Closedness of a set of operators defined using a set in Hilbert space Consider some separable Hilbert space $\mathcal{H}$ and let $T(\mathcal{H})$ be the space of trace-class operators on $\mathcal{H}$. Given a set $S \subset \mathcal{H}$ such that $\|x\| = 1$ $\forall x \in S$, define a set $S' \subset T(\mathcal{H})$ as
$$ S' = \{ \langle x, \cdot \rangle y : x, y \in  S \}. $$
I will assume for simplicity that $a x \in S$ for any $x \in S$ and $|a|=1$. I have two related question about sets of this type.
(1) If $S$ is closed (in the weak topology), then is $S'$ also closed (in the weak operator topology)?
(2) Is the convex hull defined as
$$ S'' = \mathrm{co} \{ \langle x, \cdot \rangle y : x, y \in \overline{\mathrm{co}} (S) \}$$
closed? Here, $\overline{\mathrm{co}}$ refers to the (weakly) closed convex hull.

I believe that (1) is true, and my reasoning is as follows. Consider any sequence of trace-class operators $(T_n) \in S'$ and write $T_n(\cdot) = \langle x_n, \cdot\rangle y_n$ for some $x_n, y_n \in S$. I will assume that this sequence converges weakly to some $T$ defined by $T(\cdot) =  \langle x, \cdot \rangle y$, which means that
$$ \lim_{n\to\infty} | \langle v, \langle x, w \rangle y - \langle x_n, w \rangle y_n \rangle | = 0 \; \forall v, w \in \mathcal{H}.$$ 
Since $x$ and $y$ can be completed to an orthonormal basis for the Hilbert space, for any $x_n, y_n$ we can write $x_n = \alpha_n x + x_\perp$ and $y_n = \beta_n y + y_\perp$. Choosing $v = y$ and then $w = x$ shows that this is only possible when $(x_n) \to x$ and $(y_n) \to y$ weakly, but since $S$ is closed we get $T \in S'$. The only thing I'm not sure here is if I can assume that $T$ is of the form $T(\cdot) =  \langle x, \cdot \rangle y$ or if some further justification is necessary.
As for (2), I know that convex hulls of closed sets are not generally closed, but I am not sure if the particular structure of the set in question makes this property true, or if there is a counterexample. I don't know how to approach this part, and I would appreciate any help with this problem.
 A: A limit of rank-one operators is either zero or rank-one. But I don't immediately see how you can guarantee from $\alpha_n\beta_n\to1$ that $\alpha_n\to1$ and $\beta_n\to1$. 
Here is a slightly different approach. Note that you can use sequences because both topologies are metrizable on bounded sets. Once you have the $T_n$, you have the two sequences $\{x_n\}$ and $\{y_n\}$, both made up of unit vectors. As the unit ball is weakly compact, there exist convergent subsequences $\{x_{n_j}\}$ and $\{y_{n_j}\}$, say $x_{n_j}\to x'$ and $y_{n_j}\to y'$. It follows that $T_{n_j}\to \langle x',\cdot\rangle\,y'$. But as the original sequence $\{T_n\}$ is convergent, $T=\langle x',\cdot\rangle\,y'$ and $x',y'\in \mathcal S$ since it's closed. 
The above approach does not require any assumption on $T$, as it proves that it is of the required form. 
As for question 2, the answer is no. Your $S''$ is made out of finite-rank operators, and in general it is not even norm closed. Take $S=\{x:\ \|x\|=1\}$. Fix an orthonormal basis $\{x_n\}$. Then $$T=\sum_n2^{-n}\langle \ \cdot\ ,x_n\rangle\,x_n$$ is in the norm closure of $S''$.
