Can we reconstruct norms from their induced operator norm?

Assume $(V,\|\cdot \|_V),(W\|\cdot \|_W)$ are two finite dimensional normed spaces (over $\mathbb{R}$).

Now I only give you the operator norm on $\text{Hom}(V,W)$ w.r.t $\|\cdot\|_V,\|\cdot \|_W$. What can you say about the original norms?

Of course multiplying both norms by the same scalar does not change the operator norm. So the best we can hope for is a reconstruction up to scalar multiple.

Is this always achievable? If not, is there “something intelligent” we can say about the norms?

Solutions Collecting From Web of "Can we reconstruct norms from their induced operator norm?"

Yes, you can reconstruct original norms up to common scalar multiple. It suffices to consider rank-one operators $T_{\varphi,w}x = \varphi(x)w$ where $\varphi$ is a linear functional on $V$. The key point is that the operator norm of $T_{\varphi,w}$ is $\|\varphi\|_{V^*}\|w\|_W$.

Fix one nonzero functional $\varphi$, and define $\| w\|_\# = \|T_{\varphi,w}\| $ for $w\in W$. By above, this gives a norm that agrees with the original norm on $W$ up to a scalar multiple.

Next, fix $w$ with $\|w\|_\#=1$ and define
$$\|u\|_\flat = \sup\{\varphi(u) : \|T_{\varphi,w}\|\le 1 \}$$
This is the only norm on $V$ for which the dual norm on $V^*$ satisfies $\|T_{\varphi,w}\| = \|\varphi\|_{\flat^*}\|w\|_\sharp$.

I am just adding some detailes to the solution of 1999:

Summary of the key points:

(1) $\|T_{\varphi,w}\|$ detrmines $\| \cdot \|_W, \| \cdot \|_{V*}$ up to scalar multiple.

(2) By a well known corollary of the Hahn-Banach theorem, $\| \cdot \|_{V*}$ determines $\| \cdot \|_V$

Observation: $\|T_{\varphi,w}\| = \|\varphi\|_{V^*}\|w\|_W$

Proof: $\sup_{v \in V, \|v\| = 1} \|\varphi(v)w\|_W = \sup_{v \in V, \|v\| = 1} |\varphi(v) |\cdot \|w\|_W = \|\varphi\|_{V^*}\|w\|_W$

Now fix one nonzero functional $\varphi$, and define $\| w\|_\# = \|T_{\varphi,w}\|$. By the observation above, this norm is a scalar multiple of the original norm on $W$.

That is, $\| \cdot \|_W= \alpha \| \cdot \|_\#$.

Now fix $w$ with $\|w\|_{\#} = 1$ and define $\|u\|_\flat = \sup_{\varphi \in V^*, \|T_{\varphi,w}\|=1} \varphi(u)$.

Claim: $\| \cdot \|_\flat = \frac 1\alpha \| \cdot \|_V$.

Proof:

$\|T_{\varphi,w}\| = 1 \iff \|\varphi\|_{V^*}\|w\|_W = 1 \iff \alpha \cdot \|\varphi\|_{V^*}\|w\|_\# = 1 \iff \alpha \cdot \|\varphi\|_{V^*} = 1$.

By a corollary of the Hahn-Banach theorem, $\|v\| = \sup_{f \in V^*, \|f\| = 1} f(v)$, so: $\|v\|_\flat = \sup_{\varphi \in V^*, \|T_{\varphi,w}\|=1} \varphi(v) = \sup_{\varphi \in V^*, \|\varphi\|_{V^*}=\frac 1\alpha} \varphi(v) = \frac 1\alpha \|v\|_V$.

Hence we reconstructed up to scalar multiple, as required.

Corollary: an operator norm is fully determined by its values on rank-one operators

Proof:

The operator norm w.r.t $\| \cdot \|_\#, \| \cdot \|_\flat$ is the original (given) operator norm. The conclusion follows since $\| \cdot \|_\#, \| \cdot \|_\flat$ are determined only by norms of rank-one operators $\|T_{\varphi,w}\|$.