How strong is the operator norm topology? Let $(V,\tau_V), (W,\tau_W)$ be normable topological vector spaces.
Let $||\cdot||_V, ||\cdot||_W$ be norms on $V,W$ inducing $\tau_V, \tau_W$ respectively.
Let $||\cdot||_{op}$ be the operator norm on $B(V,W)$ and $T$ be the topology on $B(V,W)$ induced by this operator norm. (Note that $T$ does not depend on norms $||\cdot||_V$ and $||\cdot||_W$, but it depends on the topologies $\tau_V$ and $\tau_W$)
Note: When $X$ is a topological space and $Y$ is a metric space, there is a (topological) hierarchy on $C(X,Y)$ such as "product topology $\subset$ compact open topology $\subset$ uniform topology $\subset$ box topology".
Likewise, how strong is $T$? I'm curious where the topology $T$ lie in the above hierarchy. Is it compatiable with the compact-open topology on $B(V,W)$? And is there a standard naming for this $T$? What would be a keyword to google the properties of this topology?
 A: The topology $T$ is usually called the norm topology (when we view $V,W$ as normed spaces, and not only as normable spaces), or the topology of uniform convergence on bounded subsets of $V$. To avoid trivialities, I assume that neither $V$ nor $W$ is the trivial vector space in the following (for then $B(V,W) = \{0\}$ and all topologies coincide).
Since every finite set is bounded, $T$ is clearly finer than the product topology (more precisely, the subspace topology induced on $B(V,W)$ by the product topology on $W^V$, which is known as the "strong operator topology").
Every compact subset of $V$ is bounded, so $T$ is finer than the topology of compact convergence (uniform convergence on compact subsets of $V$). Unless $V$ is finite-dimensional, $T$ is strictly finer than the topology of compact convergence.
The topology of compact convergence is at least as fine as the compact-open topology: A neighbourhood basis of $f\in B(V,W)$ in the compact-open topology is given by the family of finite intersections of sets of the form
$$(K,U) = \{g \in B(V,W) : g(K) \subset U\},$$
where $K\subset V$ is compact and $U\subset W$ is an open set with $f(K) \subset U$. Since $f(K)$ is compact, there is a neighbourhood $N$ of $0$ in $W$ with $f(K) + N \subset U$, and thus $M(K,N) = \{ h \in B(V,W) : h(K) \subset N\}$ is a neighbourhood of $0$ in the topology of compact convergence with $f + M(K,N) \subset (K,U)$, so the topology of compact convergence is finer than the compact-open topology.
The entire space is not bounded, so $T$ is strictly coarser than the topology of uniform convergence (which happens to coincide with the discrete topology on $B(V,W)$, and thus is not a vector space topology [except in the trivial case]).
Thus, in your scale, the topology $T$ sits strictly (unless $V$ is finite-dimensional) between the compact-open topology and the topology of uniform convergence.
