Semibounded closed-form representation and perturbation

If $H=\{0\}$, density forces $Q=\{0\}$ and the zero operator is the unique self-adjoint representation; the perturbation conclusion is the same. Hence assume $H\mathrm{\ne }\{0\}$.

Put $\mathfrak{a}(u,v)=q(u,v)+(c+1)⟨u,v⟩$ on $Q$. The lower bound gives $\mathfrak{a}[u]\ge \mathrm{\parallel }u{\mathrm{\parallel }}^2$, so $\mathfrak{a}$ is positive definite; its norm is exactly $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{q,c}$, hence $(Q,\mathfrak{a})$ is a Hilbert space. The inclusion $J:Q\to H$ is injective, bounded with $\mathrm{\parallel }J\mathrm{\parallel }\le 1$, and has dense range.

For each $f\in H$, Cauchy–Schwarz [F4] and $\mathrm{\parallel }J\mathrm{\parallel }\le 1$ make $v\mapsto ⟨Jv,f⟩$ a bounded linear functional on $(Q,\mathfrak{a})$. Apply [F2] and let $J^{\ast }f\in Q$ be its unique representing vector: $\mathfrak{a}(v,J^{\ast }f)=⟨Jv,f⟩$ for every $v\in Q$. Uniqueness makes $f\mapsto J^{\ast }f$ linear. Define $R:=J{J}^{\ast }:H\to H$. Taking $v=J^{\ast }f$ gives $\mathrm{\parallel }{J}^{\ast }f{\mathrm{\parallel }}_{q,c}^2=⟨Rf,f⟩$, and $\mathrm{\parallel }Rf\mathrm{\parallel }\le \mathrm{\parallel }{J}^{\ast }f{\mathrm{\parallel }}_{q,c}$; therefore $\mathrm{\parallel }{J}^{\ast }f{\mathrm{\parallel }}_{q,c}\le \mathrm{\parallel }f\mathrm{\parallel }$ and $\mathrm{\parallel }R\mathrm{\parallel }\le 1$.

For $f,g\in H$, the defining identity and symmetry of $\mathfrak{a}$ give $⟨Rf,g⟩=\mathfrak{a}(J^{\ast }f,J^{\ast }g)=⟨f,Rg⟩$, while $⟨Rf,f⟩=\mathrm{\parallel }{J}^{\ast }f{\mathrm{\parallel }}_{q,c}^2\ge 0$; hence $R$ is bounded, positive, self-adjoint, and $R\le I$. If $Rf=0$, then $J^{\ast }f=0$, so $⟨Jv,f⟩=0$ for dense $J(Q)$ and thus $f=0$; consequently $\ker R=\{0\}$ and $\mathrm{ran}R$ is dense.

The bounded positive operator $R$ has a positive square root by [F3]. On $\mathrm{ran}{R}^{1\mathrm{/}2}$ define $U_0(R^{1\mathrm{/}2}f)=J^{\ast }f$. This is well-defined because $R^{1\mathrm{/}2}$ is injective, and $\mathrm{\parallel }{J}^{\ast }f{\mathrm{\parallel }}_{q,c}^2=⟨Rf,f⟩=\mathrm{\parallel }{R}^{1\mathrm{/}2}f{\mathrm{\parallel }}^2$, so $U_0$ is an isometry. Both its domain and range are dense: the first because $\ker R^{1\mathrm{/}2}=0$, the second because a vector in $Q$ orthogonal to all $J^{\ast }f$ has image under $J$ orthogonal to every $f\in H$ and is therefore zero. Thus $U_0$ extends to a unitary $U:H\to (Q,\mathfrak{a})$; taking adjoints in $J^{\ast }=U{R}^{1\mathrm{/}2}$ yields $J=R^{1\mathrm{/}2}{U}^{\ast }$. Hence $J(Q)=\mathrm{ran}{R}^{1\mathrm{/}2}$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{q,c}=\mathrm{\parallel }{R}^{-1\mathrm{/}2}Ju\mathrm{\parallel }$ for every $u\in Q$.

By [F3], define $T=R^{-1}$ by the Borel function $s\mapsto s^{-1}$ on $s>0$ (its value at $0$ is immaterial because $\ker R=0$). Then $T$ is self-adjoint, $T\ge I$, $D(T)=\mathrm{ran}R$, and $D(T^{1\mathrm{/}2})=\mathrm{ran}{R}^{1\mathrm{/}2}$. Put $H_q=T-(c+1)I$. It is self-adjoint and $H_q\ge -cI$. Step 4.1 gives $D((H_q+cI{)}^{1\mathrm{/}2})=D((T-I{)}^{1\mathrm{/}2})=D(T^{1\mathrm{/}2})=J(Q)$, since the spectral weights $s-1$ and $s$ on $[1,\mathrm{\infty })$ are integrable for exactly the same vectors.

The unitary $U^{\ast }:Q\to H$ satisfies $U^{\ast }u=R^{-1\mathrm{/}2}Ju=T^{1\mathrm{/}2}Ju$ by step 4.1. Therefore $\mathfrak{a}(u,v)=⟨T^{1\mathrm{/}2}Ju,T^{1\mathrm{/}2}Jv⟩$ for all $u,v\in Q$. If $u\in D(T)$ and $v\in Q$, the spectral calculus gives $⟨T^{1\mathrm{/}2}Ju,T^{1\mathrm{/}2}Jv⟩=⟨TJu,Jv⟩$; subtracting $(c+1)⟨Ju,Jv⟩$ proves $q(u,v)=⟨H_{q}u,Jv⟩$. Conversely, if $q(u,v)=⟨g,Jv⟩$ for every $v\in Q$, then $\mathfrak{a}(u,v)=⟨g+(c+1)Ju,Jv⟩$, so the uniqueness in the Riesz construction gives $u=J^{\ast }(g+(c+1)Ju)$ and $Ju=R(g+(c+1)Ju)\in D(T)$; the representing vector is $g=H_{q}u$. This proves the stated operator domain. The resolvent $(H_q+(c+1)I{)}^{-1}=R$ is uniquely determined by the form's Riesz equation, so no other self-adjoint operator can represent $q$.

Now let $q_0$ and $v$ satisfy the perturbation hypotheses, and choose $C>b+1$. Then $(1-a)q_0[u]+(C-b)\mathrm{\parallel }u{\mathrm{\parallel }}^2\le (q_0+v)[u]+C\mathrm{\parallel }u{\mathrm{\parallel }}^2\le (1+a)q_0[u]+(C+b)\mathrm{\parallel }u{\mathrm{\parallel }}^2$. Since $1-a>0$ and $C-b>1$, this shifted form norm is bounded above and below by positive multiples of the complete norm $q_0[u]+\mathrm{\parallel }u{\mathrm{\parallel }}^2$. Also $(q_0+v)[u]\ge -b\mathrm{\parallel }u{\mathrm{\parallel }}^2$. Set $N_C=(q_0+v)[u]+C\mathrm{\parallel }u{\mathrm{\parallel }}^2$ and $N_{b+1}=(q_0+v)[u]+(b+1)\mathrm{\parallel }u{\mathrm{\parallel }}^2$. Then $N_C\ge (C-b)\mathrm{\parallel }u{\mathrm{\parallel }}^2$ and $N_{b+1}=N_C-(C-b-1)\mathrm{\parallel }u{\mathrm{\parallel }}^2\ge N_C\mathrm{/}(C-b)$, while $N_{b+1}\le N_C$. Thus the defining shift $b+1$ norm is equivalent and complete, so $q_0+v$ is closed with lower bound $-b$. Applying the representation argument in steps 1.2, 2.1, 3.1, 4.1, 5.1, and 6.1 with lower bound $-b$ gives its unique self-adjoint representing operator bounded below by $-b$.