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Define $T: \ell_\infty \rightarrow \ell_\infty$ the continuous linear operator defined by $$T(x_1,x_2,x_3,\dots) = (x_2,x_3,\dots).$$

Consider the subspace $M$ of $\ell_\infty$ defined by $$M = \{ x- T(x) : x \in \ell_\infty\}.$$

Is it true that $M$ is a closed subspace of $\ell_\infty$? I want to know if $M$ is a Banach space, but I couldn't prove or disprove that.

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  • $\begingroup$ It seems to me that M is all of $\ell^\infty$: given $y$, set $x_1=y_1+y_2,x_2=y_2+y_3$ etc. $\endgroup$ – Ian May 7 '18 at 1:39
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    $\begingroup$ @Ian In this case, $x-Tx=(y_1-y_3,y_2-y_4,\dots)$. Note that $(1,1,1,...) \notin M$. $\endgroup$ – H R May 7 '18 at 1:57
  • $\begingroup$ The sequence $(x_n)_{n\in \mathbb N}$ where $x_n= 1/n$ is also not in M. But I guess if you cut it of from some n onwards it is. That shows the non-closedness. $\endgroup$ – pcp May 7 '18 at 6:39
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The sequence $(x_n)_{n\in\mathbb N}$ where $x_n=1/n$ is not in $M$.

In the other hand, the sequences $(x_n^m)_{n\in \mathbb N}$ where $x_n^m = x_n$ for $n\leq m$ and $x_n^m = 0$ else, are in $M$.

The convergence of $ (x_n^m)_{N \in \mathbb N}$ to $(x_n)_{n\in\mathbb N}$ in $\ell^\infty$ is clear.

In other words, $M$ is not closed.

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    $\begingroup$ $y=(y_n)_n\in M\;$ iff $(\;y\in l_{\infty}$ and $\;\sup_{n\in\Bbb N}\sum_{j=1}^ny_j<\infty\;).$.................+1 $\endgroup$ – DanielWainfleet May 7 '18 at 6:59

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