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Suppose $A: \mathcal{l^\infty}\rightarrow\mathcal{l^\infty}$ is a bounded linear transformation defined by $$A(x_1, x_2, x_3, ...) = (x_1, \frac{x_1+x_2}{2}, \frac{x_1+x_2+x_3}{3},...).$$ I want to show that if $(x_1, x_2, ...)$ has a limit then so does $A(x_1, x_2, ...)$.

Proof: Suppose $x:=(x_1, x_2, ...)$ converges to $x$ under the $||\cdot||_\infty$ norm. Consider $$||Ax_n-Ax||_\infty = ||A(x_n-x)||_\infty\leq M||x_n-x||_\infty,$$ where $M>0$.

However, since $||x_n-x||_\infty\rightarrow0$, we have that $||Ax_n-Ax||_\infty\rightarrow0$, which in other words means that $A(x_1, x_2,..)$ has a limit under the $||\cdot||_\infty$ norm.

Is this proof correct? And how would I go about proving that the converse is not true?

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    $\begingroup$ How do you know that $\|A(x_n-x)\|_\infty\leq M\|x_n-x\|_\infty$? I am not saying it is wrong, just curious. Oh, it is stated in the problem that $A$ is bounded. Then your proof for the first part is correct! $\endgroup$ – N3buchadnezzar Apr 23 '17 at 10:37
  • $\begingroup$ Haha no problem! And thanks. Any clues for how to prove that the converse is not true? $\endgroup$ – Vladimir Nabokov Apr 23 '17 at 10:42
  • $\begingroup$ @N3buchadnezzar is the counter-example I've provided below valid? $\endgroup$ – man_in_green_shirt Apr 23 '17 at 15:15
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    $\begingroup$ What is $Ax_n$? $A$ is supposed to act on sequences of real numbers and $x_n$ is a real number. The exercise asks the following: Let $c$ be the space of convergent sequences. Let $\tilde{x}=(x_n)_n\in c\subseteq \ell_\infty$. Show that $A\tilde{x}$ also belongs in $c$. That is, $A$ leaves $c$ invariant: $A(c)\subseteq c$. What you proved is that if $(\tilde{x}_n)_n$ is a convergent sequence in $\ell_\infty$, (now $(\tilde{x}_n)_n$ is a sequence of sequences) then $(A\tilde{x}_n)_n$ is also convergent, but this is just the def. of sequential continuity and holds for any bounded operator. $\endgroup$ – tree detective Apr 23 '17 at 21:52
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Your proof is correct.

For the converse: define $A(x_1,x_2,...)=(0,...).$ This is linear and certainly bounded and will always have a limit, even if the sequence of $x$'s doesn't.

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