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

Question

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

This question occurred to me after seeing this question. I'm hoping that perhaps there exists something like the polarization identity, i.e. some identity one may use to "recover" the vector norm the same way the polarization identity "recovers" the inner product.

I am aware that any induced norm satisfies the inequality $$ \left|\|A^r\|\right|^{1/r} \geq \rho(A) $$ Also, there exists some invertible matrix $S$ such that $$ \left|\|SAS^{-1}\|\right| = \rho(A) $$ I'm wondering if one or both of the above might also be a sufficient condition for $\left|\|\cdot \|\right|$ to be a derived norm, and that they might somehow be used to derive some sort of identity producing the necessary vector norm.

Any input is appreciated!

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Runaway train of thought below:

Following HHO's advice (see comment below), here's a neat way to recover the vector norm (assuming that a suitable vector norm does exist):

Let $\|\cdot \|_O$ denote the operator norm. I will define a vector norm $\|\cdot \|$ as follows: arbitrarily, I set $\|e_1\| = 1$ (because $\alpha \|\cdot\|$ is a vector norm for any $\alpha>0$ and any vector norm $\|\cdot\|$ and since both of these result in the same induced norm, we may set $\|e_1\| = \alpha$ for any $\alpha>0$). From there, we may define $$ \|u\| = \left\| u e_1^* \right\|_O = \left\| \pmatrix{|&|&&|\\ u&0&\cdots&0\\ |&|&&|} \right\|_O $$ What remains to be seen is under which conditions this defines a valid vector-norm.

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In fact, the above must always be a valid vector-norm, by the definition of a matrix norm. It is necessary to check whether the vector norm produced above induces the operator norm that we started with.

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By the above, here's a neat criterion for checking whether $\|\cdot\|_O$ is an operator norm:

We can state that $\|\cdot\|_O$ is an induced (matrix) norm if and only if for all $A \in \mathbb{F}^{n\times n}$, we have $$ \|A\|_O = \max_{x \neq 0} \frac{\|Ax e_1^*\|_O}{\|xe_1^*\|_O} $$ and, presumably, $e_1$ can be replaced by any convenient $v \in \mathbb{F}^n: v^*v = 1$. I guess I answered my own question then.

Is this a known theorem? This question is now a reference request. If anyone has seen something like this, please say so.

Answer 1

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.