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We will say that ${T_i}\subset B(X,Y^*)$ converges to $T$ in W*-operator topology if $T_i(x)\rightarrow T(x)$ in W*-topology of $Y^*$( $\forall y\in Y \langle T_i(x),y\rangle \rightarrow \langle T(x),y\rangle$).

Now someone has proved the below theorem. Is it true?

BEGIN

Let $X$ and $Y$ be two arbitrary Banach spaces. Then $F (X; Y^*)$(Finite rank operator) is dense in $B(X; Y^*)$ with respect to the weak* operator topology.

Proof.

Let $T \in B(X; Y^*)$ and take a finite subset $F =\{ x_1,...,x_n\}$ of X. Assume that $x_1,...,x_m$ are linearly independent for all $m\leq n$. By the Hahn Banach theorem, for each $j\in \{1,2,...,m\}$ there is $f_j\in X^*$ such that $f_j(x_j) = 1$ and $f_j(x_i) = 0$ for all $i\in\{1,2,...,m\}-\{j\}$.For each $j\in\{1,...,m\}$ define $T_j\in B(X; Y^*)$ by $T_j(x) = f_j(x)T(x_j)$.Then $$T_j(x_j) = T(x_j); T_j(x_i) = 0\ \ (i, j\in\{1,...,m\}, i\neq j)$$ Now define $T_F = T_1 +...+ T_m$. It can be easily seen that $$T_F(x_i) = T(x_i)\ \ (i\in\{1,...,n\})$$ So $T_F = T$ on the span of $F$ and $\operatorname{rank}(T_F)\leq \dim F$. Now it is obvious that the net $(T_F)_{F\in F(X)}$$\big(F(X)=$ all finite subset of $X\big)$ converges to $T$ in the weak* operator topology, as desired.

END

If it is true then my question is this that why we can't say $T_F\rightarrow T$ in strong operator topology($T_F(x)\rightarrow T(x) \ \ \forall x\in X$)?

Daniel Fischer
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Hamid Shafie Asl
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    In the first line under "Proof" I suspect you meant to say "Assume $x_1,\ldots,x_m$ are linearly independent ($m \le n$)." – hardmath Feb 26 '14 at 11:27
  • We can assume that they are linearly independent.I think there is no any problem there. – Hamid Shafie Asl Feb 26 '14 at 15:25
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    I was trying to address two matters. One was the otherwise undefined introduction of $m$ (except to say some $m \le n$). The other is to address linear independence. Recall that $F$ is a typical finite subset of $X$. Generally such sets are not always linearly independent. Moreover the plan of the proof is to consider a net $\mathscr{F}$ upon all finite sets. For both these reasons it makes to distinguish a (maximal) linearly independent subset ${x_1,\ldots,x_m}$ of $F$, noting that the subscripts can be assumed like this because a set lacks a preferred ordering. – hardmath Feb 26 '14 at 18:49
  • A similar question is answered here http://math.stackexchange.com/questions/535645/weak-operator-topology-and-finite-rank-operators – Conifold Jun 13 '14 at 00:03

1 Answers1

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Indeed we also have $T_F \to T$ in the strong operator topology.

The strong operator topology is defined by the seminorms

$$p_F(S) = \sup \{\lVert S(x)\rVert : x \in F\},$$

where $F$ traverses the finite subsets of $X$. The construction immediately yields

$$p_F(T_F - T) = 0$$

for any linearly independent finite $F\subset X$, and it is straightforward to see $p_F(T_{F'} - T) = 0$ for any finite $F \subset X$ when $F' \subset F$ is a maximal linearly independent subset of $F$. That means every neighbourhood of $T$ in the strong operator topology contains operators of finite rank, so $F(X,Y^\ast)$ is also dense in the strong operator topology.

Daniel Fischer
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