Space of bounded linear maps induced by different norm Suppose $X,Y$ are normed space, there are two norms $\|\|_1^Y,\|\|_2^Y$ on $Y$ which induce the same topology. We can define the norm of bounded linear mappings from $X$ to $Y$ as $$||f||_i=sup\{\|f(x)\|_i, x\in X,\|x\|\leq1\}, i=1,2.$$
Suppose $Y$ is complete w.r.t. $\|\|_1^Y$, we know that the normed space of bounded linear mappings w.r.t. $||f||_1$ is complete, is it true that
1. It is complete w.r.t. $||f||_2$?
2. $\|f\|_i$ induce the same topology?
 A: First note that the norms are equivalent:
Open set generated by two equivalent norm
(remark: I use $C$ as constant for that equivalence)
Assume now $f_n$ is cauchy w.r.t. $\left\|\cdot\right\|_1$. then by $sup_{x}\left\|f_n(x)-f_m(x)\right\|_2^{Y}\leq C sup_{x}\left\|f_n(x)-f_m(x)\right\|_1^{Y}$ you get the first result. (since both spaces are complete w.r.t. their functional norms it is enough to show that the cauchy-sequences coincide)
For the second part it is enough to show, that the convergence coincides.
Let $f_n\stackrel{1}{\rightarrow} f$. then with the same reasoning as above we get $\left\|f_n-f\right\|_2\leq C\left\|f_n-f\right\|_1$ and we are done. (the other way around works the same)
A: If the two norms $\| \cdot \|^Y_1$ and $\|\cdot\|^Y_2$ induce the same topology, then there are constants $\alpha, \beta \in \Bbb K$ such that for all $x \in Y$, $\alpha \|x\|^Y_1 \le \|x\|^Y_2 \le \beta \|x\|^Y_1$.
If $f \in \mathcal L(X,Y)$, then:
$$\|f\|_2 = \sup_{\|x\| = 1}{\|f(x)\|^Y_2} \le \beta \sup_{\|x\| = 1} \|f(x)\|^Y_1 = \beta \| f\|_1$$
Similarly, we get:
$$\alpha \|f\|_1 \le \|f\|_2 \le \beta \|f\|_1 \ \ \ \ \ (\star)$$
it is easy now to prove that $(\mathcal L(X,Y), \| \cdot \|_2)$ is complete.
As to the second question, $(\star)$ shows that the two norms induce the same topology 
