Question: Given a vector space $V$, is it possible to endow it with two nonequivalent norms $\|\cdot\|_1$ and $\|\cdot\|_2$ such that any linear functional on $V$ is continuous in the sense of $\|\cdot\|_1$ if only if it is continuous in the sense of $\|\cdot\|_2$?
By nonquivalent norms I mean the induced topologies of the norms are different.
There is a similar question on topological vector spaces (link) which has been answered. Obviously there are examples that two different vector space topologies can have same continuous linear functionals. However I am primarily interested in the normed space case.
Best Answer
For $i = 1,2$, let $V_i$ denote $V$ equipped with $\|\cdot\|_i$.
From our assumption, the map $\operatorname{Id}: V_1 \to V_2$ is sequentially weakly continuous and hence it is continuous. This yields $\|\cdot\|_2 \leq C \|\cdot\|_1$.
Interchanging the role of $V_1$ and $V_2$ then yields equivalence of norms.
To see that a weakly sequentially continuous linear map must be bounded, you can proceed as follows. Suppose $T$ is weakly sequentially continuous and unbounded. Then there is a sequence $x_n \to 0$ such that $\|T x_n\| > n$. Since $x_n \to 0$ in norm, it also does so weakly so that $T x_n$ ought converge weakly, but weakly convergent sequences are norm bounded so that we reach a contradiction.