# Variations on Kaplansky Density

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 $$T(b) = a_0b$$.

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 $$k>\alpha$$).

• 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. – Matthew Daws Jun 25 '19 at 11:50
• 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 $A$). – Narutaka OZAWA Jun 25 '19 at 14:49
• @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. – mathbeginner Jul 8 at 18:50