Closed Operator on a Sobolev space

Question

I am wondering if the following differential operator $A:D(A)( \subset {\bf{H}}) \to {\bf{H}}$ defined on the sobolev space $\mathbf{H}=H_{0}^{k}(0,1)\times {{L}^{2}}(0,1)\text{ }$ is a closed operator or not \begin{align} & A(f,g)=(g,\frac{{{d}^{k}}f}{d{{x}^{k}}}) \\ & D(A)=\left\{ (f,g)\in \mathbf{H}|A(f,g)\in \mathbf{H} \right\} \\ \end{align}

I do appreciate in advance if you help me with this; please give me the name of the Theorem which deals with the closeness of such an operator.

Answer 1

I don't know of a theorem to help you in this case. However your operator is a sum of an inclusion from $L^2$ to $H_0^k$ in the first coordinate and a $k$th derivative into $L^2$ from $H^k$ in the second. The differential part is in fact a bounded operator. Therefore you only need to consider the inclusion. I think you can verify directly that the inclusion is closed - it is not such a hard thing to do, since an inclusion has straightforward action.

Is what I am saying about the decomposition clear?

$$ A = \left( \begin{array}{cc} 0 & I \\ 0 & 0 \\ \end{array} \right) + \left( \begin{array}{cc} 0 & 0 \\ \frac{d^k}{dx^k} & 0 \\ \end{array} \right)$$

Does this seem reasonable? You have to be careful with the domains of sums of unbounded operators, but in this case since the one term is bounded you will be ok.

Answer 2

I found the answer, in a book written by robinson, 2001, "infinite dimensional spaces and systems"

A $k$th order differential operator in the Sobolev space $H^k_0$ a closed operator.

Answer 3

For the closeness you have to pay attention carefully to domains, an operator can be turned into closed operator only by changing the domain of definition.