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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!

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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 $L^1$.

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  • $\begingroup$ Dear Sir can you please tell me why given operator is linear? I can't see that $\endgroup$ – MathLover May 16 at 10:26
  • $\begingroup$ @SRJ Do you know the definition of "linear"? $T(f_1+f_2)=g(f_1+f_2)=gf_1+gf_2=Tf_1+Tf_2$; similarly for scalar multiplication. $\endgroup$ – David C. Ullrich May 16 at 14:18
  • $\begingroup$ Yes Sir I was confused as you defined g(t) =1/1+t^2.how to show g is linear from here $\endgroup$ – MathLover May 17 at 1:28
  • $\begingroup$ @SRJ Of course $g$ is not linear! I didn't say it was linear. $T$ is linear. $\endgroup$ – David C. Ullrich May 17 at 13:41
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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.

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  • $\begingroup$ How can you proof that the range is dense? I’ve been thinking here, however I couldn’t get anywhere $\endgroup$ – user 242964 Aug 8 at 23:20
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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 $v_j\to v_0$.
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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.

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  • $\begingroup$ @SRJ $\sqrt{j}$ is not bounded $\endgroup$ – Martin Erhardt Aug 6 at 16:24

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