Let $T_n$ be a sequence of continuous linear operators from a Banach space $X$ to a normed linear space $Y$. Now, for all $x \in X$, $\lim_{n \rightarrow \infty} T_n(x)$ exists in $Y$. Define $T(x) = \lim_{n \rightarrow \infty} T_n (x)$ on $X$. Show that $$\|T\| \leq \liminf \|T_n\|.$$

Here's my proof:

Convergence of $T_n(x)$ implies that for all $x \in X$, $\|T_n(x)\|\leq M_x$ for some $M_x$. The uniform boundedness principle implies $T_n$ is uniformly bounded, in particular, $\liminf \|T_n\| < \infty.$ Now for all $z \in X$ with $\|z\|=1$, $$\|T(z)\|=\lim \|T_n(z)\| \leq \liminf \|T_n\| \|z\| = \liminf \|T_n\|.$$

I'm not too sure about that last line. How can I improve this proof?

  • $\begingroup$ @Seirios The assumption says $Tx=\lim T_nx$. By the reverse triangular inequality, $\|Tx\|=\lim \|T_nx\|$. $\endgroup$ – Julien Feb 25 '13 at 18:24
  • $\begingroup$ You can also think that $\|\cdot\|$ is continuous so $T_nx \to Tx$ implies $\|T_nx\| \to \|Tx\|$. In other words, $\lim \|T_nx\| = \|Tx\|$. (I know I'm late, but people who are even later than this might find this useful, as I did this question) $\endgroup$ – Ivo Terek Jul 9 '15 at 20:52

You are correct. Let me just add a tiny intermediate step to convince you you are right.

Fix $x$.

For all $n$ $$ \|T_{n}x\|\leq \|T_{n}\|\|x\|. $$

So $$ \|Tx\|=\lim \|T_nx\|=\liminf\|T_{n}x\|\leq \liminf \|T_{n}\|\|x\| $$ for all $x$.

Hence $$ \|T\|\leq \liminf \|T_n\|. $$

Note: we have $\lim \|T_nx\|=\|Tx\|$ because $\lim T_nx=Tx$ and $|\|T_nx\|-\|Tx\||\leq \|T_nx-Tx\|$.


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