I am interested in operators on a non-reflexive Banach space. Let $X$ be a Banach space and let $L(X)$ be the algebra of operators acting on $X$. We may embed $L(X)$ into $L(X^{\ast\ast})$ by $\Phi(T)=T^{\ast\ast}$; this is an algebra homomorphism. Is the image $\Phi(L(X))$ dense in $L(X^{**})$ in the sense of some of the classical operator topologies like WOT, SOT etc.?
Nice question, but why do you ask about the density of the embedding into $L(X^{∗∗})$ instead of into $L(X^∗)$? – Bill Johnson Nov 1 '12 at 1:29
If $X^*$ has the bounded approximation property, then the embedding of $L(X)$ into $L(X^*)$ is strongly dense. This is an immediate consequence of the principle of local reflexivity, which you can find in text books. – Bill Johnson Nov 1 '12 at 1:48
Your original question has a negative answer for every non reflexive space. Just consider any operator in $L(X^{**})$ that maps some point in $X$ to a point not in $X$. There are plenty of rank one operators that do that. Operators in $\Phi(L(X))$ map $X$ into $X$. – Bill Johnson Nov 1 '12 at 1:56
Yes, SOT. Incidentally, in my comment about density of $L(X)$ in $L(X^{*})$, I was thinking about taking limits of uniformly bounded nets. If you really only care about SOT density, then no condition is needed. Given $T$ in $L(X^{*})$ and a finite dimensional subspace $F$ of $X^{*}$, choose a basis $(f_n)$ for $f$ and elements $(x_n)$ in $X$ s.t. $(x_n,f_n)$ is biorthogonal and consider $S:=\sum_n x_n\otimes f_n$. This finite rank operator is clearly weak$^*$ continuous and agrees with $T$ on $F$. – Bill Johnson Nov 1 '12 at 13:42