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Let $E$ and $F$ be two Banach spaces. Let $(T_n)$ be a sequence of continuous linear operators from $E$ into $F$ such that for every $x \in E, T_n(x)$ converges (as $n \to \infty$) to a limit denoted by $T(x)$. Then we have

(a) $\sup \Vert T_n\Vert < \infty$

(b)$ T \in L(E,F)$,

(c) $\Vert T\Vert_{L(E,F)} \le \liminf_{n\to \infty} \Vert T_n\Vert_{L(E,F)}.$

Why do we need $F$ to be Banach space instead of a normed space in this corollary of the Uniform boundary theorem?

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  • $\begingroup$ I think it should be $\|T\|\leq\liminf_{n}\|T_{n}\|$? $\endgroup$ – user284331 Jan 10 '18 at 2:43
  • $\begingroup$ Uniform boundedness principle, not "Uniform boundary theorem"? $\endgroup$ – Robert Israel Jan 10 '18 at 2:50
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The following is taken from An Introduction to Banach Space, Robert E. Megginson, page 45, 46:

Let $\mathcal{F}$ be a nonempty family of bounded linear operators from a Banach space $X$ into a normed space $Y$. If $\sup\{\|Tx\|: x\in\mathcal{F}\}$ is finite for each $x\in X$, then $\sup\{\|T\|: T\in\mathcal{F}\}$ is finite.

And as a corollary:

Let $(T_{n})$ be a sequence of bounded linear operators from a Banach space $X$ into a normed space $Y$ such that $\lim_{n}T_{n}x$ exists for each $x\in X$. Define $T:X\rightarrow Y$ by the formula $Tx=\lim_{n}T_{n}x$. Then $T$ is a bounded linear operator from $X$ into $Y$.

So I think the completeness of $Y$ is not needed.

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We do not need $F$ to be a Banach space. Given $E$, $F$, $T_n$ and $T$ as specified but with $F$ a normed space, you can simply replace $F$ by its completion $\widetilde{F}$, and get the same situation with $T_n$ and $T$ mapping $E$ into $\widetilde{F}$. The theorem for operators from $E$ into $\widetilde{F}$ then implies the theorem for operators from $E$ into $F$.

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  • $\begingroup$ Thanks! This point of view helps me see the connection much more clearly . $\endgroup$ – Xiao Jan 10 '18 at 5:09

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