Let $\Delta$ be the operator on $L^2(0, \infty)$ defined as follows: $\Delta \phi:= \phi''$, with domain $D(\Delta):=C^\infty_0(0, \infty)$. Is $\Delta$ closed or closable? In the case, what is its closure?
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does $C_0^\infty(0,\infty)$ denote smooth functions with compact support? – Yurii Savchuk Sep 09 '13 at 13:23
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Yes, ************ – BGA Sep 09 '13 at 19:12
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The answer is yes and its closure is the Laplacian on $H^1_0(0, \infty) \cap H^2(0, \infty)$. To see this simply note that the graph of the closure of an operator is, by definition, the operator whose graph is the closure of the graph of the original operator under the graph norm $\|f\|_\Delta = \|f\|_2 + \|\Delta f\|_2$. More explicitly, if we let $G(T)$ denote the graph of $T$ we have $$G(\overline{\Delta }) = \overline{G(\Delta)}= \operatorname{cl} \{(f, f''): f \in C^\infty_0(0,\infty)\} = \{(f, f''): f \in H^1_0(0, \infty) \cap H^2(0, \infty)\}$$ since the graph norm $\|\cdot \|_\Delta$ is equivalent to the Sobolev norm.
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Thank you brom. I have to say that I must study better Sobolev spaces in order to fully undertand your answer. – BGA Sep 10 '13 at 18:12