[Math] Relationship between different topologies of bounded operators on a Hilbert space

functional-analysisgeneral-topologyhilbert-spacesoperator-theory

I am self-studying functional analysis. Given that $B(H)$ are the bounded operators on a Hilbert space, $H$.

I would like to ask how to formally prove that

  1. the weak topology is weaker than the strong topology,
  2. the strong topology is weaker than the strong-* topology,
  3. the strong-$*$ topology is weaker than the norm-topology.

We would also like to prove that "weaker" can be replaced by "strictly weaker" by providing sequences that converge in the weaker topology but not in a stronger topology. Finally, if the weak and norm topologies coincide does it imply $B(H)$ is finite dimensional ?

My idea for a proof is as follows : We define a set of semi-norms for the different topologies and show that convergence in a stronger topology implies convergence in the weaker topologies. But I am unsure about how to prove strictly weaker. Also, I am being unable to form an intuitive picture of what it means for two topologies to coincide.

Any help would appreciated and I would like to thank you for helping me.

Best Answer

The most usual way to show that a topology $A$ is stronger than a topology $B$ is to show that a convergent net in $A$ is also convergent in $B$. You show that the two topologies are equal by showing that both are stronger than the other.

Since all the topologies you want to consider are linear, it is enough to consider convergence at $0$.

Here, suppose that $T_j\to 0$ in norm, i.e. $\|T_j\|\to0$. For any $x\in H$, $$ \|Tx\|+\|T^*x\|\leq\|T\|\,\|x\|+\|T^*\|\,\|x\|=2\|T\|\,\|x\|\to0; $$ so norm convergence implies strong* convergence.

From $\|Tx\|\leq\|Tx\|+\|T^*x\|$ we get that strong* convergence implies strong convergence.

If $T_j\to0$ strongly, this means that $T_jx\to0$ for all $x\in H$. Then $$ |\langle T_jx,y\rangle|\leq\|T_jx\|\,\|y\|\to0, $$ so strong convergence implies weak convergence.

So far we have shown the "weaker than" implications. Now we need to see that they are strict.

Fix an orthonormal basis $\{e_j\}$ of $H$ and let $E_j$ be the projection onto the span of $e_j$ (i.e. $E_jx=\langle x,e_j\rangle e_j$). Then for any $x=\sum_j\alpha_je_j$, $$ \|E_jx\|+\|E_j^*x\|=2\|E_jx\|=|\alpha_j|\to0, $$ so $E_j\to0$ in the strong* topology. But $\|E_j\|=1$ for all $j$, so $\{E_j\}$ does not converge in norm.

Now consider the left-shift operator $T$ given by $Te_1=0$, $Te_j=e_{j-1}$ for $j>1$ (we don't need to assume $H$ separable here, just use a well-ordering of the index set). If you are familiar with it, $T$ is the adjoint of the unilateral shift. Put $T_n=T^n$, $n\in\mathbb N$. Then $$ \|T_nx\|^2=\sum_{k>n}|\alpha_k|^2\to0\ \ \text{ as }n\to\infty; $$ so $T_n\to0$ in the strong topology. But $T^*$ is an isometry, so $$ \|T_nx\|+\|T_n^*x\|\geq\|T_n^*x\|=\|x\| $$ for all $n$, so $\{T_n\}$ does not converge in the strong* topology.

Finally, we need a net that converges weakly but not strongly. Here we can use the unilaterial shift $S$ ($T^*$ above). If $x=\sum\alpha_ke_k$, $y=\sum_j\beta_je_j$, then $$ |\langle S^nx,y\rangle|=|\sum_k\alpha_k\beta_{k+n}|\leq\sum_k|\alpha_k|\,|\beta_{k+n}|\leq\left(\sum_k|\alpha_k|^2\right)^{1/2}\,\left(\sum_k|\beta_{k+n}|^2\right)^{1/2}\to0 $$ (the second sum goes to zero because of the $n$). So $S^n\to0$ weakly but not strongly (recall that $S$ is an isometry).

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