Let $X$ be a Banach space and consider $B(X)$, the set of all bounded linear maps on $X$. By the W-topology on $B(X)$ we mean the topology induced by the semi-norms 
$$B(X)\to [0,\infty): T\to |\langle Tx,x^\*\rangle|$$
where $x\in X$ and $x^\*\in X^\*$. 


The algebraic tensor product $X\otimes X^\*$ may be considered as a subspace of $B(X)$. Does the following hold?


Under the **W-topology**, $B(X)$ is the closure of $X\otimes X^\*$. 



A dense subset in $B(X)$ under the weak operator topology
I think yes, for the following general reason: If $E$ is a real vector space and $F$ a linear space of linear forms of $E$, then a linear subspace $V$ of $E$ is $\sigma(E,F)$-dense in $E$ if and only if no $f\in F\setminus\{0\}$ vanishes identically on $V$. (Equivalently, any proper $\sigma(E,F)$-closed linear subspace of $E$ is included in $\ker f$ for some $f\in F\setminus\{0\}$. This follows from the Hahn-Banach theorem, and from the fact that the $\sigma(E,F)$-continuous linear forms on $E$ are precisely the elements of $F$).


In your situation, $E:=B(X)$, $F:=X\otimes\_{alg} X^\*$, seen as a linear space of linear forms on $B(X)$ (that is, $f=\sum\_{i=1}^m x\_i\otimes x^\*\_i$ corresponds to the linear form $T\mapsto\langle T,f\rangle:= \sum\_{i=1}^m \langle Tx\_i, x^\*\_i\rangle$ on $B(X)$), and $V:=X\otimes\_{alg} X^\*$, seen as the subset of $B(X)$ of finite rank operators (that is, $T=\sum\_{j=1}^m y\_j\otimes y^\*\_j$ corresponds to the finite rank operator $u\mapsto Tu:=\sum\_{j=1}^m y\_j\langle u,y^\*\_j \rangle$ on $X$). 


Thanks to the above general fact, to prove that $V$ is $\sigma(E,F)$-dense in $E$ it is sufficient to prove: for any $f\in F\setminus\{0\}$ there exists $T\in V$ such that $\langle T, f\rangle\neq 0$.


Indeed, let $f\in F$ be a non-zero linear form. Let $f=\sum\_{i=1}^m x\_i\otimes x^\*\_i$ be a representation of $f$ with minimum $m$. This implies that $x\_1,\dots,x\_m$ are linearly independent elements of $X$ and $x\_1^\*,\dots,x\_m^\*$ are linearly independent elements of $X^\*$. Then there are $y\_1^\*,\dots,y\_m^\*$ in $X^\*$ such that $\langle x\_i,y^\*\_j\rangle=\delta\_{ij}$, and there are $y\_1,\dots,y\_m$ in $X$ such that $\langle y\_i,x^\*\_j\rangle=\delta\_{ij}$. Then the operator $T=\sum\_{j=1}^m y\_j\otimes y^\*\_j$ is such that $\langle T,f\rangle:=\sum\_{i=1}^m \langle Tx\_i, x^\*\_i\rangle=\sum\_{ij}\langle y\_j,x^\*\_i \rangle\langle x\_i,y^\*\_j \rangle=\sum\_{ij}\delta\_{ij}=m>0$, that is, no non-zero element of $F$ vanishes identically on $V$. 


