I have $H=L^2(0,2)$ and $Aw=w^{(4)}, D(A)=H^4(0,2)\cap H^2_0(0,2)$, this operator is non-negative and self-adjoint because $A$ is monotone maximal, $R(I+A)=H$.

A form $t(u,v)=(Au,v)_{H}$ is closable, symmetric, definite dense, non-negative, by the first and second representation theorem for sesquilinear forms (Kato, Perturbation theory for linear operators), $D(\overline{t})=D(A^{1/2})=H^2_0(0,2)$, where $\overline{t}$ is a small closed extension of $t$.

By the Spectral theorem, exists $A^{1/2}$, $A^{1/2}\geq 0$ the only operator that $(A^{1/2})^2=A$. Let's $Bu=-u^{(2)}$, with $D(B)=H^2_0(0,2)$; then $B\geq0$ and $B(Bu))=u^{(4)}$; it's $B=A^{1/2}$? according to example 10.4 of Schmüdgen,Unbounded Self-adjoint Operators on Hilbert Space; $B$ is not self-adjoint. I need suggestions to calculate $A^{1/2}$.



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