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Timeline for Density of adjoint operators

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Nov 1, 2012 at 14:54 comment added Bill Johnson Do you mean SOT limits but with the operators uniformly bounded? That is what I meant. Uniform limits usually mean limits in the operator norm, so that density is equivalent to the reflexivity of $X$.
Nov 1, 2012 at 13:58 comment added Slavoj Žižek That's interesting if it can be done any better than SOT. Are there any reasonable assumptions for $X$ to consider uniform limits?
Nov 1, 2012 at 13:42 comment added Bill Johnson 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 $F$.
Nov 1, 2012 at 11:49 comment added Slavoj Žižek Thank you, Professor. Just to clarify, strongly means in the sense of SOT?
Nov 1, 2012 at 1:56 comment added Bill Johnson 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 $X$.
Nov 1, 2012 at 1:48 comment added Bill Johnson 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.
Nov 1, 2012 at 1:29 comment added Bill Johnson Nice question, but why do you ask about the density of the embedding into $L(X^{∗∗})$ instead of into $L(X^∗)$?
Nov 1, 2012 at 0:07 history asked Slavoj Žižek CC BY-SA 3.0