Let $E, F$ be two Banach Spaces. Let $\{ T_{n} \}$ be a sequence of continuous linear operators from $E$ into $F$ such that:

For all $x \in E: T_{n}x \rightarrow Tx,$ some limit in $F$.

Then the following 3 properties are satisfied:

(a) $^{\text{sup}}_{n \in \mathbb{N}} ||T_{n}|| < \infty $

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

(c) $ ||T|| \leq ^{\text{lim inf}}_{n \rightarrow \infty} ||T_{n}|| $

Here $L(E,F)$ is the space of continuous linear functions from $E$ to $F$ with the norm $|| T || = \ ^{\ \text{ sup}}_{|x| \leq 1} |Tx|$.

I have proven properties (a) and (b) easily, but (c) is a problem. I am under the assumption that the author of the textbook I am reading has written the property this way because $^{\text{lim}}_{n \rightarrow \infty} ||T_{n}||$ does not necessarily exist, since if it did, the lim sup would just equal the limit.

I know that $|T_{n}x| \leq ||T_{n}|| \cdot |x|,$ for all $n \in \mathbb{N}, x \in E$, but am unable to see how I can use this without a limit for $||T_{n}||$.

  • $\begingroup$ Why not take liminf both sides and by continuity of absolute value we get $|Tx| \leq \liminf\limits_{n\to \infty} ||T_n|| |x|$? Now it follows by definition. $\endgroup$ – James Yang May 11 at 14:50

Fix $x\in E$ or norm one: then $\left\vert Tx\right\rvert_F=\lim_{n\to +\infty}\left\vert T_nx\right\rvert_F$ hence taking the $\liminf_n$ in the inequality $\left\vert T_nx\right\rvert_F\leqslant \lVert T_n\rVert $, we get that $$ \left\vert Tx\right\rvert_F\leqslant \liminf _{n\to +\infty}\lVert T_n\rVert . $$ Since this estimate is valid for all $x$ of norm one, we see that the operator norm of $T$ does not exceed $\liminf _{n\to +\infty}\lVert T_n\rVert$.


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