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.

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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.

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And what about 3.? Do you have at least a little comment? Thanks very much. –  Diego Silvera Dec 28 '12 at 13:52
${\cal L}(X,Y)$ 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