Assume $(V,∥∥_V),(W,∥∥_W)$ are both finite dimensional normed spaces. We have the induced operator norm on ${\rm Hom}(V,W)$.

It turns out that the operator norm is induced by an inner product iff both $V,W$ are inner product spaces and at least one of $V,W$ has dimension 1. (This is proved here).

Are there any other obstructions for a norm on ${\rm Hom}(V,W)$ to be realized as an operator norm for some suitable norms on $V,W$?

(The main interest is in the case where $\dim V>1,\dim W>1$. Take a norm on the Hom space which does not come from an inner product. Is it realizable?)

  • $\begingroup$ I'm not sure to really understand your question... Do you know that $\sup_{\|x\|_V=1}\|Ax\|_W$ is a norm on $Hom(V,W)$ for any norms $\|\cdot\|_V,\|\cdot\|_W$? $\endgroup$
    – Surb
    Jun 4 '15 at 10:04
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    $\begingroup$ Yes. I am asking about the reverse direction. Assume you are given some norm on $Hom(V,W)$. Can we find suitable norms on $V$,$W$ such that the operator norm they induce will be the original (given) norm. $\endgroup$ Jun 4 '15 at 10:07
  • $\begingroup$ Yes, but it is not an operator norm. $\endgroup$ Jun 4 '15 at 10:17

Operator norms are the same thing as injective tensor norms or, equivalently, smallest dual cross norms on $\operatorname{Hom}(V, W) \simeq V^\ast \otimes W$. This means that the dual norm $\Vert \cdot \Vert^\ast$ on $V \otimes W^\ast$ has to satisfy the following:

$$\Vert x \Vert^\ast = \inf_{x = \sum_i y_i \otimes z_i} \sum_i \Vert y_i \Vert \Vert z_i \Vert$$

where the infimum is over all possible representations of a tensor as a combination of rank-ones. This follows immediately from the definition of the operator norm, i.e. that it's determined by testing against rank-ones.

In particular, the unit ball of $\Vert \cdot \Vert^\ast$ is the convex hull of its rank-one vectors. Thus, for example, it cannot be strictly convex (unless everything is of rank one, which means that $\min(\dim V, \dim W) = 1$). This generalizes the Euclidean case.

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    $\begingroup$ Yes indeed. It is a general fact that operator norms have rather flat unit sphere. For instance, the norm on ${\bf M}_n(\mathbb R)$ induced by $\ell_p$ is constant over the $(n-1)^2$-dimensional convex set of bistochastic matrices ! $\endgroup$ Jun 4 '15 at 13:11
  • $\begingroup$ @Alexander: can you please elaborate more? I agree that operator norm is determined by its value on rank-one operators. I am not sure why an operator norm is a crossnorm. (I do not see why the condition holds for the dual norm of the operator norm). I am also not sure why the condition you gave is sufficient or necessary for the norm to be an operator norm. $\endgroup$ Jun 28 '15 at 14:02

Here is a non trivial constraint : ${\rm Hom}(V,W)$ contains (is spanned by) rank one morphisms $$v\mapsto\ell(v)w,\qquad\ell\in V',w\in W.$$ If a given norm over ${\rm Hom}(V,W)$ is induced, then $$\|w\otimes\ell\|=\|w\|_W\|\ell\|_*.$$ This yields the necessary condition $$\|w_1\otimes\ell_1\|\cdot\|w_2\otimes\ell_2\|=\|w_1\otimes\ell_2\|\cdot\|w_2\otimes\ell_1\|,\qquad\forall w_1,w_2\in W,\ell_1,\ell_2\in V'.$$ This raises the question whether this is the only restriction.

  • $\begingroup$ You are right. Do you see immediately that ypur additional constraint really decreases the space of possible realizable norms? i.e is it trivial to find a norm which is not induced by inner product, and does not satisfy this requirement. such a norm will be outside the realizable space. $\endgroup$ Jun 4 '15 at 10:24
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    $\begingroup$ This condition is satisfied by the Frobenius/Hilbert-Schmidt norm which seems not to be an operator norm. There should be another obstruction hanging around. $\endgroup$ Jun 4 '15 at 12:57
  • $\begingroup$ Yes, but the Frobenius norm is induced by an inner product. As I stated in the question, we already know that (except for trivial dimensions) these kind of norms cannot be realized as operator norms. For all we know, all norms that come from inner products satisfy the condition Denis suggested, so the space of optional candidates for realization does not really decrease. Saying it differently, I want to make sure Denis's condition is indeed restrictive. $\endgroup$ Jun 6 '15 at 21:50

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