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A Hilbert space $ \mathscr l^2 $ is defined as a space with the scalar product $ (x,y)=\sum_{i=1}^\infty x_iy_i$ over $ \mathbb R $, where $x_i$ and $y_i$ are sequences.

Then there is a space $L(\mathscr l^2)$, which is a space of all bounded linear operators $\mathscr l² \rightarrow \mathscr l²$. A weak operator topology is defined on this space, as an initial topology with a function that induces it: $$ w_{x,y} : L(\mathscr l²)\rightarrow \mathbb R : w_{x,y}(A)=(Ax, y), x,y \in \mathscr l²$$ It is to be shown that the multiplication of the operators in form $$ L(\mathscr l²)\times L(\mathscr l²) \rightarrow L(\mathscr l²):(A,B) \mapsto AB $$

is not continuous.

Well, I've encountered many proofs of this property, but given that we have certain translations, in this case left $ A_n(x)(k) = x(k+n) $ and right $ B_n(x)(k)=x(k-n)$ if $ k\gt n$ and 0 otherwise, I am not very sure of how I can use that in my proof.

Also, what does this given function $w_{x,y}$ do in general, as in how can I actually use this given property to show the problem with multiplication of the operators?

Milena
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  • What do you want exactly, a proof that the multiplication of operators isn't continuous? Or a particular proof of this using those sequences of operators you have defined? – Demophilus Nov 02 '17 at 01:44
  • @Demophilus I was looking for a proof of the incontinuity using the sequences of operators, because I'm not very clear how should I apply them in this example. – Milena Nov 02 '17 at 07:45

1 Answers1

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The weak operator topology is such that if $(A_i)_i$ is a net in $L(l^2)$ that converges in the weak operator topology if and only if for all $x,y \in l^2$ we have that the net $\big( (A_ix,y) \big)_i$ converges in $\mathbb{C}$.

Thus an easy way to prove that the multiplication operator isn't weak continuous, is to find a pair of nets $(A_i)_i$, $(B_i)_i$ so that both converge weakly to zero but $(A_i B_i)_i$ doesn't converge weakly to zero. Note that your $(A_n)_n$ and $(B_n)_n$ you have defined in your question precisely satisfy these requirements. Both converge weakly to zero (it is sufficient to prove this for one, since they are adjoints of each other). But $A_n$ is also the left inverse of $B_n$, i.e. $A_n B_n = \text{Id}$. So the sequence $(A_n B_n)_n$ can't possibly weakly converge to zero, and there we have our counterexample.

Demophilus
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  • Thanks! It makes a lot of sense now! The only thing that is still unclear to me, is in which way does my function $ w_{x,y}$ contribute to this part of the proof? Is it important for it? Because I can't seem to find any connection between this function and the whole continuity story. – Milena Nov 02 '17 at 11:09
  • You can prove that a net $(A_i)i$ converges to $A$ in the initial topology if and only if $(w{xy}(A_i-A))_i$ converges to zero for all $x,y \in l^2$. It's not that hard, you simply have to use the definition of initial topology. – Demophilus Nov 02 '17 at 11:15