Recover a norm-closed ideal by a norm-closed subset of positive elements in a $C^*$-Algebra The following questions is from Exercise $4$, $\S 46$ in A Course in Operator Theory written by J. Conway:

Given a von Neumann Algebra $\mathcal{A}\subseteq\mathbb{B}(H)$, let $\mathcal{A}_{+}$ denote the set of its positive elements. Given a norm closed subset $\mathcal{I}_0 \subseteq \mathcal{A}_{+}$, show that there is a norm closed ideal $\mathcal{I}$ such that $\mathcal{I}_{+} = \mathcal{I}_0$ iff the following conditions hold: 1. $\mathcal{I}_0$ is a convex cone. 2. whenever $A\in\mathcal{I}_0$ and $T\in\mathcal{A}_{+}$ such that $T\leq A$, then $T\in\mathcal{I}_0$. 3. $UAU^*\in\mathcal{I}_0$ whenever $A\in\mathcal{I}_0$ and $U$ is a unitary in $\mathcal{A}$.

$\mathcal{I}$ must be the (norm-closed) ideal generated by $\mathcal{I}_0$, i.e. $\mathcal{I} = \overline{\mathcal{A}\mathcal{I}_0\mathcal{A}}$, and hence one direction is obvious. One of my attempts is trying to show $UAV\in\mathcal{I}_0$ whenever $A\in\mathcal{I}_0$ and both $U, V$ are unitaries of $\mathcal{A}$. This is because I can approximate $\mathcal{I}$ by linear combination of unitaries but I failed. Knowing what positive elements $\mathcal{I}_0$ contains will help a lot  (for instance, it would be great if I know $\mathcal{I}_0$ contains even one invertible element), but I mainly do not know how to use the third condition.
 A: $\def\cI{\mathcal I}$ $\def\cA{\mathcal A}$
(this is an exercise about C$^*$-algebras; the fact that Conway put it in a section about homomorphisms and ideals of von Neumann algebras is puzzling, and I have to assume that he had another approach in mind. The argument below does not require that $\cA$ is a von Neumann algebra, only that it is a C$^*$-algebra)
We will need the (crude) inequality
$$\tag1
X^*Y+Y^*X\leq X^*X+Y ^*Y.
$$
This follows straightforwardly from expanding $0\leq (X-Y)^*(X-Y)$ and removing the $1/2$. We also need to note that
$$\tag2
B\in\cI_0\implies B^2\in\cI_0.
$$
This follows from $B^2\leq\|B\|\,B\in\cI_0$.
$\implies$) We have that $\cI_+=\cI_0$. Then positive linear combinations of positive elements is positive. Ideals are hereditary, which gives 2; this follows from the fact that if $0\leq T\leq A$, then there exists a contraction  $C$ such that $T^{1/2}=CA^{1/2}$, and this gives $T=CAC^*\in\cI_+$. If $A\in\cI_+$ then $UAU^*\geq0$ and $UAU^*\in\cI$ since $\cI$ is an ideal.
$\impliedby$) Let $\cI=\overline{\cA\cI_0\cA}$. If $B\in\cI_0$ and $U\in\cA$ is a unitary, there exists $B'\in\cI_0$ with $UBU^*=B'$; we may write this as $UB=B'U$. As $\cA$ is the span of its unitaries, we get that
$$
\cA\cI_0=\cI_0\cA=\cA\cI_0\cA.
$$
Let $Y\in\cA\cI_0$. We can write $Y=\sum_jA_jB_j$, with $A_1,\ldots,A_m\in\cA$ and $B_1,\ldots,B_m\in \cI_0$. Then, using $(1)$ and $(2)$,
\begin{align}
YY^*&=\sum_{k,j}B_kA_k^*A_jB_j=\sum_{k}B_kA_k^*A_kB_k
+\sum_{k<j}B_kA_k^*A_jB_j+B_jA_j^*A_kB_k\\[0.3cm]
&\leq\sum_{k}B_kA_k^*A_kB_k
+\sum_{k}B_kA_k^*A_kB_k+\sum_jB_jA_j^*A_jB_j\\[0.3cm]
&\leq3\sum_{k}B_kA_k^*A_kB_k\leq3\sum_{k}\|A_k^*A_k\|\,B_k^2\in\cI_0.
\end{align}
Hence $YY^*\in\cI_0$.
Now suppose that $X\in\cI_+$. Since $\cI$ is a C$^*$-algebra, $X^{1/2}\in \cI$. Then the exists a sequence $\{Y_n\}\subset\cA\cI_0$ with $Y_n\to X^{1/2}$. Then
$$
X=\lim_nY_nY_n^*\in\overline{\cI_0}=\cI_0.
$$
