Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading about functional analysis and I found the definition of the operator norm, if you have $(X,\|\|_1)$ and $(Y,\|\|_2)$ normed spaces then the set $\mathcal{L}_{\|\|_1,\|\|_2}(X,Y) := \{T:X \to Y \text{ linear }: \sup\{ \|T(v)\|_2: \|v\|_1 = 1 \} < \infty \}$ has a norm defined by $\|\|_1$ and $\|\|_2$ and son on. My questions are:

  1. If $\mathcal{L}_{\|\|_1,\|\|_2}(X,Y) = \mathcal{L}_{\|\|'_1,\|\|_2}(X,Y)$ then, can I ensure that $\|\|_1$ and $\|\|'_1$ are equivalents? Note that this generalizes the fact that all the norms in $\mathbb{R}^n$ are equivalents, because any linear operator is continuos with any norm in $\mathbb{R}^n$. Similarly with the other side,

  2. If $\mathcal{L}_{\|\|_1,\|\|_2}(X,Y) \subseteq \mathcal{L}_{\|\|'_1,\|\|_2}(X,Y)$, can I say something? And like before with the other side,

  3. If I have a subspace $Z$ of $\mathbb{L}(X,Y)$ then there exist norms such that $Z = \mathcal{L}_{\|\|_1,\|\|_2}(X,Y)$.

I apologize if my questions are not interesting, thank for your help.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

1) Yes. Consider operators of the form $T(x) = \phi(x) y$ where $\phi$ is a bounded linear functional on $(X, \|\cdot\|_1)$ and $ y \in Y$ is fixed. Then $\|T\| = \|\phi\| \|y\|$. So if ${\cal L}_{\|\cdot\|_1,\|\cdot\|_2}(X,Y) ={\cal L}_{\|\cdot\|'_1,\|\cdot\|_2}(X,Y) $, $X$ must have the same bounded linear functionals in the two norms. Now the dual space $X^*$ is a Banach space. Let $B$ be the unit ball of $X$ in the $\|\cdot\|_1$ norm, but use the norm on $X^*$ corresponding to the $\|\cdot\|'_1$ norm. The norm of $x \in X$, considered as a linear functional on $X^*$, is then the $\|\cdot\|'_1$ norm. For each $\phi \in X^*$, $\{|\phi(x)|: x \in B\}$ is bounded. Therefore by the uniform boundedness principle, $\{\|x\|'_1: x \in B\}$ is bounded. Thus for some constant $C$, $\| \cdot \|'_1 \le C \| \cdot \|_1$. Similarly in the other direction.

share|improve this answer
    
And what about 3.? Do you have at least a little comment? Thanks very much. –  Diego Silvera Dec 28 '12 at 13:52
1  
${\cal L}(X,Y)$ has to include at least the rank-$1$ operators from my answer above. A general subspace doesn't. –  Robert Israel Dec 28 '12 at 19:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.