Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $L^{p'}$

share|improve this question
    
Your grammar, punctuation, and TeXing were poor. You should take pains to make your question readable before posting. –  Bill Johnson May 22 '12 at 6:30
    
Arg, Bill - I had it all fixed up a moment ago :) –  David Roberts May 22 '12 at 6:30
    
Sorry our edits overlapped. I think it is OK now. I'm done. –  Bill Johnson May 22 '12 at 6:33
    
That's ok. It's your area of expertise, I shall leave it in your more experienced hands. –  David Roberts May 22 '12 at 6:36
    
Sorry for that,I'm trying to do a better job. –  user23078 May 22 '12 at 6:45

1 Answer 1

$\mathcal{F}L^{p}$ 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.

share|improve this answer
    
thanks,and I also learned the nontrivial fact that $L^p$ is not isomorphic to $L^{p'}$ if $p \ne p'$ from mathoverflow.net/questions/79713/lp-mathbbr-vs-lq-mathbbr –  user23078 May 22 '12 at 15:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.