I am trying to prove the following "contiuity-type" result.

Let $X,Y$ normed linear spaces. Let $\{T_n\} \to T \in \mathcal{L}(X,Y)$ and $\{u_n\} \to u \in X$. Show that $\{T_n(u_n)\} \to \{T(u)\} \in Y$

Any help will be most appreciated! Thank you very much in advance.

  • $\begingroup$ I think I have proven something like this in the case of real valued functions and sequences of real numbers. Is the proof the same? $\endgroup$ – user44069 Feb 21 '13 at 5:29
  • $\begingroup$ Is the proof the same as what? Could you please elaborate? Have you tried something to solve this problem, starting with using the definition of $\{T_n(u_n)\}\to T(u)$? $\endgroup$ – Jonas Meyer Feb 21 '13 at 5:30
  • $\begingroup$ Replacing the operators by functions from $\mathbb{R} \to \Bbb R$ and the sequences by sequences of real numbers, will the proof of this statement be the same as the statement for which I am asking for help? $\endgroup$ – user44069 Feb 21 '13 at 5:31
  • $\begingroup$ For arbitrary functions $\mathbb R\to\mathbb R$? Under what topology? You have not made a precise statement or indicated what the analogous proof is, so I don't know how to answer whether it would be similar. For this problem, I suggest using the triangle inequality. $\endgroup$ – Jonas Meyer Feb 21 '13 at 5:33

Try using the inequality

$$\|T_n(u_n)-T(u)\|=\|T_n(u_n)-T_n(u)+T_n(u)-T(u)\|\leq \|T_n\|\|u_n-u\|+\|T_n-T\|\|u\|,$$ and note that you will want to use the fact that convergent sequences are bounded.

(Some text copied from this answer.)


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