Minimal and maximal domains of the interval Laplacian

Density follows from [F3]. For $f,g\in C_c^{\mathrm{\infty }}(0,L)$, integration by parts gives $⟨-f^{\mathrm{\prime }\mathrm{\prime }},g⟩=⟨f,-g^{\mathrm{\prime }\mathrm{\prime }}⟩$, since both functions and their derivatives vanish near the endpoints. Thus $T_c$ is dense and symmetric.

Suppose $u\in L^2(0,L)$ has distributional second derivative $D^{2}u=h\in L^2(0,L)$. By [F6], $h\in L^1(0,L)$. Define $v(x)={\int }_0^{x}h(t)dt$. Then $\mathrm{\mid }v(x)\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_{L^1}\le \sqrt{L}\mathrm{\parallel }h{\mathrm{\parallel }}_2$, so $v\in L^2(0,L)$. For every $\phi \in C_c^{\mathrm{\infty }}(0,L)$, choose a measurable representative of $h$ and define $G(t,x):={\mathbf{1}}_{\{0<t<x<L\}}h(t){\phi }^{\mathrm{\prime }}(x),(t,x)\in (0,L{)}^2\mathrm{.}$ The function $q(t,x)=\mathrm{\mid }h(t)\mathrm{\mid }\mathrm{\mid }{\phi }^{\mathrm{\prime }}(x)\mathrm{\mid }$ is nonnegative and product-measurable. By [F6], $h\in L^1(0,L)$. The test derivative ${\phi }^{\mathrm{\prime }}$ is continuous with compact support in $(0,L)$, hence belongs to $L^2(0,L)$; [F6] then gives ${\phi }^{\mathrm{\prime }}\in L^1(0,L)$. Tonelli [F8], applied only to this nonnegative product, gives ${\int }_{(0,L{)}^2}qd(\lambda \times \lambda )=({\int }_0^L\mathrm{\mid }h(t)\mathrm{\mid }dt)({\int }_0^L\mathrm{\mid }{\phi }^{\mathrm{\prime }}(x)\mathrm{\mid }dx)<\mathrm{\infty }\mathrm{.}$ Since $\mathrm{\mid }G\mathrm{\mid }\le q$, this establishes $G\in L^1((0,L{)}^2)$ without using Fubini to prove its own hypothesis. Its sections are integrable for every parameter because $h,{\phi }^{\mathrm{\prime }}\in L^1(0,L)$. Fubini [F9] therefore gives ${\int }_0^{L}v(x){\phi }^{\mathrm{\prime }}(x)dx={\int }_0^L{\int }_0^{x}h(t){\phi }^{\mathrm{\prime }}(x)dtdx={\int }_0^{L}h(t)({\int }_t^L{\phi }^{\mathrm{\prime }}(x)dx)dt=-{\int }_0^{L}h(t)\phi (t)dt,$ where [F7] applied to real and imaginary parts gives ${\int }_t^L{\phi }^{\mathrm{\prime }}(x)dx=\phi (L)-\phi (t)=-\phi (t)$, since $\phi$ vanishes near $L$. Hence $Dv=h$ weakly and $v\in H^1$. The distribution $Du-v$ has derivative zero. For every test function $\phi$ of integral zero, its primitive $\psi (x)={\int }_0^x\phi (t)dt$ is compactly supported in $(0,L)$, and hence $(Du-v)(\phi )=(Du-v)(D\psi )=-(D(Du-v))(\psi )=0$. Obtain a test function $\eta$ with integral one as follows. The constant function $1$ lies in $L^2(0,L)$ with norm $\sqrt{L}$. By the density in [F3], choose $\chi \in C_c^{\mathrm{\infty }}(0,L)$ with $\mathrm{\parallel }\chi -1{\mathrm{\parallel }}_2<\sqrt{L}\mathrm{/}2$; then $\mathrm{\parallel }\chi {\mathrm{\parallel }}_2>\sqrt{L}\mathrm{/}2$. Set $\eta =\mathrm{\mid }\chi {\mathrm{\mid }}^2\mathrm{/}\mathrm{\parallel }\chi {\mathrm{\parallel }}_2^2$. It is a smooth compactly supported test function and ${\int }_0^L\eta =1$. For arbitrary $\phi$, $\phi -({\int }_0^L\phi )\eta$ has integral zero, so $(Du-v)(\phi )=(Du-v)(\eta ){\int }_0^L\phi$. Therefore $Du-v$ is a constant distribution, $Du=v+c\in L^2$, and $u\in H^2(0,L)$ by [F1].

The Hermite cubic polynomials $h_{00}(s)=2s^3-3s^2+1$, $h_{01}(s)=-2s^3+3s^2$, $h_{10}(s)=s^3-2s^2+s$, and $h_{11}(s)=s^3-s^2$ satisfy the endpoint data $(h_{00}(0),h_{00}(1),h_{00}^{\mathrm{\prime }}(0),h_{00}^{\mathrm{\prime }}(1))=(1,0,0,0)$, $(h_{01}(0),h_{01}(1),h_{01}^{\mathrm{\prime }}(0),h_{01}^{\mathrm{\prime }}(1))=(0,1,0,0)$, $(h_{10}(0),h_{10}(1),h_{10}^{\mathrm{\prime }}(0),h_{10}^{\mathrm{\prime }}(1))=(0,0,1,0)$, and $(h_{11}(0),h_{11}(1),h_{11}^{\mathrm{\prime }}(0),h_{11}^{\mathrm{\prime }}(1))=(0,0,0,1)$. Consequently, for arbitrary $a_0,a_L,d_0,d_L\in \mathbb{C}$, the polynomial $p(x)=a_0{h}_{00}(x\mathrm{/}L)+a_L{h}_{01}(x\mathrm{/}L)+L{d}_0{h}_{10}(x\mathrm{/}L)+L{d}_L{h}_{11}(x\mathrm{/}L)$ has $p(0)=a_0$, $p(L)=a_L$, $p^{\mathrm{\prime }}(0)=d_0$, and $p^{\mathrm{\prime }}(L)=d_L$. It belongs to $H^2(0,L)$, proving that $\gamma$ is onto.

If $g\in H^2(0,L)$, [F2] applied with a compactly supported test function shows $⟨T_{c}f,g⟩=⟨f,-D^{2}g⟩$; thus $g\in D(T_c^{\ast })$ and $T_c^{\ast }g=-D^{2}g$. Conversely, if $g\in D(T_c^{\ast })$ and $h=T_c^{\ast }g$, testing the adjoint identity against every $f\in C_c^{\mathrm{\infty }}(0,L)$ gives distributionally $D^{2}g=-h\in L^2$. Step 1.2 then gives $g\in H^2(0,L)$. Hence $T_c^{\ast }=T_{\max }$ with exactly the claimed domain.

By [F5] and step 2.1, $T_{\min }=T_c^{\ast \ast }=T_{\max }^{}$. If $u\in D(T_{\max }^{})$ and $h=T_{\max }^{}u$, testing against $C_c^{\mathrm{\infty }}(0,L)\subset H^2(0,L)$ gives distributionally $D^{2}u=-h$. Step 1.2 implies $u\in H^2(0,L)$. The adjoint identity for all $g\in H^2(0,L)$ and [F2] now reads $⟨{\mathrm{\Gamma }}_{1}g,{\mathrm{\Gamma }}_{0}u⟩-⟨{\mathrm{\Gamma }}_{0}g,{\mathrm{\Gamma }}_{1}u⟩=0$. By step 1.3 the two traces of $g$ can be prescribed independently: first set ${\mathrm{\Gamma }}_{0}g=0$ and vary ${\mathrm{\Gamma }}_{1}g$ to obtain ${\mathrm{\Gamma }}_{0}u=0$; then set ${\mathrm{\Gamma }}_{1}g=0$ and vary ${\mathrm{\Gamma }}_{0}g$ to obtain ${\mathrm{\Gamma }}_{1}u=0$. Conversely, if $u\in H^2$ has both traces zero, [F2] gives $⟨T_{\max }g,u⟩=⟨g,-D^{2}u⟩$ for every $g\in H^2$, so $u\in D(T_{\max }^{})$. Therefore $D(T_{\min })=\{u\in H^2:{\mathrm{\Gamma }}_{0}u={\mathrm{\Gamma }}_{1}u=0\}$ and its action is $-D^{2}u$.

The preceding steps establish density, symmetry, the exact adjoint, the closure domain, and trace surjectivity as stated.