Let $X$ be a separable Banach space with (norm $1$) Schauder-Basis $\{e_n\}_{n\in\mathbb N}$. Denote for $x\in X$ with $|\cdot|_x$ the seminorm on $\mathcal L(X)$ given by $|A|_x = \|A x\|$. Consider:
$$\|\cdot\|_1:=\sum_n 2^{-n} |\cdot|_{e_n} \tag{1}$$
This defines a norm and the topology of this norm is
- Weaker than the uniform topology: You have $\|A\|≥|A|_{e_n}$ and then $\|A\|=\sum_n 2^{-n} \|A\|≥\sum_n 2^{-n} |A|_{e_n}$.
- Strictly weaker than uniform topology in the case of $X$ a Hilbert space and $e_n$ ONB: Denote $\pi_n$ the orthogonal projection onto span of $e_n$, then $\sum_{n=N}^\infty \pi_n$ converges to zero in this topology but not in uniform norm.
- Stronger than the strong operator topology.
- Becomes equal to the strong operator topology upon restriction to bounded subsets.
For 3 and 4 see here. I don't know the relation to the $\sigma$-strong topology (generated by seminorms $\sqrt{\sum_n |\cdot|^2_{x_n}}$ whenever $\sum_n |x_n|^2$ converges), but it looks like this is weaker than the $\sigma$-strong topology.
Does this topology have a name? Is it remarkable that it has a norm?
A side question, if $\{x_n\}_n$ is a dense countable subset of $X$, then $\|\cdot\|_2:=\sum_n 2^{-n} \frac{|\cdot|_{x_n}}{\|x_n\|}$ gives another norm. I think it is equivalent to the above, but calculation looks unhappy. Is there a reason to expect it to be inequivalent to $\|\cdot\|_1$?
Remark: The norm is not submultiplicative, for example on a Hilbert space with $L$ the left shift, $R$ the right shift you have
\begin{gather}\|L^k\|_1 = \sum_{n=k+1}^\infty 2^{-n}=2^{-k} \qquad \|R^k\|_1=\sum_{n=1}^\infty 2^{-n}=1\\ \|L^kR^k\|_1=\|\mathbb 1\|_1 =1 \not≤\|L^k\|_1\cdot\|R^k\|_1=2^{-k} \end{gather}
This makes it a bit less interesting.