If $x_j \to x $ and $||Tx_j ||\le k $, show that $||Tx|| \le k $ for a continuous linear operator $T$

Question

Let $T $ be a continuous linear operator. Suppose ${x _j } $ is a sequence in some Banach space $X $, with limit $x $, such that $||Tx _j || \le k $. Show that $||Tx ||\le k $

Well I suppose that I should use that $T $ and the norm are continuous functions, so that the composition is to. Now put $\phi (x) = ||Tx ||$, then $\lim \phi (x _j) = \phi (x) $. And since $\phi (x _j) \le k $ we must have $\sup _{x _j } \phi (x _j ) \le k $

Correct?

Thanks in advance!

Answer 1

Yes. It's ok. You better use $limsup$ and $liminf$. But the limit exists,so it's ok.

With this you can see the mechanics of the solution.

$|Tx|=|T(x-x_j)+Tx_j|\leq \|T\|\cdot|x-x_j|+|Tx_j|\leq \epsilon+k$ for $j$ big enough and arbitrary $\epsilon$