Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose I have a bounded linear operator from a space $X$ to $Y$, both Banach. I know that if D(T) and Ran(T) are closed, then the graph G(T) is closed in $X\times Y$. However is the converse true ? Does G(T) closed imply that Ran(T) is closed ? More generally, is $A, B$ closed if and only if $A\times B$ is closed ? Thank you.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

Every bounded operator has a closed graph, but not necessarily a closed range. Consider for instance the inclusion operator from $C([0,1])$ to $L^1([0,1])$. It is bounded, but its range is $C([0,1])$ which is not closed in $L^1$.

I don't see how your second question ("more generally") is related to this, but the answer is affirmative assuming $A,B$ are both nonempty. Use the fact that the projection maps on a product space are continuous.

share|improve this answer
    
Thank you. I was being led astray by the fact that while the graph is a cartesian product, the two sets are not independent unlike the case of $A \times B$ I mentioned above. –  me10240 Mar 15 '13 at 6:06
    
@me10240: The graph $G(T)$ is a subset of the Cartesian product $X \times Y$, but $G(T)$ is not itself the Cartesian product of any two sets (except in the trivial case $T=0$, when $G(T) = D(T) \times \{0\}$). I think I know what you mean by "independent", but the Cartesian product has a simple definition: $A \times B$ is the set of all the ordered pairs $(a,b)$ where $a \in A$ and $b \in B$. –  Nate Eldredge Mar 15 '13 at 12:46
    
Ah! I was making a mistake there too. Also, the book by Cloud and Vorowich suggested (pg 154) that $G(T) = D(T) \times Ran(T)$ which is where my confusion originated. This seems to be clearly wrong as per the definition of cartesian product. Please correct me if I havent understood it right. –  me10240 Mar 16 '13 at 14:05
    
@me10240: Yes, if it says $G(T) = D(T) \times \operatorname{Ran}(T)$ that is clearly an error. –  Nate Eldredge Mar 16 '13 at 14:51
add comment

Thanks to Nate for his answers, I think I understand it now. I am summarizing it here for the benefit of other readers.

Firstly, $D(T)$ closed, $\mathrm{Ran}(T)$ closed implies $D(T) \times \mathrm{Ran}(T)$ is closed and vice versa provided the sets are non-empty.

However, the graph of the operator $G(T) = \{(x, Tx)| x \in D(T)\}$ being closed does NOT imply that $\mathrm{Ran}(T)$ is closed.

Finally, the book "Functional Analysis" by Cloud & Vorowich has an error on page -154. They state that $G(T) = D(T) \times \mathrm{Ran}(T)$. This is false. $G(T)$ is in fact a subset of $D(T) \times \mathrm{Ran}(T)$

share|improve this answer
add comment

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.