Semibounded forms, variational spectra and Birman–Schwinger

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 $-b$.

For every $f\in H$, [F2] gives $\mathrm{\parallel }{P}_{g}f\mathrm{\parallel }=\mathrm{\mid }⟨f,g⟩\mathrm{\mid }\mathrm{\parallel }g\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}^2\mathrm{\parallel }f\mathrm{\parallel }$. Also $⟨P_{g}f,h⟩=⟨f,g⟩⟨g,h⟩=⟨f,P_{g}h⟩$, so $P_g$ is bounded and self-adjoint, including $g=0$. Thus $\lambda P_g$ is symmetric on $D(H_0)$ with relative bound zero; [F1] implies that $H_{\lambda }$ is self-adjoint on $D(H_0)$.

Fix $z\in \rho (H_0)$ and write $R_0=R_0(z)$, $m=⟨R_{0}g,g⟩$, and $X=R_0{P}_g$. Then $Xf=⟨f,g⟩R_{0}g$ and $X^{2}f=⟨f,g⟩⟨R_{0}g,g⟩R_{0}g=mXf$; hence $X^2=mX$. If $d=1+\lambda m\mathrm{\ne }0$, direct multiplication gives $(I+\lambda X{)}^{-1}=I-\lambda X\mathrm{/}d$.

Define $S=(I-\lambda X\mathrm{/}d)R_0$. Its range is contained in $D(H_0)$. Since $P_g{R}_0{P}_g=m{P}_g$, on $D(H_0)$ one has $S(H_{\lambda }-zI)=(I-\lambda X\mathrm{/}d)(I+\lambda X)=I$. On $H$ one has $(H_{\lambda }-zI)S=I-\lambda P_g{R}_0\mathrm{/}d+\lambda P_g{R}_0-{\lambda }^{2}m{P}_g{R}_0\mathrm{/}d=I$, using $d=1+\lambda m$. Thus $S$ is a bounded two-sided inverse from $H$ onto $D(H_{\lambda })$, and its expansion is the stated resolvent formula.

Let $E<\inf \sigma (H_0)$ be real. By [A1], $R_0(E)$ exists. If $H_{\lambda }u=Eu$ with $u\mathrm{\ne }0$, then $u=-\lambda ⟨u,g⟩R_0(E)g$. The coefficient $⟨u,g⟩$ cannot be zero, because that would give $u=0$; applying $⟨\cdot ,g⟩$ therefore yields $1+\lambda ⟨R_0(E)g,g⟩=0$. Conversely, if this scalar is zero, then $u=R_0(E)g\mathrm{\ne }0$ and $(H_{\lambda }-EI)u=g+\lambda ⟨R_0(E)g,g⟩g=0$. Every eigenvector satisfies the same scalar-multiple formula for $R_0(E)g$, so the eigenspace is one-dimensional.

Define $q_0$ on $Q_0=W^{1,2}$ by the displayed weak-derivative pairing. By [F2], for $u,v\in Q_0$, $q_0(u,v)={\int }_{{\mathbb{R}}^3}4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}(\xi )\overline{\widehat{v}(\xi )}d\xi \mathrm{.}$ Thus $q_0$ is sesquilinear, symmetric and nonnegative. Moreover $q_0[u]+\mathrm{\parallel }u{\mathrm{\parallel }}_2^2=\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,2}}^2$, which is complete by [F3], so $q_0$ is closed by [F4]. Since Schwartz functions lie in $Q_0$ and are dense in $L^2$ by [F5], $Q_0$ is dense.

If $u\in D(H_0)=W^{2,2}$ and $v\in Q_0$, Plancherel [F2] and Cauchy–Schwarz [F6] give $q_0(u,v)=\int 4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}\bar{\widehat{v}}=⟨H_{0}u,v⟩$; both factors in this pairing are in $L^2$ by the $W^{2,2}$ and $W^{1,2}$ hypotheses.

Conversely, suppose $u\in Q_0$ and $f\in H$ satisfy $q_0(u,v)=⟨f,v⟩$ for every $v\in Q_0$. Testing with Schwartz $v$ and using Plancherel gives $\int (4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}(\xi )-\widehat{f}(\xi ))\overline{\widehat{v}(\xi )}d\xi =0$. The Fourier transform maps Schwartz space onto itself by [F5], and complex conjugation preserves Schwartz space, so the locally integrable function $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}-\widehat{f}$ pairs to zero with every compactly supported smooth test function. It is locally integrable because $\widehat{u},\widehat{f}\in L^2$ and the polynomial weight is bounded on compact sets. Injectivity of the regular-distribution map [F7] gives $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}=\widehat{f}$ almost everywhere; hence $\mathrm{\mid }\xi {\mathrm{\mid }}^2\widehat{u}\in L^2$. By [F3], $u\in W^{2,2}=D(H_0)$, and [F1] then gives $H_{0}u=f$. Thus the form-associated operator domain and action are exactly those of $H_0$.

Steps 1.2-1.3 show that the nonnegative self-adjoint operator $H_0$ is the operator represented by the dense closed form $q_0$. Uniqueness in [F4] therefore gives $Q(H_0)=D(H_0^{1\mathrm{/}2})=Q_0$, completing the Statement.

By Choice [A1], [F1] and the spectral calculus give, for $u\in Q(A)$, $q_A[u]=\int \lambda d{E}_u(\lambda )$; the lower bound ensures this integral is finite after the stated shift. The form norm in the Statement is exactly $\mathrm{\parallel }(A+(c+1)I{)}^{1\mathrm{/}2}u\mathrm{\parallel }$.

Fix $t<\alpha$. Since $A\ge -cI$, the spectral points in $(-\mathrm{\infty },t]$ lie in the compact interval $[-c,t]$. If this interval contained infinitely many spectral points, compactness of the interval and closedness of $\sigma (A)$ would give an accumulation point $s\le t$ in $\sigma (A)$. Such an accumulation point is essential by [F2], contradicting $s\le t<\alpha$. Thus there are finitely many spectral points below $t$, each isolated with finite multiplicity. By [F3], $E_A((-\mathrm{\infty },t))$ is supported on their eigenspaces, so it has finite rank.

Now assume the form-domain inclusion is compact. The map $(A+(c+1)I{)}^{-1}:H\to Q(A)$ is bounded because $\mathrm{\parallel }(A+(c+1)I{)}^{1\mathrm{/}2}(A+(c+1)I{)}^{-1}f\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }$ by [F3]. Composing with the compact inclusion $Q(A)\hookrightarrow H$ shows that this resolvent is compact by [F5]. It is positive, self-adjoint and injective. By [F4], its nonzero eigenspaces have finite dimension and span $H$. On each such subspace, $A=(A+(c+1)I)-(c+1)I$ has eigenvalue $r^{-1}-(c+1)$ when the resolvent eigenvalue is $r>0$. The compact spectral theorem says these $r$ can accumulate only at zero; therefore the resulting eigenvalues of $A$ have finite multiplicity and can accumulate only at $+\mathrm{\infty }$. This proves the compact-embedding conclusion.

Define ${\nu }_k=\inf \{t\in \mathbb{R}:\mathrm{rank}{E}_A((-\mathrm{\infty },t))\ge k\}$, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. If there are at least $k$ eigenvalues below $\alpha$, their finite-rank spectral projections show that ${\nu }_k$ is the $k$th such eigenvalue counted with multiplicity. If fewer exist, no $t<\alpha$ has rank at least $k$. When $\alpha <\mathrm{\infty }$, closedness of ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$ gives $\alpha \in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$, so every $t>\alpha$ contains an interval about $\alpha$ inside $(-\mathrm{\infty },t)$ whose spectral projection has infinite rank by [F2]; hence ${\nu }_k=\alpha$. When $\alpha =+\mathrm{\infty }$, the spectrum consists only of discrete eigenvalues; if fewer than $k$ exist, their finite-dimensional eigenspaces span $H$ by [F3], so $\dim H<k$, no $k$-dimensional subspace exists, and ${\nu }_k=+\mathrm{\infty }$. Thus ${\nu }_k$ has exactly the target value in every case.

Let $L\subset Q(A)$ have dimension $k$, and set $m_L={\sup }_{u\in L,\mathrm{\parallel }u\mathrm{\parallel }=1}{q}_A[u]$. By finite-dimensional Gram--Schmidt, choose an orthonormal basis $e_1,\dots ,e_k$ of $L$. For a unit vector $u={\sum }_i{a}_i{e}_i$, each $\mathrm{\mid }{a}_i\mathrm{\mid }\le 1$ and $q_A[u]={\sum }_{i,j}{a}_i\overline{a_j}{q}_A(e_i,e_j)$, so $\mathrm{\mid }{q}_A[u]\mathrm{\mid }\le {\sum }_{i,j}\mathrm{\mid }{q}_A(e_i,e_j)\mathrm{\mid }<\mathrm{\infty }$ uniformly on the unit sphere. Thus $m_L<\mathrm{\infty }$. For any $t>m_L$, if $\mathrm{rank}{E}_A((-\mathrm{\infty },t))<k$, the projection restricted to $L$ has a nonzero kernel; choose a unit $u\in L$ with $E_A((-\mathrm{\infty },t))u=0$. Its spectral measure is supported in $[t,\mathrm{\infty })$, so [F1] gives $q_A[u]\ge t>m_L$, a contradiction. Hence $\mathrm{rank}{E}_A((-\mathrm{\infty },t))\ge k$, so ${\nu }_k\le t$ for every $t>m_L$ and therefore ${\nu }_k\le m_L$. Taking the infimum over $L$ yields ${\nu }_k\le {\lambda }_k(A)$.

If ${\nu }_k<\mathrm{\infty }$ and $t>{\nu }_k$, monotonicity of the interval projections gives some $s<t$ with $\mathrm{rank}{E}_A((-\mathrm{\infty },s))\ge k$. Choose a $k$-dimensional subspace $L$ of this range. Its spectral support is bounded below by $-c$ and above by $s$, hence $L\subset Q(A)$; [F1] also gives $q_A[u]\le s\mathrm{\parallel }u{\mathrm{\parallel }}^2$ on $L$. Thus ${\lambda }_k(A)\le s<t$; letting $t$ decrease to ${\nu }_k$ gives ${\lambda }_k(A)\le {\nu }_k$. If ${\nu }_k=+\mathrm{\infty }$, Step 2.1 shows $\dim H<k$, so the variational infimum is also $+\mathrm{\infty }$. Combining with Step 3.1 gives ${\lambda }_k(A)={\nu }_k$, proving the min-max identity and its spectral interpretation.

Steps 1.1-4.1 establish the form min-max characterization, including the essential-spectrum threshold and the empty-spectrum convention; Step 1.3 proves the complete discrete eigenbasis under compact form-domain embedding.

Spherical coordinates give $\mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_1={\int }_0^{\mathrm{\infty }}r{e}^{-\kappa r}dr={\kappa }^{-2},\mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_2^2=\frac{1}{4\pi }{\int }_0^{\mathrm{\infty }}{e}^{-2\kappa r}dr=\frac{1}{8\pi \kappa }\mathrm{.}$ Thus both norms are finite; the point value at the origin does not affect the $L^p$ class.

For $\rho =\mathrm{\mid }\xi \mathrm{\mid }>0$, [F1] and the elementary integral ${\int }_0^{\mathrm{\infty }}{e}^{-\kappa r}\sin (ar)dr=a\mathrm{/}({\kappa }^2+a^2)$ for $a>0$ give ${\widehat{G}}_{\kappa }(\xi )=\frac{1}{2\pi \rho }{\int }_0^{\mathrm{\infty }}{e}^{-\kappa r}\sin (2\pi \rho r)dr=\frac{1}{{\kappa }^2+4{\pi }^2{\rho }^2}\mathrm{.}$ At $\xi =0$, direct integration gives ${\widehat{G}}_{\kappa }(0)={\int }_0^{\mathrm{\infty }}r{e}^{-\kappa r}dr={\kappa }^{-2}$, the same formula.

Put $M=\mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_1>0$ and $A(x)=\int G_{\kappa }(y)\mathrm{\mid }f(x-y){\mathrm{\mid }}^{2}dy$. Tonelli and translation invariance [F2, F3] give $\int A(x)dx=M\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$, so $A(x)<\mathrm{\infty }$ for almost every $x$. For each such $x$, $\int G_{\kappa }(y)\mathrm{\mid }f(x-y)\mathrm{\mid }dy\le (M+A(x))\mathrm{/}2<\mathrm{\infty }$ by $2t\le 1+t^2$, so define $b(x)=\int G_{\kappa }(y)f(x-y)dy$. Expanding the nonnegative integral $\int G_{\kappa }(y)\mathrm{\mid }f(x-y)-b(x)\mathrm{/}M{\mathrm{\mid }}^{2}dy$ gives $0\le A(x)-\mathrm{\mid }b(x){\mathrm{\mid }}^2\mathrm{/}M$, hence $\mathrm{\mid }b(x){\mathrm{\mid }}^2\le MA(x)$. Integrating yields $\mathrm{\parallel }{G}_{\kappa }\ast f{\mathrm{\parallel }}_2^2\le M^2\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. Therefore convolution is defined almost everywhere and gives a bounded operator on $L^2$.

If $f$ is Schwartz, then $G_{\kappa }\ast f\in L^1$ by Fubini and translation invariance [F3], and lies in $L^2$ by step 2.2. Fubini in the ordinary Fourier integral gives the integral-transform identity $\widehat{G_{\kappa }\ast f}={\widehat{G}}_{\kappa }\widehat{f}$. Since $G_{\kappa }\ast f\in L^1\cap L^2$, [F4] identifies its integral transform with its Plancherel transform almost everywhere; the same interface identifies $\widehat{f}$ with ${\mathcal{F}}_{2}f$ almost everywhere. Thus ${\mathcal{F}}_2(G_{\kappa }\ast f)={\widehat{G}}_{\kappa }{\mathcal{F}}_{2}f$ as $L^2$ classes. Since $\mathrm{\mid }{\widehat{G}}_{\kappa }(\xi )\mathrm{\mid }\le \mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_1$ by the integral definition of the Fourier transform, multiplication by ${\widehat{G}}_{\kappa }$ is bounded on $L^2$. Schwartz density and boundedness of convolution therefore extend the identity to all $f\in L^2$.

By [A1], $-\mathrm{\Delta }$ is the multiplier $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$. Therefore $(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2+{\kappa }^2{)}^{-1}$ is the bounded inverse multiplier of $-\mathrm{\Delta }+{\kappa }^2$; its product with the latter multiplier is $1$, and its range lies in the operator domain because $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2+{\kappa }^2)\le 1$. Step 3.1 identifies the same multiplier for convolution, so convolution is exactly $(-\mathrm{\Delta }+{\kappa }^2{)}^{-1}$.