Small note

One might have wondered why B(X,Y) contains only linear bounded operators, and not linear operators of any and every kind. This has a very specific reason. Unless we fix x\in X, we cannot construct a cauchy sequence \{T_1,T_2,\dots\} of linear operators. And we do not want to fix x\in X, as we want to define the limit \lim\limits_{n\to\infty} T_n x for every x\in X.

Now if we allow ourselves to take all x\in X, knowing that T_i (\alpha x)=|\alpha|T_i(x), we coud convert any sequence of linear operators into a divergent one. Hence, we need to construct sequences of the kind \{T_i(x)/\|x\|\}. This requires that we use bounded linear operators. 

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics.

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