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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.

Martin Sleziak
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me10240
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2 Answers2

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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.

Nate Eldredge
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  • 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 06:06
  • @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
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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 $D(T) \times \mathrm{Ran}(T)$

me10240
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