Comparison of Bounded Operator Topologies
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:
$$
tau_{rm weak}subset tau_{rm strong}subsettau_{rm uniform}
$$
general-topology functional-analysis
asked Nov 20 at 1:14
user2379888
23719
add a comment |
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:
$$
tau_{rm weak}subset tau_{rm strong}subsettau_{rm uniform}
$$
general-topology functional-analysis
asked Nov 20 at 1:14
user2379888
23719
add a comment |
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:
$$
tau_{rm weak}subset tau_{rm strong}subsettau_{rm uniform}
$$
general-topology functional-analysis
asked Nov 20 at 1:14
user2379888
23719
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:
$$
tau_{rm weak}subset tau_{rm strong}subsettau_{rm uniform}
$$
general-topology functional-analysis
general-topology functional-analysis
asked Nov 20 at 1:14
user2379888
23719
asked Nov 20 at 1:14
user2379888
23719
asked Nov 20 at 1:14
user2379888
23719
asked Nov 20 at 1:14
user2379888
23719
asked Nov 20 at 1:14
user2379888
23719
23719
add a comment |
add a comment |
1 Answer
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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 $0$.
answered Nov 20 at 2:44
Aweygan
13.4k21441
add a comment |
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1 Answer
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1 Answer
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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 $0$.
answered Nov 20 at 2:44
Aweygan
13.4k21441
add a comment |
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 $0$.
answered Nov 20 at 2:44
Aweygan
13.4k21441
add a comment |
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 $0$.
answered Nov 20 at 2:44
Aweygan
13.4k21441
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 $0$.
answered Nov 20 at 2:44
Aweygan
13.4k21441
answered Nov 20 at 2:44
Aweygan
13.4k21441
answered Nov 20 at 2:44
Aweygan
13.4k21441
answered Nov 20 at 2:44
Aweygan
13.4k21441
13.4k21441
add a comment |
add a comment |
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