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Let ${\cal{B}}(X,Y)$ be the set of all bounded linear operators from normed linear spaces $X$ to $Y$.

Hunter and Nachtegaele say that if bounded linear operators $(T_{n})$ satisfy $\displaystyle\lim_{n \rightarrow \infty} ||T_{n}-T|| = 0$ for some $T \in {\cal{B}}(X,Y)$, then the $T_{n}$ converge uniformly to $T$.

My question is: Is it necessary to assume that $T \in {\cal{B}}(X,Y)$? If we just assume that $T$ is some linear operator from $X$ to $Y$, would we get automatically that it is bounded if we have that limit?

(I'm just thinking about how uniform convergence of continuous functions gives something still continuous and about how continuity of linear operators is equivalent to boundedness, so I'm not sure why the authors make a point to specify that $T \in {\cal{B}}(X,Y)$ when they define uniform convergence here.)

Thank you in advance for any insight!

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Yes, if $T:X\to Y$ is a linear operator and $\exists S\in B(X,Y)$ such that $$ \alpha := \sup_{\|x\| = 1} \|S(x) - T(x)\| < \infty $$ Then for any $x\in X,\|x\| = 1$, you have $$ \|T(x)\| \leq \|S(x) - T(x)\| + \|S(x)\| \leq \alpha + \|S\| $$ and so $T \in B(X,Y)$.

This is applicable in your situation with $S=T_1$. However, if $T \notin B(X,Y)$ a priori, it does not make sense to write down an expression $\|T_n - T\|$, which is why the authors assume $T\in B(X,Y)$.

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  • $\begingroup$ That makes sense. Thank you! $\endgroup$ – J. Anadis Nov 30 '15 at 12:24

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