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I consider a sequence of elements $\{f_n\}_n$ in a normed space $X$ such as $\Vert f_n-f\Vert_X\to 0, n\to\infty$. Let $\{Q_n\}_n$ a complex sequence with $Q_n\to Q$ as $n\to\infty$. Does $Q_nf_n$ converge to $Qf$ in $X$? I can write $$\Vert Q_nf_n-Qf\Vert_X=\Vert Q_nf_n-Qf_n+Qf_n-Qf\Vert_X\leq$$ $$\leq \Vert f_n\Vert_X\vert Q_n-Q\vert+|Q|\Vert f_n-f\Vert_X$$ and I have the thesis.

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You still have to show that $||f_n||_X$ is bounded. If it blows up, then your first term will not go to zero.

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  • $\begingroup$ But by continuity of norm $\Vert f_n\Vert\to\Vert f\Vert$ $\endgroup$
    – Sue
    Apr 22, 2013 at 7:23
  • $\begingroup$ A norm convergent sequence is necessarily bounded. $\endgroup$
    – Sue
    Apr 22, 2013 at 7:46
  • $\begingroup$ Yes, it's not hard to show, but it should be mentioned. $\endgroup$
    – N.U.
    Apr 22, 2013 at 7:47
  • $\begingroup$ Yes, I forgot to underline this fact in my post; it can be seen as a consequence of "principle of uniform boundedness". $\endgroup$
    – Sue
    Apr 22, 2013 at 7:50

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