Can the "inducing" vector norm be deduced or "recovered" from an induced norm?

Let $V$ be a topological vector space. Let $O(V)$ be the space of continuous operators in $V$, endowed with the strong operator topology. Assume that $O(V)$ can be given a structure of normed space with the norm $\| \cdot \|$, with the norm generating the strong operator topology of $O(V)$.

Let $f \in V^* \setminus \{ 0 \}$ be a linear and continuous functional. It is easy to show that $v \otimes f$ is a continuous linear operator, since $f$ is so. Define $\| v \| _f = \| v \otimes f \|$. Again, it is easy to show that this is a norm.

Let us show that the topology of $\| \cdot \| _f$ is the same as the original one of $V$. Let $(v_i) _{i \in I}$ be a net convergent to $v$ in the topology of $V$. Then

$$\begin{align} v_i \underset {i \in I} \to v \Leftrightarrow \\ f(u) v_i \underset {i \in I} \to f(u) v \; \forall u \in V \Leftrightarrow \\ (v_i \otimes f) (u) \underset {i \in I} \to (v \otimes f) (u) v \; \forall u \in V \Leftrightarrow \\ v_i \otimes f \underset {i \in I} \to v \otimes f \Leftrightarrow \\ \|v_i \otimes f - v \otimes f \| \underset {i \in I} \to 0 \Leftrightarrow \\ \|(v_i - v) \otimes f \| \underset {i \in I} \to 0 \Leftrightarrow \\ \| v_i - v \| _f \underset {i \in I} \to 0 \; . \end{align}$$

Finally, let the operator norm $\| \cdot \| _1$ induced on $O(V)$ by $\| \cdot \| _f$ be given by $\| U \| _1 = \sup \{ M>0 | \; \| Uv \| _f \le M \| v \| _f \}$, as usual. Let $U_i \to U$ in this norm. We have $\| (U_i - U)v \| _f \le \|U_i - U \| _1 \| v \| _f$, so $\| (U_i - U)v \| _f \to 0$, so $U_i v \to Uv$ in the original topology of $V$, so $U_i \to U$ in the topology of $O(V)$ and, since this was assumed to be given by $\| \cdot \|$, we obtain $\| U_i - U \| \to 0$, so the topology of $\| \cdot \| _1$ is stronger than the topology of $\| \cdot \|$. I suspect, but cannot prove, that the two topologies do not coincide.

Furthermore, if $\| \cdot \|$ has the supplementary property that $\| UV \| \le \| U \| \| V \|$ then we can obtain even more: since $Uv \otimes f = U \circ (v \otimes f)$, then $\| Uv \| _f = \| Uv \otimes f \| = \| U \circ (v \otimes f) \| \le \| U \| \| v \otimes f \| = \| U \| \| v \| _f$, so $\| \cdot \| \le \| \cdot \| _1$. Again, I believe that the opposite inequality does not hold, but I do not know how to prove it.