L^2 potential infinitesimal kinetic bound

Maximal free multiplier and its actual domain: put $a(\xi )=4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$, and define $M_a$ on $D(M_a)=\{f\in L^2:af\in L^2\}$. This domain is dense because $1_{a\le n}f\in D(M_a)$ and these cutoffs converge to any $f\in L^2$ by dominated convergence. It is symmetric because $a$ is real. If $g\in D(M_a^{\ast })$ and $h=M_a^{\ast }g$, test the adjoint identity against every $L^2$ vector supported on $\{a\le n\}$; multiplication by $a$ is bounded there, and the identity gives $1_{a\le n}h=a{1}_{a\le n}g$. These sets exhaust the space, so $ag=h\in L^2$ and $g\in D(M_a)$. Conversely, $ag\in L^2$ gives the adjoint identity by Cauchy-Schwarz. Thus $M_a=M_a^{\ast }$ including domains. By unitary transport and [F1], $H_0={\mathcal{F}}_2^{\ast }{M}_a{\mathcal{F}}_2$ is self-adjoint on precisely $W^{2,2}$ and agrees with the weak Laplacian. Its explicit inverse at $z=\pm i\mu$, $\mu >0$, is $Q_z={\mathcal{F}}_2^{\ast }{M}_{(a-z{)}^{-1}}{\mathcal{F}}_2$; one has $\mathrm{\parallel }{Q}_z\mathrm{\parallel }\le {\mu }^{-1}$ and $\mathrm{\parallel }{H}_0{Q}_z\mathrm{\parallel }\le 1$. At $z=-s$, $s>0$, the same formula gives $\mathrm{\parallel }{Q}_{-s}\mathrm{\parallel }\le s^{-1}$ and $\mathrm{\parallel }{H}_0{Q}_{-s}\mathrm{\parallel }\le 1$. Each inverse maps onto the exact free domain, as multiplying by $a-z$ verifies both inverse equations.

Fourier sup interpolation, with the representative justified: write $\widehat{u}={\mathcal{F}}_{2}u$. By [F2], ${\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le R}\mathrm{\mid }\widehat{u}\mathrm{\mid }\le \sqrt{4\pi \mathrm{/}3}{R}^{3\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_2$ and ${\int }_{\mathrm{\mid }\xi \mathrm{\mid }>R}\mathrm{\mid }\widehat{u}\mathrm{\mid }\le \sqrt{4\pi }{R}^{-1\mathrm{/}2}\mathrm{\parallel }\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}{\mathrm{\parallel }}_2=\sqrt{4\pi }{R}^{-1\mathrm{/}2}\mathrm{\parallel }{H}_{0}u{\mathrm{\parallel }}_2\mathrm{/}(4{\pi }^2)$. In particular $\widehat{u}\in L^1$. The absolutely convergent inverse integral $v(x)=\int \widehat{u}(\xi )e^{2\pi ix\cdot \xi }d\xi$ defines a bounded continuous function. For any compact smooth test function $\phi$, Fubini gives $\int v\bar{\phi }=\int \widehat{u}\bar{\widehat{\phi }}=⟨u,\phi ⟩$ by Plancherel; absolute integrability for Fubini follows from $\mathrm{\parallel }\widehat{u}{\mathrm{\parallel }}_1\mathrm{\parallel }\phi {\mathrm{\parallel }}_1<\mathrm{\infty }$. Thus $v-u$ has zero distribution and [F2] gives $v=u$ almost everywhere. Taking the essential supremum of its inverse integral yields (a), with $C=\max \{\sqrt{4\pi \mathrm{/}3},\sqrt{4\pi }\mathrm{/}(4{\pi }^2)\}$, and $\mathrm{\parallel }{H}_{0}u{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_2$.

Full unbounded potential bound and symmetry: $V$ is finite almost everywhere and $\mathrm{\parallel }Vu{\mathrm{\parallel }}_2\le \mathrm{\parallel }V{\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $u\in W^{2,2}$, so $D(H_0)\subseteq D(M_V)$. Step 2.1 gives $\mathrm{\parallel }Vu{\mathrm{\parallel }}_2\le C\mathrm{\parallel }V{\mathrm{\parallel }}_2{R}^{-1\mathrm{/}2}\mathrm{\parallel }{H}_{0}u{\mathrm{\parallel }}_2+C\mathrm{\parallel }V{\mathrm{\parallel }}_2{R}^{3\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Given $\eta >0$, choose $R=\max \{1,(C\mathrm{\parallel }V{\mathrm{\parallel }}_2\mathrm{/}\eta {)}^2\}$ and set $b_{\eta }=C\mathrm{\parallel }V{\mathrm{\parallel }}_2{R}^{3\mathrm{/}2}$; then the asserted relative estimate holds, including $V=0$, and the relative bound is $0$. Realness of $V$ gives $⟨Vu,w⟩=⟨u,Vw⟩$ for $u,w\in D(M_V)$: the two equal integrands are absolutely integrable by the $L^2$ Cauchy-Schwarz bounds. In particular the restriction of $M_V$ to $D(H_0)$ is symmetric, even when $V$ is unbounded. The maximal multiplier itself has no artificial bounded-potential hypothesis.

Two nonreal Neumann inverses on the exact domain: fix $\eta =1\mathrm{/}4$ in step 3.1 and write $b=b_{1\mathrm{/}4}\ge 0$. For $z=\pm i\mu$ with $\mu >4b$, step 1.1 gives $\mathrm{\parallel }{M}_V{Q}_z\mathrm{\parallel }\le 1\mathrm{/}4+b\mathrm{/}\mu <1\mathrm{/}2$. Hence $L_z=I+M_V{Q}_z$ has the norm-convergent inverse ${\sum }_{n\ge 0}(-M_V{Q}_z{)}^n$, whose norm is at most $2$; multiplication of partial geometric sums and passage to the norm limit verify both inverse identities. On $D(H_0)$ the exact factorization is $H-z=L_z(H_0-z)$. Therefore $X_z=Q_z{L}_z^{-1}$ maps $L^2$ into $D(H_0)$ and is a two-sided inverse of $H-z:D(H_0)\to L^2$, with $\mathrm{\parallel }{X}_z\mathrm{\parallel }\le 2\mathrm{/}\mu$. Both signs use the same estimate; the order $Q_z{L}_z^{-1}$ is essential. This is an explicit domain argument, not an invocation of Kato-Rellich.

All sufficiently negative real inverses: for every $s>4b$, the explicit $Q_{-s}=(H_0+s{)}^{-1}$ from step 1.1 similarly satisfies $\mathrm{\parallel }{M}_V{Q}_{-s}\mathrm{\parallel }\le 1\mathrm{/}4+b\mathrm{/}s<1\mathrm{/}2$. The same geometric series gives the inverse of $I+M_V{Q}_{-s}$ and the factorization $H+s=(I+M_V{Q}_{-s})(H_0+s)$ on $D(H_0)$. Thus every real $\lambda =-s<-4b$ lies in the resolvent set, with bounded inverse $X_{\lambda }=Q_{-s}(I+M_V{Q}_{-s}{)}^{-1}$ and norm at most $2\mathrm{/}s$. This establishes a whole negative half-line of resolvent points; a single negative resolvent point alone would not prove a lower bound.

Exact adjoint domain: $H$ is densely defined because $D(H)=D(H_0)$ is dense, and it is symmetric by steps 1.1 and 3.1. Let $y\in D(H^{\ast })$ and choose $\mu >4b$. Step 4.1 supplies $x\in D(H_0)$ solving $(H-i\mu )x=(H^{\ast }-i\mu )y$. Symmetry gives $H^{\ast }x=Hx$, so $w=y-x$ satisfies $(H^{\ast }-i\mu )w=0$. For every $v\in D(H)$ the adjoint identity implies $⟨(H+i\mu )v,w⟩=0$; since $H+i\mu$ is onto by the other sign of step 4.1, $w=0$. Hence $y=x\in D(H)$ and $H^{\ast }y=Hy$. Together with symmetry this proves $H=H^{\ast }$ including the exact domain $D(H_0)$, for every real $L^2$ potential. No extra separability of a potential range or boundedness of $V$ was introduced.

Negative spectral support and lower bound under CC: apply [F3] to the self-adjoint $H$ of step 5.1 on the fixed-dense-sequence space $L^2({\mathbb{R}}^3)$. Put $L=4b+1$. For each $\lambda <-L$, step 4.2 gives a bounded inverse $X_{\lambda }$ and hence $\mathrm{\parallel }(H-\lambda )v\mathrm{\parallel }\ge c_{\lambda }\mathrm{\parallel }v\mathrm{\parallel }$ on $D(H)$, where $c_{\lambda }=\mathrm{\parallel }{X}_{\lambda }{\mathrm{\parallel }}^{-1}>0$. For $I_{\lambda }=(\lambda -c_{\lambda }\mathrm{/}2,\lambda +c_{\lambda }\mathrm{/}2)$, a vector in the range of $E_H(I_{\lambda })$ lies in $D(H)$ because that interval is bounded; its spectral norm formula gives $\mathrm{\parallel }(H-\lambda )v\mathrm{\parallel }\le (c_{\lambda }\mathrm{/}2)\mathrm{\parallel }v\mathrm{\parallel }$, so $v=0$ and $E_H(I_{\lambda })=0$. Consider all rational-endpoint intervals $J\subset (-\mathrm{\infty },-L)$ with $E_H(J)=0$. They form a countable collection and cover $(-\mathrm{\infty },-L)$: at each $\lambda$ choose rational endpoints enclosing it within the open spectral hole $I_{\lambda }\cap (-\mathrm{\infty },-L)$. This existence argument chooses no uncountably indexed family; taking all such rational intervals is canonical. Countable subadditivity of each scalar spectral measure yields $E_H((-\mathrm{\infty },-L))=0$. Therefore for $u\in D(H)$, $⟨Hu,u⟩=\int td{\mu }_u(t)\ge -L\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$; the integral is absolutely convergent by its second-moment domain condition and Cauchy-Schwarz. This proves boundedness below. Parts (a), (b) and (c) are respectively steps 2.1, 3.1 and 5.1 together with this step.