Vrftsjtryk

I'm looking at the topologies on the set of bounded operators between Banach spaces, $L(X,Y)$, and while I see how uniform convergence implies strong convergence implies weak convergence, I'm struggling to see why we have that the weak operator topology is weaker than the strong operator topology which in turn is weaker than the uniform operator topology:
$$
tau_{rm weak}subset tau_{rm strong}subsettau_{rm uniform}
$$

The best case to see why these topologies differ is in Hilbert space. Consider the case that $X=Y=ell^2$, and let ${e_n}_{ninmathbb N}$ be an orthonormal basis for $ell^2$.

Let $P_nin L(ell^2)$ be the orthogonal projection onto $operatorname{span}{e_1,ldots,e_n}$. Then ${P_n}$ converges strongly to the identity operator $I$, but $|P_n-I|=1$ for all $n$. This shows that the strong operator topology is strictly weaker than the norm topology.

To show that the weak operator topology is strictly weaker than the strong operator topology, it suffices to show that the adjoint operation $Tmapsto T^*$ is continuous in the weak operator topology but not in the strong operator topology.

Showing that the adjoint operation is continuous in the weak operator topology is very simple (just take a weakly converging net, and show the adjoint converges weakly by rearranging). To show that this is not continuous in the strong operator topology, consider the operators $T_n:ell^2toell^2$ defined by $T_n(x)=langle x,e_nrangle e_1.$ Then $|T_n(x)|=|langle x,e_nrangle|to 0$ for any $xinell^2$, so ${T_n}$ converges strongly to $0$. But $T_n^*(x)=langle x,e_1rangle e_n$, and $|T_n^*(e_1)|=1$ for all $n$, so $T_n^*$ does not converge strongly to $0$.