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 $T(b) = a_0b$.
 A: 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 $k>\alpha$). 
