Could you please help me to find some classical (counter)examples in functional analysis? Let $X$ and $Y$ be some normed spaces over $\mathbb{C}$. By $\mathcal{B}(X,Y)$ we denote the space of bounded linear operators, which is endowed with three topologies: WOT,SOT and NT (weak operator topology, strong operator topology and the topology with the operator norm resp.). One can easily prove that WOT$\subset$SOT$\subset$NT. But how to prove that WOT$\ne$SOT when $Y$ is of infinite dimension?

The last inequlity means that there exists a sequence $\{T_n\}\subset \mathcal{B}(X,Y)$, an operator $T\in\mathcal{B}(X,Y)$ such that $f T_nx\to f Tx$ for each $f\in Y^*$ but $T_nx_0$ doesn't converge to $Tx_0$ for some $x_0\in X$.

  • $\begingroup$ Do you know of a sequence of vectors in a Hilbert space that does not converge but converges weakly? You might be able to do something with that. $\endgroup$ – DisintegratingByParts Jun 18 '14 at 23:27

Take $\ell^2(\mathbb{R})$. A sequence of vectors that converges weakly but not strongly is $\{x_i(k)\}_{i=1}^\infty$ with $x_i(k) = \delta_{ik}$, that is $$ x_1 = (1,0,0,0,\dots)\\ x_2 = (0,1,0,0,\dots)\\ x_3 = (0,0,1,0,\dots)\\ \vdots $$ In order to show that this weakly converges, note (by the Riesz representation theorem) that every functional can be expressed in the form $\sum_{i=1}^\infty a_i e_i$ where $a_i \in \mathbb{R}$ and $e_i(y) = \langle y,x_i \rangle = x(i)$.

It follows that the sequence of operators $\langle \cdot,x_i\rangle x_i$ converges in the weak, but not the strong, operator topology. In fact, we may simply state $\{e_i\}_{i=1}^\infty$ is a satisfactory sequence of operators for the same reason, so it would seem that the finiteness of $Y$ doesn't matter.

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