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Let $\Omega$ be a bounded Lipschitz domain in two dimensions. Consider the partial differential operator $$\dfrac{\partial}{\partial x}: H^1(\Omega)\to L^2(\Omega),\quad u(x,y)\to \dfrac{\partial u}{\partial x}.$$ Is the range of $\frac{\partial}{\partial x}$ closed in $L^2(\Omega)$ ?

Note: Obviously, the operator is not injective and is not bounded from below, i.e., we do not have for $c>0$ $$||\dfrac{\partial u}{\partial x}||_{L^2}\geq c ||u||_{H^1}\quad \text{or even}\quad ||\dfrac{\partial u}{\partial x}||_{L^2}\geq c ||u||_{L^2}.$$ Hence it is easily seen that the answer from the link in "possible duplicate" does not apply here. An example where the answer from the "possible duplicate" link does not work can be found in my comment.

booksee
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    Possible duplicate of When is the image of a linear operator closed? – Vlad Mar 31 '17 at 06:27
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    @Vlad This is a question about a specific operator, the link you gave is a question about the general situation. – daw Mar 31 '17 at 11:56
  • I have voted to Leave Open (from Review), because I think there is a little bit to be said here about the boundedness of this particular linear operator in view of the proposed hypotheses. Even so, it would be helpful if the OP gave an indication of their level of knowledge of and/or interest in the problem (context), so that Readers are better able to supply a cogent response. – hardmath Mar 31 '17 at 15:56
  • @Vlad The operator in my post is not bounded from below, so the answer in your link does not apply. – booksee Mar 31 '17 at 16:32
  • The reference @Vlad gave was an if and only if, so no, your operator does not have closed range. – Matt Mar 31 '17 at 16:41
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    @Matt Note that in that link, the operator is also injective. The operator here does not have to be injective, so the "only if" part does not apply. – booksee Mar 31 '17 at 17:19
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    @Matt If you want a counter example where the operator has closed range but is neither bounded from below nor injective. Consider the gradient operator from $H^1(\Omega)$ to $L^2(\Omega)^d$ in $d$ dimensions. – booksee Mar 31 '17 at 17:24
  • @hardmath I see, good point. – Vlad Mar 31 '17 at 17:43

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The range contains $C_c^{\infty}(\Omega)$, so if it was closed it would be $L^2(\Omega)$. But it clearly does not contain step functions such as $f(x,y)=1_{y>y_0}$ (the preimages of this function are not absolutely continuous along almost all lines parallel to the $y$ axis, which clashes with the ACL characterization of Sobolev spaces)

Bananach
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  • In fact, the motivation of my question is to check if $\frac{\partial}{\partial x}$ is surjective or not. I'm glad you found it, with a nice example. – booksee Apr 01 '17 at 03:24