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Suppose that $X$ and $Y$ are two normed spaces over the same field ($\mathbb{R}$ or $\mathbb{C}$). Show that the range of a bounded linear operator $T \colon X \to Y$ need not be closed in $Y$.

Kreyszig gives the following hint:

Consider the operator $T \colon \ell^\infty \to \ell^\infty$ defined by $$Tx := y = (\eta_j), \, \mbox{ where } \, \eta_j := \frac{\xi_j}{j} \, \mbox{ for all } x := (\xi_j) \in \ell^\infty.$$

How do we characterise the range of this operator and show that the range is not closed in $\ell^\infty$?

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  • $\begingroup$ The range contains the standard unit vectors; so, if the range were closed, it would contain $c_0$. Now note $(1/\sqrt j)$ is not in the range. $\endgroup$ Commented Dec 15, 2014 at 18:14
  • $\begingroup$ David, I'm unable to follow your argument. So could you please elaborate? $\endgroup$ Commented Dec 16, 2014 at 4:25
  • $\begingroup$ Is the inverse of the bounded linear operator $T \colon \ell^\infty \to \ell^\infty$ defined by $Tx \colon= y$, wehre $y \colon= (\xi_j/j)$ for all $x \colon= (\xi_j) \in \ell^\infty$, bounded? $\endgroup$ Commented Dec 18, 2014 at 4:04

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Let $e_n$ be the element of $\ell_\infty$ whose $m$'th coordinate is $1$ if $m=n$ and $0$ otherwise. The closed linear span of $\{e_n\mid n\in \Bbb N\}$ in $\ell_\infty$ is the space $c_0$ of sequences that tend to $0$.

For your operator, we have $T(je_j)=e_j$; so the range of $T$ contains each $e_j$. Since the range of a linear operator is a linear space, it follows that the range of $T$ contains the linear span of the $e_j$. So, if the range of $T$ is closed, it must contain the space $c_0$ (in fact it would be equal to $c_0$).

But this is not the case: the vector $y=(1/\sqrt j)$ is in $c_0$ but is not on the range of $T$. If $Tx=y$, then the $j$'th coordinate of $x$ would have to be $\sqrt j$. The sequence $(\sqrt j)\notin\ell_\infty$.

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  • $\begingroup$ David, what do you mean by the term "closed linear span"? And how do we show that the closed linear span of the set $\{ e_n: n \in \mathbb{N} \}$ is the space $c_0$ of all sequences converging to zero? $\endgroup$ Commented Dec 17, 2014 at 8:42
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    $\begingroup$ @SaaqibMahmuud The "closed linear span" is the closure of the linear span. The linear span (set of finite linear combinations) of the $e_n$ is the set of sequences that are eventually $0$. It's not hard to prove that this set is dense in $c_0$. $\endgroup$ Commented Dec 17, 2014 at 9:00
  • $\begingroup$ Yes, the linear span of the set $\{e_n \colon n \in \mathbb{N}\}$ is indeed the set of all sequences that are eventually zero, but how to show that this set is dense in $c_0$? Or, how to show that the closure of the set of all sequences that are eventually zero is the set $c_0$ in $\ell^\infty$? $\endgroup$ Commented Dec 18, 2014 at 3:52
  • $\begingroup$ @SaaqibMahmuud If $x\in c_0$, then the sequence $(y_n)$ with $y_n$ defined by $y_n(i)=x(i)$, $i\le n$, $y_n(i)=0$, $i>0$ converges to $x$. $\endgroup$ Commented Dec 18, 2014 at 7:12

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