# Does norm equivalence imply norm equivalence of induced operator norms?

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?

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 $$\left\lVert T\right\rVert_{1,1}\leqslant 1/c\left\lVert T\right\rVert_{2,1}\leqslant C/c\left\lVert T\right\rVert_{1,1}$$.
• Thank you, Davide! Any idea about how the norms $\|T\|_{1,2}$ and $\|T\|_{2,1}$ relate to $\|T\|_{1,1}$ and $\|T\|_{2,2}$? Oct 22, 2018 at 13:18
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 $$c\|T\|_1\leq \|T|_2\leq C\|T\|_1$$.