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The closed range theorem tells us that given two banach spaces X,Y,and a closed densely defined linear operator T:$X \to Y$. We have the following equivalence $R(T)$ is closed in $Y \iff R(T^{*})$ is closed in $X^{*} \iff R(T)=N(T^{*})^{\perp}$. This gives a complete characterisation, but I don't know whether it's convenient or not for application.

Question 1: Can anyone pose some examples to use this theorem ?

When we consider the case that $T$ is a bounded operator, it's true that if the range $TX$ has finite codimension in $Y$, then $TX$ is closed. So it seems that there are lots of bounded operators whose range is not closed. For instance, when $X=L^{p},1 \leq p<2$, $Y=L^{p'}$, where p' is the dual index. $T$ represents the fourier transform $\mathcal{F}$. Then

Question 2: Is $\mathcal{F}L^{p}$ closed in $L^{p'}$? I believe the range is not closed. BTW, I only know that the map is not surjective, if it is not closed, then we can see that actually the range is much smaller than $L^{p'}$

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  • $\begingroup$ Your grammar, punctuation, and TeXing were poor. You should take pains to make your question readable before posting. $\endgroup$ May 22, 2012 at 6:30
  • $\begingroup$ Arg, Bill - I had it all fixed up a moment ago :) $\endgroup$ May 22, 2012 at 6:30
  • $\begingroup$ Sorry our edits overlapped. I think it is OK now. I'm done. $\endgroup$ May 22, 2012 at 6:33
  • $\begingroup$ That's ok. It's your area of expertise, I shall leave it in your more experienced hands. $\endgroup$ May 22, 2012 at 6:36
  • $\begingroup$ Sorry for that,I'm trying to do a better job. $\endgroup$
    – user23078
    May 22, 2012 at 6:45

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$\mathcal{F}L^{p}$ is not closed in $L^{p'}$ unless $p=2$. $\mathcal{F}$ is one to one and for $1<p $ the space $\ell_p$ embeds isomorphically into $L^p$ but not into $L^{p'}$. Alternatively, $\mathcal{F}$ has dense range but $L^p$ is not isomorphic to $L^{p'}$. For $p=1$ use the fact that the closure of $\mathcal{F}L^{1}$ is a $C(K)$ space and thus contains a subspace isomorphic to $c_0$ while $c_0$ does not embed into $L^1$. Or see this MO question.

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