partial isometry, one has $y_1p \approx_{\epsilon_1}vp$ and $y_1p \approx_{\epsilon_1} qy_1$. Thus for $a':=ap^\perp=q^\perp a p^\perp$ (which has $\|a'\|\le k+1-\alpha$) and $$y_2:= qvp + q^\perp((1-\beta)\frac{y_1}{\|y_1\|}+\beta\frac{a'}{\|a'\|})p^\perp$$ one has $\| y_2 \|\le 1$ and $y_2\approx_{\epsilon_1}y_1$. Moreover, since \begin{align*} \|q^\perp(y_2-a)p^\perp\|&\le\|y_1-\frac{y_1}{\|y_1\|}\|+\|(1-\beta)q^\perp y_1p^\perp+\frac{\beta}{\|a'\|}a' - a'\|\\ &\le \epsilon_1+(1-\beta)(k+\epsilon_1)+\beta(\|a'\|-1)\\ &\le k+2\epsilon_1-\alpha\beta = k, \end{align*} one has $\|y_2-a\|\le k$ (assuming $k>\alpha$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great! So the strategy is to use the trick to move to the bidual $A^{**}$, apply Hahn-Banach to get within $\epsilon>0$, and then some functional calculus arguments. BTW I edited to make it $\|a\| \leq 1+k$ in Lemmas 3 and 4. $\endgroup$ – Matthew Daws Jun 25 '19 at 11:50
  • 2
    $\begingroup$ I feel the answer to the second question is NO; probably the solution $x$ to the norm inequality may not extend from $\pi(A)''$ to $A^{**}$ (which is necessarily if there is a solution in $A$). $\endgroup$ – Narutaka OZAWA Jun 25 '19 at 14:49
  • $\begingroup$ @Narutaka OZAWA, would you mind introducing some topics about $C^*$-algebras and unsolved questions. I am very upset and I don't know what topic to choose when I am ready to write a paper. $\endgroup$ – mathbeginner Jul 8 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.