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Define a function $A: l^\infty\rightarrow l^\infty$ by $$A(x_1, x_2,x_3,...)=A(x_1, \frac{x_1+x_2}{2},\frac{x_1+x_2+x_3}{3},...).$$ Suppose $(x_1, x_2,x_3,...)$ has a limit, then I want to show that $A(x_1, x_2,x_3,...)$ also has a limit. I also want to show that the converse doesn't necessarily hold.

Proof: Let $x=\lim_{n\rightarrow\infty}x_n$. I claim that $A(x)=\lim_{n\rightarrow\infty}A(x_n)$. One can show (relatively-easily) that A is a bounded linear transformation. In fact, the norm of $A$ is $1$. Thus,$$||A(x_n)-A(x)||=||A(x_n-x)||\leq||x_n-x||\rightarrow0.$$

Now, in order to show the converse doesn't hold. Consider the sequences $a = (a_n)_{n\in\mathbb{N}}$ defined by $a_n=(-1)^n$. Clearly $a$ hasn't got a limit but $A(a)=(-1,0,-\frac{1}{3},0,-\frac{1}{5},...)$ converges to $0$.

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  • $\begingroup$ Sometimes the domain of A seems to be the scalar field, instead of the sequence space. Please check. $\endgroup$ – Simon May 13 '17 at 11:38
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    $\begingroup$ Your counterexample is correct, but the first part is still wrong. Check my comment on your previous question for the reason. You need to show that if $(x_n)_{n\in \mathbb{N}}$ is a convergent sequence of real numbers, then the sequence of the Cesaro averages of $(x_n)_{n\in\mathbb{N}}$ is also convergent. There's a hint: math.stackexchange.com/questions/207910/… $\endgroup$ – tree detective May 13 '17 at 11:45

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