Closed positive forms: representation and the exact square-root domain

Programme exposition: OpenAI Codex (AI). Course selection and prerequisite bindings: GPT-6.1 Sol (OpenAI), Ultra. Integration and bounded proof check: GPT-6 Astra (OpenAI), Ultra, October 2026.

A closed positive form specifies an energy and its finite-energy domain. This lesson recovers its nonnegative self-adjoint operator, proves that the energy domain is exactly the square-root domain, and identifies which vectors have an operator value. Hilbert spaces may have arbitrary dimension, including zero.

Barry Simon treats the form-space construction in A canonical decomposition for quadratic forms with applications to monotone convergence theorems. Zoltán Sebestyén and Zsigmond Tarcsay give a complementary treatment of representation and operator domains in Basic representation theorems of forms. The proof here uses a bounded energy resolvent and a common dense subspace of two Hilbert spaces of finite-energy vectors.

Prerequisites and conventions

Inner products are linear in the first variable. A positive operator means a nonnegative operator; injectivity is an additional condition. Operator equalities include equality of domains.

• Hilbert spaces and compact operators, Sections 1–3, proves completion, orthogonal projection, the Riesz representation theorem, bounded forms and adjoints. These statements impose no separability condition.
• Spectral calculus, the bounded and unbounded constructions, supplies the bounded Borel calculus and the closedness, adjoint and product rules for measurable functions. Powers and actual ranges proves the inverse-domain and square-root identities used below. These proofs do not use closed-form representation or polar decomposition.
• Integration and convergence, Theorems 2.1–2.2, supplies scalar monotone and dominated convergence. Only sequential scalar convergence is used here.

For completeness, the bounded adjoint works between different Hilbert spaces. If $T:V\to H$ is bounded, Riesz representation on $V$ gives the unique $T^*y\in V$ such that $\langle Tx,y\rangle_H=\langle x,T^*y\rangle_V$ for all $x\in V$. Uniqueness makes $T^*$ linear, and Cauchy–Schwarz gives $\|T^*y\|\leq\|T\|\|y\|$. Taking the supremum of the same pairing over unit balls gives $\|T^*\|=\|T\|$; this also holds for zero spaces. Reversing the pairing gives $T^{**}=T$. Moreover, a vector $y$ is orthogonal to $\operatorname{ran}T$ exactly when $T^*y=0$. Orthogonal projection therefore gives $\overline{\operatorname{ran}T}=(\ker T^*)^\perp$. Applying this identity to $T^*$ gives $\overline{\operatorname{ran}T^*}=(\ker T)^\perp$.

A form includes its domain

A nonnegative sesquilinear form on $H$ consists of a linear subspace $D(q)\subseteq H$ and a sesquilinear map $q:D(q)\times D(q)\to\mathbb C$, linear in the first variable, with $q[x]:=q(x,x)\geq0$. The reality of $q[x+y]$ and $q[x+iy]$, after expansion, gives $q(y,x)=\overline{q(x,y)}$. Positivity then gives

Indeed, apply positivity to $q[x+t y]$; when $q[y]>0$, minimizing over $t\in\mathbb C$ proves the inequality, and when $q[y]=0$, an arbitrarily large scalar with a suitable phase forces the cross term to vanish. The same expansion determines the form from its diagonal by

Conversely, the diagonal axioms already contain the whole sesquilinear form. Let $D$ be a complex vector space and suppose that $Q:D\to[0,\infty)$ satisfies

Define first the real polarization

This map is symmetric. Two uses of the parallelogram identity give

Putting $u=x+z$, $v=x-z$ in the first identity and using the second shows $B(u+v,y)=B(u,y)+B(v,y)$. Thus $B$ is additive in each variable. For rational $r$, it follows that $B(rx,y)=rB(x,y)$. Positivity of

for rational $r$ yields

If $r_n\in\mathbb Q$ tends to a real number $t$, additivity, (QF.3), and the first identity in (QF.1) give

Hence $B(tx,y)=tB(x,y)$, so $B$ is real bilinear. The invariance $Q(ix)=Q(x)$ implies

Therefore

is complex linear in its first variable, conjugate linear in its second, and Hermitian. Moreover $B(ix,x)=0$, so $s(x,x)=Q(x)$. Thus $s$ is the unique positive sesquilinear form with diagonal $Q$. The argument is algebraic: no topology or density assumption on $D$ is hidden in the polarization step.

The same statement covers an extended-valued diagonal on a larger vector space. If $Q:V\to[0,\infty]$ satisfies (QF.1), with $0\cdot\infty=0$, then

is a complex subspace. Indeed, homogeneity handles scalar multiples, while the parallelogram identity shows that $Q(x+y)<\infty$ whenever $Q(x)$ and $Q(y)$ are finite. Formula (QF.4) then gives the associated sesquilinear form on this exact finite-value domain.

The form norm and its inner product are

The form is closed if $D(q)$ is complete for this norm. It is densely defined if $\overline{D(q)}=H$. These conditions are independent.

It is useful to extend the diagonal by

For two forms, $q_1\leq q_2$ means $\widetilde q_1[x]\leq\widetilde q_2[x]$ for every $x\in H$. Equivalently,

Thus an increasing family of forms can have decreasing domains. Statements about this order never mean only an inequality on an unspecified common core.

Representation with the exact square-root domain

Theorem. For each densely defined closed nonnegative form $q$ on $H$, there is a unique nonnegative self-adjoint operator $A$ on $H$ such that

Moreover,

and the vector $y$ in this formula equals $Ax$. Conversely, every nonnegative self-adjoint $A$ gives a densely defined closed form by these formulas.

Proof. Equip $V=D(q)$ with the energy inner product $$\langle x,v\rangle_V=\langle x,v\rangle_H+q(x,v).$$ Closedness makes $V$ a Hilbert space. The inclusion $j:V\to H$ is an injective contraction with dense range. Its bounded adjoint exists by the Hilbert-space argument above. Set $B=jj^*$. For $f\in H$, $$0\leq\langle Bf,f\rangle_H=\|j^*f\|_V^2\leq\|f\|^2. \tag{QF.R1}$$ If $Bf=0$, this identity gives $j^*f=0$; density of $jV$ then gives $f=0$. Thus $B$ is an injective positive contraction. In particular its spectral projection at zero is zero, even if zero belongs to its spectrum.

For $u=Bf$, its preimage in $V$ is $j^*f$. The adjoint identity gives, for every $v\in V$, $$\langle u,v\rangle_H+q(u,v)=\langle f,v\rangle_H. \tag{QF.R2}$$ The spectral calculus of $B$ defines $$A=B^{-1}-I,\qquad a(t)=t^{-1}-1\quad(0<t\leq1). \tag{QF.R3}$$ This is a nonnegative self-adjoint operator. Its domain is exactly $\operatorname{ran}B$: the inverse-domain assertion is proved in SK-07, and the inequalities $$(t^{-1}-1)^2\leq t^{-2}\leq 2(t^{-1}-1)^2+2$$ show that the two square-integrability conditions coincide. Equation (QF.R2) already proves $q(u,v)=\langle Au,v\rangle$ for $u\in D(A)$, $v\in V$.

We now recover the entire form domain. Put $W=D(B^{-1/2})$, with norm $\|x\|_W=\|B^{-1/2}x\|$. It dominates the Hilbert norm because $0<B\leq I$. It is complete: if $x_n$ is Cauchy in this norm, then $x_n\to x$ in $H$ and $B^{-1/2}x_n\to y$ in $H$; closedness of the spectral operator gives $x\in W$ and $B^{-1/2}x=y$.

The common subspace $D(A)=\operatorname{ran}B$ is dense in each of $V$ and $W$, with the same norm on that subspace. Indeed, $\operatorname{ran}j^*$ is dense in $V$, since its orthogonal complement is $\ker j=0$. Its image under $j$ is $\operatorname{ran}B$, and for $x=Bf$, $$\|x\|_V^2=\|j^*f\|_V^2 =\langle Bf,f\rangle =\|B^{1/2}f\|^2 =\|B^{-1/2}x\|^2=\|x\|_W^2. \tag{QF.R4}$$ For density in $W$, take $P_n=E_B([1/n,1])$. If $x\in W$, then $x_n=P_nx\in\operatorname{ran}B$, because $t^{-1}\mathbf1_{[1/n,1]}(t)$ is bounded. Moreover $$\|x-x_n\|_W^2 =\int_{(0,1/n)}t^{-1}\,d\langle E_B(t)x,x\rangle\longrightarrow0. \tag{QF.R5}$$ This is dominated convergence for the finite integral defining $W$.

For clarity, density and equal norms here give equality of actual subsets of $H$. Approximate any $x\in V$ by a sequence in $\operatorname{ran}B$ in its $V$-norm. The sequence is Cauchy in $W$, hence has a limit there, and both limits equal $x$ in $H$. Thus $V\subseteq W$. Conversely the spectral approximants in (QF.R5) are Cauchy in $V$; completeness and their Hilbert-space limit put $x\in W$ in $V$. Their limiting norms agree. No countable basis or cofinal subset of a later directed set is involved.

Consequently, $$D(q)=D(B^{-1/2}),\qquad \|x\|^2+q[x]=\int_{(0,1]}t^{-1}\,d\langle E_B(t)x,x\rangle. \tag{QF.R6}$$ Since the scalar spectral measure is finite, subtracting $\|x\|^2$ gives $$D(q)=D(A^{1/2}),\qquad q[x]=\|A^{1/2}x\|^2. \tag{QF.R7}$$ Polarization proves the sesquilinear identity in the theorem.

For the converse graph criterion, suppose $x\in V$ and $q(x,v)=\langle y,v\rangle$ for all $v\in V$. Then $$\langle x,v\rangle_V=\langle x+y,jv\rangle_H =\langle j^*(x+y),v\rangle_V.$$ Riesz uniqueness in $V$ gives $x=j^*(x+y)$ there. Applying $j$ gives $x=B(x+y)$, so $x\in D(A)$ and $Ax=y$. Density makes $y$ unique. This proves the exact graph formula in both directions.

If a nonnegative self-adjoint $C$ represents the same form, the spectral form-pairing identity in SK-07 gives $q(x,v)=\langle Cx,v\rangle$ for $x\in D(C)$, $v\in D(q)$. The graph criterion therefore gives $C\subseteq A$. Adjoints reverse this inclusion: if $z\in D(A^*)$, then the identity $\langle Ax,z\rangle=\langle x,A^*z\rangle$ restricts to $D(C)$, so $z\in D(C^*)$ with the same value. Hence $A=A^*\subseteq C^*=C$, and $A=C$.

Finally, a nonnegative self-adjoint $A$ has a closed densely defined square root by SK-05–SK-07. The graph map $x\mapsto(x,A^{1/2}x)$ identifies the form norm with the norm on its closed graph in $H\oplus H$, which is complete. This gives the converse form and finishes the proof. $\square$

The operator domain requires a Hilbert-space vector representing the form functional; the square-root domain requires only finite energy. Equation (QF.R4) also proves that the operator domain is dense in the form norm.

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