Operator norm is the unique norm up to strong equivalence that implies pointwise convergence

Question

Let $(E,||\cdot||_E)$ and $(F,||\cdot||_F)$ be normed vector spaces. Let $||\cdot||$ be the operator norm in the space of bounded linear maps from $E$ to $F$ given by $$||A||=\sup_{v\neq 0} \frac{||Av||_F}{||v||_E}$$ Let $||\cdot||'$ be another norm in the same space such that for every sequence $(A_n)_{n\ge 1}$ of linear maps and every $A$ $$||A_n-A||'\to 0\Rightarrow ||A_nv-Av||_F\to 0\quad \forall v\in E$$ I would like to find out whether $||\cdot||'$ must be strongly equivalent to $||\cdot||$, i.e., if there exist $C_1,C_2\in (0,\infty)$ such that $$C_1\le \frac{||A||'}{||A||}\le C_2\quad \forall A\neq 0$$ I know that if $E$ and $F$ are Banach spaces then this claim is true because by the open mapping theorem it is enough to verify the second inequality and if it were not true one would be able to build a sequence $(B_n)$ such that $||B_n||'\to 0$ but $\sup_n ||B_n||=\infty$ and a contradiction follows by the uniform boundedness theorem. However, I have no clue if it holds in the following cases:

1. $E$ is not a Banach space and $F$ is a Banach space;
2. $E$ is a Banach space and $F$ is not a Banach space;
3. Neither $E$ nor $F$ is a Banach space.

Since the aforementioned theorems are not both valid in those cases, the problem seems to be harder, so I would appreciate some hint. Thank you in advance.

Answer 1

This is not true even for Banach spaces $E$ and $F$. Take, e.g., $E=c_0$ and $Y=\mathbb R$ so that operator norm on $L(c_0,\mathbb R)=\ell^1$ is the usual norm of $\ell^1$. Using a Hamel basis of $\ell^1$ you can find a discontinuous linear map $\varphi:\ell^1\to\mathbb R$ so that $\|x\|'=\|x\|+|\varphi(x)|$ is a strictly finer norm which is not equivalent to $\|\cdot\|$ (because $\varphi$ is continuous w.r.t. $\|\cdot\|'$ but discontinuous w.r.t. $\|\cdot\|$).