Let $X$ be a Hilbert space and $\mathcal{B}(X)=\mathcal{B}(X,X)$ be the set of bounded linear operators on $X$.
The strong operator topology $\tau_S$ is the topology on $\mathcal{B}(X)$ induced by seminorms of the form $T\mapsto ||Tx||$ as $x$ varies in $X$.
The uniform operator topology $\tau_u$ is the topology induced by the operator norm.
The uniform operator topology is stronger than the strong operator topology which means it has more open sets.
Is it possible to give a simple explicit example of an open set that is in $\tau_u$ but not in $\tau_s$? For the cases:
- $X = \mathbb{R}^3$
- $X = L^2(\mathbb{R})$
For $X=L^2(\mathbb{R})$ take the ball $\{T\in B(L^2(\mathbb{R})):\ \|T\|<1\}$.
Any strong open set contains an open set of the form
$$U=\{T\in B(L^2(\mathbb{R})):\ \|T(x_1)\|<\epsilon_1,...,\|T(x_n)\|<\epsilon_n\}$$
For certain fixed $x_1,...,x_n\in L^2(\mathbb{R})$ and $\epsilon_1,...,\epsilon_n>0$.
But these sets go outside $U$. Just take an operator that vanishes on $x_1,...,x_n$ but has a very large norm. For example, there is a $y\in L^2(\mathbb{R}$ orthogonal to all $x_1,...,x_n$. Define $T(x)=M\cdot P_{(y)}(x)$, where $P_{(y)}$ is the orthogonal projection onto the space generated by $y$ and $M$ is a scalar with large absolute value.
For $X=\mathbb{R}^3$ a set $U$ as above in which $x_1,x_2,x_3$ is a basis can be placed inside a uniform ball by taking $e_1,e_2,e_3$ small. Therefore both topologies coincide in this case.