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I've been trying to get some intuition on what it means for a bounded linear operator to have closed range. Can anyone give some simple examples of such an operator that does not have closed range?

Thanks!

Martin Sleziak
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Presquevu
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4 Answers4

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Take the bounded linear operator $$T: l^\infty \ni (x_j) \mapsto \left(\dfrac{x_j}{j}\right)\in l^\infty.$$

Then $x^{(n)}=(\sqrt{1},\sqrt{2},\dots,\sqrt{n},0,0,\dots)$ is in $l^\infty$ with $Tx^{(n)}=(1/\sqrt{1},\dots,1/\sqrt{n},0,\dots)$.

We have $Tx^{(n)} \to y=(1/\sqrt{j})$ in $l^\infty$, but $y \notin T(l^\infty)$ as $(\sqrt{j})\notin l^\infty$. So $T(l^\infty)$ is not closed.

Judge Rinder
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    @SRJ $\sqrt{j}$ is not bounded – Martin Erhardt Aug 06 '19 at 16:24
  • Great answer :) thanks. This happened to be a question from a practice exam I was struggling with. –  Dec 07 '20 at 01:56
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    can't you also more simply take $x^{(n)}=(1,2,...,n,0,0,...)$ to show the same thing? – glS Aug 15 '24 at 10:51
  • @glS You cannot. If you define $x^{(n)}$ that way then $Tx^{(n)}$ doesn't have limit points in $\ell^\infty$. (The candidate is the sequence of just 1s, but each $Tx^{(n)}$ is always 1 away in the $\ell^\infty$ norm.) You need to take the square root so that the difference of $Tx^{(n)}$ is $1/\sqrt{n}$, which we can make arbitrarily small. – beeclu Jan 28 '25 at 16:18
  • @beeclu ah, you're right! thanks – glS Jan 29 '25 at 07:26
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Define $T:L^1(\Bbb R)\to L^1(\Bbb R)$ by $Tf = gf$, where $$g(t)=\frac{1}{1+t^2}.$$Functions with compact support are dense in $L^1$, hence $T(L^1)$ is dense in $L^1$. But you can easily find an example showing that $T(L^1)\ne L^1$; hence $T(L^1)$ is not closed in $L^1$.

Andrea
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David C. Ullrich
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Take an operator $T\colon \ell_\infty \to c_0$ given by

$$T(\xi_n)_{n=1}^\infty = (\frac{\xi_n}{n})_{n=1}^\infty\quad (\xi_n)_{n=1}^\infty\in \ell_\infty).$$

Certainly $T$ is injective and has dense range. However, it cannot have closed range as it would be an isomorphism which is impossible as $\ell_\infty$ is non-separable.

This operator is actually compact. Compactness is a handy condition thay prevents operators whose range is not finite-dimensional to have closed range.

Tomasz Kania
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This is an example from Booss Bleecker. Let $\mathcal{H}:=l^2(\mathbb{N}_{\geq1})$ and $\{\delta_x\}_{x\in\mathbb{N}_{\geq1}}$ be the orthonormal basis of $\mathcal{H}$. Define $A:\mathcal{H}\to\mathcal{H}$ by: $$ A:=\sum_{x\in\mathbb{N}_{\geq1}}\frac{1}{x}\delta_x<\delta_x,\cdot>$$

Verify that:

  1. $A\in\mathcal{B}(\mathcal{H})$ (with $||A||=1$)
  2. $v\in im(A)$ if and only if $\sum_{x\in\mathbb{N}_{\geq1}}x^2|<\delta_x,v>|^2 < \infty$.
  3. With $v_0 := \sum_{x\in\mathbb{N}_{\geq1}}\frac{1}{x^{\frac{3}{2}}}\delta_x$ and $\{v_j\}_{j\in\mathbb{N}_{\geq1}}$ with $v_j := \sum_{x\in\mathbb{N}_{\geq1}}\frac{1}{x^{\frac{3}{2}+\frac{1}{j}}}\delta_x$ we have $v_0\notin im(A)$ yet $v_j \in im(A)$ and $v_j\to v_0$.
PPR
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