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Let $X$ be a complex vector space and let $\|\cdot\|_1$ and $\|\cdot\|_2$ denote norms on $X$ which are equivalent, i.e. there exist constants $c,C > 0$ such that for all $x \in X$ $$ c \|x\|_1 \leq \|x\|_2 \leq C \|x\|_1. $$

Each norm induces a operator norm $\|\cdot\|_{i,i}$ on $L(X)_{i,i}$, $i = 1,2$ via $$ \| T\|_{i,i} := \sup_{x \in X} \frac{\|T x\|_i}{\|x\|_i}. $$

Are also these operator norms equivalent (maybe we need to know more about $X$ like completeness?)?

Can anything be said about the "mixed" operator norms $\| T\|_{1,2}$ and $\|T \|_{2,1}$ w.r.t their comparability?

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2 Answers 2

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For $x\neq 0$, the following inequalities take place: $$ \frac{\|T x\|_1}{\|x\|_1} \leqslant \frac{\|T x\|_2/c}{\|x\|_2/C}= \frac Cc \frac{\|T x\|_2}{\|x\|_2} $$ hence $\left\lVert T\right\rVert_{1,1}\leqslant (C/c)\left\lVert T\right\rVert_{2,2}$. By a similar reasoning, $\left\lVert T\right\rVert_{2,2}\leqslant (c/C)\left\lVert T\right\rVert_{1,1}$.

Similarly, $$ \frac{\|T x\|_1}{\|x\|_1} \leqslant \frac{\|T x\|_2/c}{\|x\|_1}= \frac 1c \frac{\|T x\|_2}{\|x\|_1} $$ hence $\left\lVert T\right\rVert_{1,1}\leqslant 1/c\left\lVert T\right\rVert_{2,1}\leqslant C/c\left\lVert T\right\rVert_{1,1}$.

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  • $\begingroup$ Thank you, Davide! Any idea about how the norms $\|T\|_{1,2}$ and $\|T\|_{2,1}$ relate to $\|T\|_{1,1}$ and $\|T\|_{2,2}$? $\endgroup$
    – el_tenedor
    Oct 22, 2018 at 13:18
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You have: $c\|T(x)\|_1\leq \|T(x)\|_2\leq C\|T(x)\|_1$ implies that $cSup_{\|x||=1}\|T(x)\|_1\leq Sup_{\|x\|=1}\|T(x)\|_2\leq CSup_{\|x\|=1}\|T(x)\|_1$ which is equivalent to $c\|T\|_1\leq \|T|_2\leq C\|T\|_1$.

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