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.?

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`$L(X^{∗∗})$`

instead of into`$L(X^∗)$`

? – Bill Johnson Nov 1 '12 at 1:29`$X^*$`

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`$L(X^{**})$`

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 1:56stronglymeans in the sense of SOT? – Slavoj Žižek Nov 1 '12 at 11:49`$L(X^{*})$`

, 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