An equivalent formulation of 0-closed sesquilinear forms

Abstract

In 1970, McIntosh introduced the so-called 0-closed sesquilinear forms and proved a corresponding representation theorem. In this paper, we give a simple equivalent formulation of 0-closed sesquilinear forms. The main underlying idea is to consider minimal pairs of non-negative dominating forms.

1 Introduction

Let ${\mathcal {H}}$ be a Hilbert space with inner product $\langle \cdot ,\cdot \rangle$ and norm $\Vert \cdot \Vert$. Let ${\mathcal {D}}_1,{\mathcal {D}}_2$ be two subspaces of ${\mathcal {H}}$. A sesquilinear form (or, more simply, a form) ${\mathfrak {t}}$ on ${\mathcal {D}}_1\times {\mathcal {D}}_2$ is a map ${\mathfrak {t}}:{\mathcal {D}}_1\times {\mathcal {D}}_2\rightarrow {\mathbb {C}}$ which is linear in the first component and anti-linear in the second one. If ${\mathcal {D}}_1={\mathcal {D}}_2={\mathcal {D}}$, we write ${\mathfrak {t}}[f]:={\mathfrak {t}}(f,f)$ for $f\in {\mathcal {D}}$. A sesquilinear form ${\mathfrak {s}}$ on ${\mathcal {D}}\times {\mathcal {D}}$ is said to be non-negative if ${\mathfrak {s}}[f]\ge 0$ for every $f\in {\mathcal {D}}$; strictly-positive if there exists $c>0$ such that ${\mathfrak {s}}[f]\ge c\Vert f\Vert ^2$ for every $f\in {\mathcal {D}}$. If ${\mathfrak {s}}$ is non-negative, we denote by $\ker ({\mathfrak {s}})$ the subspace $\{f\in {\mathcal {D}}:{\mathfrak {s}}[f]=0\}$.

Given a sesquilinear form ${\mathfrak {t}}$ on ${\mathcal {D}}_1\times {\mathcal {D}}_2$, with ${\mathcal {D}}_2$ is dense in ${\mathcal {H}}$, it is possible to construct an operator T with domain

$$\begin{aligned} D(T)=\{f \in {\mathcal {D}}_1:\exists h\in {\mathcal {H}}, {\mathfrak {t}}(f,g)=\langle h , g\rangle , \forall g \in {\mathcal {D}}_2\} \end{aligned}$$

(1.1)

and defined as $Tf=h$, for all $f\in D(T)$, where h is the element in (1.1). The operator T is called associated to ${\mathfrak {t}}$ and then the following representation holds

$$\begin{aligned} {\mathfrak {t}}(f,g)=\langle Tf , g\rangle , \quad \forall f\in D(T), g\in {\mathcal {D}}_2. \end{aligned}$$

(1.2)

In the last decades, several theorems about the representation (1.2) have been given [1,2,3,4,5, 9, 11, 13,14,15,16,17]. The topic of the representation is connected to the Lebesgue decomposition (see [6,7,8, 12, 19]) as motivated in [8].

One of the classical representation theorems has been given for the so-called closed sesquilinear forms [14, Ch. VI]. We recall that a non-negative¹ sesquilinear form ${\mathfrak {s}}$ on ${\mathcal {D}}\times {\mathcal {D}}$ is closed if, for any sequence of vectors $\{f_n\}$ of ${\mathcal {H}}$ such that $f_n\rightarrow f$ and ${\mathfrak {s}}[f_n-f_m]\rightarrow 0$, one has $f\in {\mathcal {D}}$ and ${\mathfrak {s}}[f_n-f]\rightarrow 0$. The representation theorem for closed sesquilinear forms is useful, for instance, to define the Friedrichs extension of densely defined positive operators [14, Ch. VI] and a special sum of two operators [18].

In this paper, we specifically focus on 0-closed forms introduced and treated in [16, 17] by McIntosh in 1970. Hence, first of all, we recall the definition.

Let ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ be dense subspaces of ${\mathcal {H}}$. A sesquilinear form ${\mathfrak {t}}$ on ${\mathcal {D}}_1\times {\mathcal {D}}_2$ is called 0-closed [16, 17] if

• ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ can be made into Hilbert spaces ${\mathcal {H}}_1$ and ${\mathcal {H}}_2$ continuously embedded in ${\mathcal {H}}$ with inner products $\langle \cdot ,\cdot \rangle _1$ and $\langle \cdot ,\cdot \rangle _2$ and norms $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert _2$, respectively;

• ${\mathfrak {t}}$ is bounded with respect to $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert _2$, i.e., there exists $C>0$ such that

  $$\begin{aligned} |{\mathfrak {t}}(f,g)|\le C\Vert f\Vert _1\Vert g\Vert _2, \quad \forall f\in {\mathcal {H}}_1,g\in {\mathcal {H}}_2; \end{aligned}$$

  (1.3)

• the bounded operator $A:{\mathcal {H}}_1\rightarrow {\mathcal {H}}_2$ satisfying

  $$\begin{aligned} {\mathfrak {t}}(f,g)=\langle Af , g\rangle _2, \quad \forall f\in {\mathcal {H}}_1,g\in {\mathcal {H}}_2, \end{aligned}$$

  (1.4)

  is a bijection,²

For 0-closed forms, the following representation theorem holds.

Theorem 1.1

([16, 17]). Let ${\mathfrak {t}}$ be a 0-closed form on ${\mathcal {D}}_1\times {\mathcal {D}}_2$, where ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ are dense subspaces of ${\mathcal {H}}$. Then the operator T associated to ${\mathfrak {t}}$ is densely defined, closed, and 0 belongs to the resolvent set of T. Moreover, also the sesquilinear form ${\mathfrak {t}}^*$ given by

$$\begin{aligned} {\mathfrak {t}}^*(f,g)=\overline{{\mathfrak {t}}(g,f)}, \quad \forall f\in {\mathcal {D}}_2,g\in {\mathcal {D}}_1, \end{aligned}$$

is 0-closed and its associated operator is $T^*$.

The main scope of this paper is to prove an equivalent formulation of 0-closed forms (Theorem 2.4). In particular, we will employ the concept of minimal pairs of non-negative sesquilinear forms which dominate a given sesquilinear form (Definition 2.1). The auxiliary results Lemma 2.3 and Proposition 3.1 give some characterizations of minimal pairs assuming that the non-negative sesquilinear forms are closed.

2 The equivalent formulation

Throughout the paper, we denote by $\ker (S)$ and R(S) the kernel and the range of an operator $S:D(S) \rightarrow {\mathcal {H}}_2$, respectively. We use the symbol ${\mathcal {B}}({\mathcal {H}}_1,{\mathcal {H}}_2)$ for the set of bounded operators $S:{\mathcal {H}}_1 \rightarrow {\mathcal {H}}_2$. Firstly, we introduce the set of pairs of dominating forms. Let ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ be subspaces of ${\mathcal {H}}$ and ${\mathfrak {t}}$ a sesquilinear form on ${\mathcal {D}}_1\times {\mathcal {D}}_2$. We denote by ${\mathcal {M}}({\mathfrak {t}})$ the set of pairs $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ of non-negative sesquilinear forms such that

• ${\mathfrak {s}}_1:{\mathcal {D}}_1\times {\mathcal {D}}_1\rightarrow {\mathbb {C}}$ and ${\mathfrak {s}}_2:{\mathcal {D}}_2\times {\mathcal {D}}_2 \rightarrow {\mathbb {C}}$;

• for every $f\in {\mathcal {D}}_1$ and $g\in {\mathcal {D}}_2$, one has

  $$\begin{aligned} |{\mathfrak {t}}(f,g)|\le {\mathfrak {s}}_1[f]^\frac{1}{2}{\mathfrak {s}}_2[g]^\frac{1}{2}. \end{aligned}$$

  (2.1)

Definition 2.1

We say that a pair $({\mathfrak {s}}_1,{\mathfrak {s}}_2)\in {\mathcal {M}}({\mathfrak {t}})$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ if, for every $({\mathfrak {p}},{\mathfrak {q}})\in {\mathcal {M}}({\mathfrak {t}})$ such that ${\mathfrak {p}}\le {\mathfrak {s}}_1$ and ${\mathfrak {q}}\le {\mathfrak {s}}_2$, we have ${\mathfrak {p}}={\mathfrak {s}}_1$ and ${\mathfrak {q}}={\mathfrak {s}}_2$.

Remarks 2.2

• 1. If ${\mathfrak {s}}$ is a non-negative sesquilinear form on ${\mathcal {D}}\times {\mathcal {D}}$, then trivially $({\mathfrak {s}},{\mathfrak {s}})$ belongs to ${\mathcal {M}}({\mathfrak {s}})$ by the Cauchy-Schwarz inequality

  $$\begin{aligned} |{\mathfrak {s}}(f,g)|\le {\mathfrak {s}}[f]^\frac{1}{2}{\mathfrak {s}}[g]^\frac{1}{2}, \end{aligned}$$

  and it is also minimal in ${\mathcal {M}}({\mathfrak {s}})$.

• 2. If ${\mathfrak {t}}$ is a sesquilinear form and $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$, then also $(\alpha {\mathfrak {s}}_1,\alpha ^{-1}{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ for any $\alpha >0$. Anyway, even in the non-negative case, there might exist other less trivial minimal pairs. To make an example, let ${\mathcal {H}}={\mathbb {C}}^2$ and ${\mathfrak {s}}$ the non-negative form defined as follows

  $$\begin{aligned} {\mathfrak {s}}(f,g)=2f(1)\overline{g(1)}+2f(2)\overline{g(2)}, \quad \forall f,g\in {\mathbb {C}}^2, \end{aligned}$$

  where $f=(f(1),f(2))$ and $g=(g(1),g(2))$. The pair $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ made with

  $$\begin{aligned} {\mathfrak {s}}_1(f,g)=4f(1)\overline{g(1)}+f(2)\overline{g(2)}, \quad \forall f,g\in {\mathbb {C}}^2, \end{aligned}$$

  and

  $$\begin{aligned} {\mathfrak {s}}_2(f,g)=f(1)\overline{g(1)}+4f(2)\overline{g(2)}, \quad \forall f,g\in {\mathbb {C}}^2, \end{aligned}$$

  is in ${\mathcal {M}}({\mathfrak {s}})$. Indeed, by the Cauchy-Schwarz inequality, for every $f,g\in {\mathbb {C}}^2$,

  $$\begin{aligned} |s(f,g)|&=|2f(1)\cdot \overline{g(1)}+f(2)\cdot 2\overline{g(2)}|\\&\le (4|f(1)|^2+|f(2)|^2)^\frac{1}{2}(|g(1)|^2+4|g(2)|^2)^\frac{1}{2}\\&={\mathfrak {s}}_1[f]^\frac{1}{2}{\mathfrak {s}}_2[g]^\frac{1}{2}. \end{aligned}$$

  Moreover, $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {s}})$. Indeed, let $({\mathfrak {p}},{\mathfrak {q}})\in {\mathcal {M}}({\mathfrak {s}})$ with ${\mathfrak {p}}\le {\mathfrak {s}}_1$ and ${\mathfrak {q}}\le {\mathfrak {s}}_2$. We have

  $$\begin{aligned} 2={\mathfrak {s}}((1,0),(1,0))\le {\mathfrak {p}}[(1,0)]^\frac{1}{2}{\mathfrak {q}}[(1,0)]^\frac{1}{2}\le {\mathfrak {s}}_1[(1,0)]^\frac{1}{2}{\mathfrak {s}}_2[(1,0)]^\frac{1}{2}=2, \end{aligned}$$

  so ${\mathfrak {p}}[(1,0)]{\mathfrak {q}}[(1,0)]=4$. Since ${\mathfrak {p}}\le {\mathfrak {s}}_1$ and ${\mathfrak {q}}\le {\mathfrak {s}}_2$, we have ${\mathfrak {p}}[(1,0)]={\mathfrak {s}}_1[(1,0)]=2$ and ${\mathfrak {q}}[(1,0)]={\mathfrak {s}}_2[(1,0)]=1$. In the same way, we can prove that ${\mathfrak {p}}[(0,1)]={\mathfrak {s}}_1[(0,1)]=1$ and ${\mathfrak {q}}[(0,1)]={\mathfrak {s}}_2[(0,1)]=2$. Thus the non-negative forms ${\mathfrak {p}}$ and ${\mathfrak {q}}$ are completely determined knowing ${\mathfrak {p}}((1,0),(0,1))$ and ${\mathfrak {q}}((1,0),(0,1))$. In particular, denoting by ${\mathfrak {s}}_1-{\mathfrak {p}}$ the sequilinear form given by the difference between ${\mathfrak {s}}_1$ and ${\mathfrak {p}}$, we have

  $$\begin{aligned} |{\mathfrak {p}}((1,0),(0,1))|&= |({\mathfrak {s}}_1-{\mathfrak {p}})((1,0),(0,1))|\\&\le ({\mathfrak {s}}_1-{\mathfrak {p}})[(1,0)]^\frac{1}{2}({\mathfrak {s}}_1-{\mathfrak {p}})[(0,1)]^\frac{1}{2}=0, \end{aligned}$$

  where the first line holds because ${\mathfrak {s}}_1((1,0),(0,1))=0$ and the second line is valid by the Cauchy-Schwarz inequality since ${\mathfrak {s}}_1-{\mathfrak {p}}$ is non-negative by the hypothesis on ${\mathfrak {p}}$. In conclusion, ${\mathfrak {p}}={\mathfrak {s}}_1$. With similar arguments ${\mathfrak {q}}={\mathfrak {s}}_2$, i.e., $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {s}})$.

In the end of Section 3, we will give another example, but in infinite dimension, showing several minimal pairs. Now we are going to prove a preliminary result (Lemma 2.3) which will be useful in the proof of our main claim, i.e., Theorem 2.4. We start with a particular representation of a sesquilinear form ${\mathfrak {t}}$ on ${\mathcal {D}}_1\times {\mathcal {D}}_2$ determined by a given pair of closed non-negative sesquilinear forms $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ in ${\mathcal {M}}({\mathfrak {t}})$. Let

$$\begin{aligned} \widetilde{{\mathcal {D}}_1}=\ker ({\mathfrak {s}}_1)^\perp \cap {\mathcal {D}}_1 \quad \text {and} \quad \widetilde{{\mathcal {D}}_2}=\ker ({\mathfrak {s}}_2)^\perp \cap {\mathcal {D}}_2, \end{aligned}$$

(2.2)

where $\ker ({\mathfrak {s}}_1)^\perp$ and $\ker ({\mathfrak {s}}_2)^\perp$ are the orthogonal complements of $\ker ({\mathfrak {s}}_1)$ and $\ker ({\mathfrak {s}}_2)$ in ${\mathcal {H}}$, respectively. The restrictions $\widetilde{{\mathfrak {s}}_1}$, $\widetilde{{\mathfrak {s}}_2}$ of ${\mathfrak {s}}_1$, ${\mathfrak {s}}_2$ on $\widetilde{{\mathcal {D}}_1}$, $\widetilde{{\mathcal {D}}_2}$, respectively, are closed non-negative sesquilinear forms with $\ker (\widetilde{{\mathfrak {s}}_1})=\{0\}$ and $\ker (\widetilde{{\mathfrak {s}}_2})=\{0\}$. Now let

$$\begin{aligned}&\langle f , f'\rangle _1=\widetilde{{\mathfrak {s}}_1}(f,f') \;\text { and }\; \langle g , g'\rangle _2=\widetilde{{\mathfrak {s}}_2}(g,g'), \quad f,f'\in \widetilde{{\mathcal {D}}_1},g,g'\in \widetilde{{\mathcal {D}}_2}, \end{aligned}$$

(2.3)

$$\begin{aligned}&\quad \Vert f\Vert _1=\widetilde{{\mathfrak {s}}_1}[f]^\frac{1}{2}\;\text { and }\; \Vert g\Vert _2=\widetilde{{\mathfrak {s}}_2}[g]^\frac{1}{2}, \quad f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}. \end{aligned}$$

(2.4)

Since $\widetilde{{\mathfrak {s}}_1}$ and $\widetilde{{\mathfrak {s}}_2}$ are closed forms, $\widetilde{{\mathcal {D}}_1}$ and $\widetilde{{\mathcal {D}}_2}$ are complete with respect to the inner products $\langle \cdot ,\cdot \rangle _1$ and $\langle \cdot ,\cdot \rangle _2$ in (2.3), respectively. We denote by ${\mathcal {H}}_1$ and ${\mathcal {H}}_2$ the Hilbert spaces made in this way. By (2.1), the restriction of ${\mathfrak {t}}$ on $\widetilde{{\mathcal {D}}_1}\times \widetilde{{\mathcal {D}}_2}$ can be considered as a bounded sesquilinear form with respect to the norms $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert _2$ in (2.4) induced by $\widetilde{{\mathfrak {s}}_1}$ and $\widetilde{{\mathfrak {s}}_2}$, respectively. Then, by Riesz’s theorem, there exists an operator $B\in {\mathcal {B}}({\mathcal {H}}_1,{\mathcal {H}}_2)$ such that

$$\begin{aligned} {\mathfrak {t}}(f,g)=\langle Bf , g\rangle _2, \quad \forall f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}. \end{aligned}$$

(2.5)

In particular, again by (2.1), B has norm³$\Vert B\Vert$ less or equal 1. We have the following characterization.

Lemma 2.3

Let ${\mathfrak {t}}$ be a sesquilinear form on ${\mathcal {D}}_1\times {\mathcal {D}}_2$, where ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ are subspaces of ${\mathcal {H}}$, and let $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ be a pair of closed non-negative sesquilinear forms in ${\mathcal {M}}({\mathfrak {t}})$. Then $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ if and only if the operator B in (2.5) is a unitary operator.

Proof

Let us assume that $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$. By (2.5),

$$\begin{aligned} |{\mathfrak {t}}(f,g)|\le \Vert Bf\Vert _2 \Vert g\Vert _2=\Vert Bf\Vert _2 {\mathfrak {s}}_2[g]^\frac{1}{2}, \quad \forall f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}. \end{aligned}$$

(2.6)

Let $I:{\mathcal {H}}\rightarrow {\mathcal {H}}$ be the identity operator, $O_1:{\mathcal {H}}\rightarrow {\mathcal {H}}$ and $O_2:{\mathcal {H}}\rightarrow {\mathcal {H}}$ the orthogonal projections onto $\ker ({\mathfrak {s}}_1)^\perp$ and $\ker ({\mathfrak {s}}_2)^\perp$, respectively. We note that $\ker ({\mathfrak {s}}_1),\ker ({\mathfrak {s}}_2)$ are closed since ${\mathfrak {s}}_1,{\mathfrak {s}}_2$ are closed forms and that for $f\in {\mathcal {D}}_1,g\in {\mathcal {D}}_2$, we have $(I-O_1)f\in \ker ({\mathfrak {s}}_1)\in {\mathcal {D}}_1,(I-O_2)g\in \ker ({\mathfrak {s}}_2)\in {\mathcal {D}}_2$, so $O_1f\in \widetilde{{\mathcal {D}}_1},O_2g\in \widetilde{{\mathcal {D}}_2}$. Let ${\mathfrak {p}}$ be the sesquilinear form defined as follows

$$\begin{aligned} \mathfrak {p}(f,g)=\langle {BO_1 f,BO_1 g\rangle _2,}\quad \forall f,g \in \mathcal {D}_1. \end{aligned}$$

Hence, the inequality (2.6) can be rewritten as

$$\begin{aligned} |{\mathfrak {t}}(f,g)|\le {\mathfrak {p}}[f]^\frac{1}{2}{\mathfrak {s}}_2[g]^\frac{1}{2}, \quad \forall f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}. \end{aligned}$$

(2.7)

Considering that ${\mathfrak {t}}(f,g)=0$ if $f\in \ker ({\mathfrak {s}}_1)$ or $g\in \ker ({\mathfrak {s}}_2)$ and that ${\mathfrak {p}}[O_1f]={\mathfrak {p}}[f],{\mathfrak {s}}_2[O_2g]={\mathfrak {s}}_2[g]$ for every $f\in {\mathcal {D}}_1,g\in {\mathcal {D}}_2$, the inequality (2.7) extends by linearity for every $f\in {\mathcal {D}}_1$ and $g\in {\mathcal {D}}_2$. In other words, $({\mathfrak {p}},{\mathfrak {s}}_2)\in {\mathcal {M}}({\mathfrak {t}})$.

As said before, $\Vert B\Vert \le 1$ and then ${\mathfrak {p}}[f]^\frac{1}{2}=\Vert BO_1f\Vert _2\le \Vert O_1f\Vert _1={\mathfrak {s}}_1[O_1f]^\frac{1}{2}={\mathfrak {s}}_1[f]^\frac{1}{2}$ for every $f\in {{\mathcal {D}}_1}$. Since $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$, we have $\Vert Bf\Vert _2={\mathfrak {p}}[f]^\frac{1}{2}={\mathfrak {s}}_1[f]^\frac{1}{2}=\Vert f\Vert _1$ for every $f\in \widetilde{{\mathcal {D}}_1}$, i.e., B is an isometry. Because

$$\begin{aligned} |{\mathfrak {t}}(f,g)|\le \Vert f\Vert _1\Vert B^*g\Vert _1, \quad \forall f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}, \end{aligned}$$

holds too, in the same way, we have that $B^*$ is an isometry. In conclusion, B is unitary.

Now assume that B is a unitary operator. Let $({\mathfrak {p}},{\mathfrak {q}})\in {\mathcal {M}}({\mathfrak {t}})$ be such that ${\mathfrak {p}}\le {\mathfrak {s}}_1$ and ${\mathfrak {q}}\le {\mathfrak {s}}_2$. Then there exist $P\in {\mathcal {B}}({\mathcal {H}}_1)$ and $Q\in {\mathcal {B}}({\mathcal {H}}_2)$ satisfying

$$\begin{aligned} \Vert P\Vert \le 1, \quad \Vert Q\Vert \le 1, \end{aligned}$$

(2.8)

${\mathfrak {p}}[f]=\Vert Pf\Vert _1^2$ for every $f\in \widetilde{{\mathcal {D}}_1}$, and ${\mathfrak {q}}[g]=\Vert Qg\Vert _2^2$ for every $g\in \widetilde{{\mathcal {D}}_2}.$ By [14, Ch. VI, Lemma 3.1], there exists $R\in {\mathcal {B}}({\mathcal {H}}_{\mathfrak {p}},{\mathcal {H}}_{\mathfrak {q}})$⁴ such that

$$\begin{aligned} \Vert R\Vert \le 1, \end{aligned}$$

(2.9)

$$\begin{aligned} {\mathfrak {t}}(f,g)=\langle RPf , Qg\rangle _2, \quad \forall f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}. \end{aligned}$$

Therefore, by uniqueness of the associated operator, $B=Q^*RP$. Moreover,

$$\begin{aligned} \Vert f\Vert _1=\Vert Bf\Vert _2=\Vert Q^*RPf\Vert _2\le \Vert Pf\Vert _1\le \Vert f\Vert _1, \quad \forall f\in \widetilde{{\mathcal {D}}_1}, \end{aligned}$$

by (2.8) and (2.9), i.e., ${\mathfrak {p}}[f]=\Vert Pf\Vert _1^2=\Vert f\Vert _1^2={\mathfrak {s}}_1[f]$ for every $f\in \widetilde{{\mathcal {D}}_1}$. Furthermore, on $\ker ({\mathfrak {s}}_1)$ (which, as said before, is closed), ${\mathfrak {s}}_1$ and ${\mathfrak {p}}$ are null. Hence, by linearity, ${\mathfrak {p}}={\mathfrak {s}}_1$. Working in a similar way with $B^*$, we also find that ${\mathfrak {q}}={\mathfrak {s}}_2$. Thus $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$. $\square$

Now we are ready to prove the main result of the paper.

Theorem 2.4

Let ${\mathfrak {t}}$ be a sesquilinear form on ${\mathcal {D}}_1\times {\mathcal {D}}_2$, where ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ are dense subspaces of ${\mathcal {H}}$. The following statements are equivalent.

• (i) ${\mathfrak {t}}$ is 0-closed;

• (ii) there exists a pair $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ of closed strictly-positive sesquilinear forms minimal in ${\mathcal {M}}({\mathfrak {t}})$.

Proof

(i)$\implies$(ii) Let $\langle \cdot ,\cdot \rangle _1$ and $\langle \cdot ,\cdot \rangle _2$ be the inner products which make ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ two Hilbert spaces based on the definition of 0-closed forms. Let $A=U|A|=|A^*|U$ be the polar decomposition of the bounded operator A in (1.4). In particular, since A is bijective, U is a unitary operator and $|A|^\frac{1}{2}:{\mathcal {H}}_1\rightarrow {\mathcal {H}}_1$, $|A^*|^\frac{1}{2}:{\mathcal {H}}_2\rightarrow {\mathcal {H}}_2$ are bijective self-adjoint positive operators. By [10, Theorem 2.7], we also have $A=|A^*|^\frac{1}{2}U |A|^\frac{1}{2}$. Let us define

$$\begin{aligned} {\mathfrak {s}}_1(f,f')= & {} \langle |A|^\frac{1}{2}f , |A|^\frac{1}{2}f'\rangle _1, \quad f,f'\in {\mathcal {D}}_1, \\ {\mathfrak {s}}_2(g,g')= & {} \langle |A^*|^\frac{1}{2}g , |A^*|^\frac{1}{2}g'\rangle _2, \quad g,g'\in {\mathcal {D}}_2. \end{aligned}$$

The sesquilinear forms ${\mathfrak {s}}_1$ and ${\mathfrak {s}}_2$ are strictly-positive and closed forms because $|A|^\frac{1}{2}$, $|A^*|^\frac{1}{2}$ are bijective and $\langle \cdot ,\cdot \rangle _1$, $\langle \cdot ,\cdot \rangle _2$ are strictly-positive and closed forms. Moreover,

$$\begin{aligned} |{\mathfrak {t}}(f,g)|=|\langle |A^*|^\frac{1}{2}U |A|^\frac{1}{2}f , g\rangle |=|\langle U |A|^\frac{1}{2}f , |A^*|^\frac{1}{2}g\rangle |\le {\mathfrak {s}}_1[f]^\frac{1}{2}{\mathfrak {s}}_2[g]^\frac{1}{2}, \end{aligned}$$

i.e., $({\mathfrak {s}}_1,{\mathfrak {s}}_2)\in {\mathcal {M}}({\mathfrak {t}})$. Finally, $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ by Lemma 2.3 because the operator B constructed as in (2.5) coincides with U which is a unitary operator.

(ii)$\implies$(i) Let

$$\begin{aligned} \langle f , f'\rangle _1={\mathfrak {s}}_1(f,f') \;\text { and }\; \langle g , g'\rangle _2={\mathfrak {s}}_2(g,g'), \quad f,f'\in {\mathcal {D}}_1,g,g'\in {\mathcal {D}}_2. \end{aligned}$$

Since ${\mathfrak {s}}_1$ and ${\mathfrak {s}}_2$ are closed strictly-positive sesquilinear forms, ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ turn into Hilbert spaces continuously embedded in ${\mathcal {H}}$ with inner products $\langle \cdot ,\cdot \rangle _1$ and $\langle \cdot ,\cdot \rangle _2$, respectively. Moreover, ${\mathfrak {t}}$ is bounded with respect to the norms of these spaces. Finally, the last condition required in the definition of 0-closed forms is satisfied as a consequence of Lemma 2.3. $\square$

3 A supplementary result

In this section, we prove another characterization of minimal forms concerning a different representation in comparison to (2.5).

As previously, let ${\mathfrak {t}}$ be a sesquilinear form on ${\mathcal {D}}_1\times {\mathcal {D}}_2$ and $({\mathfrak {s}}_1,{\mathfrak {s}}_2)\in {\mathcal {M}}({\mathfrak {t}})$ with ${\mathfrak {s}}_1$ and ${\mathfrak {s}}_2$ non-negative and closed. By Kato’s second representation theorem [14, Theorem VI.2.23], there exist positive self-adjoint operators $H_1$ and $H_2$ in ${\mathcal {H}}$ with $D(H_1)={\mathcal {D}}_1$ and $D(H_2)={\mathcal {D}}_2$ such that ${\mathfrak {s}}_1[f]=\Vert H_1 f\Vert ^2$ and ${\mathfrak {s}}_2[g]=\Vert H_2 g\Vert ^2$ for every $f\in {\mathcal {D}}_1$ and $g\in {\mathcal {D}}_2$. We write $\overline{R(H_1)}$ and $\overline{R(H_2)}$ for the closures of $R(H_1)$ and $R(H_2)$ in ${\mathcal {H}}$, respectively. By (2.1) and [14, Ch. VI, Lemma 3.1], there exists a bounded operator $Q\in {\mathcal {B}}(\overline{R(H_1)},\overline{R(H_2)})$ with $\Vert Q\Vert \le 1$ such that

$$\begin{aligned} {\mathfrak {t}}(f,g)=\langle QH_1f , H_2g\rangle , \quad f\in {\mathcal {D}}_1,g\in {\mathcal {D}}_2. \end{aligned}$$

Moreover, $\ker (H_1)=\ker ({\mathfrak {s}}_1)$, $\ker (H_2)=\ker ({\mathfrak {s}}_2)$, so the restrictions $\widetilde{H_1}$ and $\widetilde{H_2}$ of $H_1$ and $H_2$ on $\widetilde{{\mathcal {D}}_1}$ and $\widetilde{{\mathcal {D}}_2}$ (which are defined in (2.2)), respectively, are injective and we can also write

$$\begin{aligned} {\mathfrak {t}}(f,g)=\langle Q\widetilde{H_1}f , \widetilde{H_2}g\rangle , \quad f\in \widetilde{{\mathcal {D}}_1},g\in \widetilde{{\mathcal {D}}_2}. \end{aligned}$$

(3.1)

Proposition 3.1

Let ${\mathfrak {t}}$ be a sesquilinear form on ${\mathcal {D}}_1\times {\mathcal {D}}_2$, where ${\mathcal {D}}_1$ and ${\mathcal {D}}_2$ are dense subspaces of ${\mathcal {H}}$ and let $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ be a pair of closed non-negative sesquilinear forms in ${\mathcal {M}}({\mathfrak {t}})$. Then $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ if and only if Q in (3.1) is a unitary operator.

Proof

Let $H_1$ and $H_2$ be the operators introduced above. The operators Q and B of (2.5) are connected by the following relation

$$\begin{aligned} Q=\widetilde{H_2}B\widetilde{H_1}^{-1} \text { on } R(H_1). \end{aligned}$$

Comparing the representations (2.5) and (3.1) for ${\mathfrak {t}}^*$ instead of ${\mathfrak {t}}$, we also have

$$\begin{aligned} Q^*=\widetilde{H_1}B^*\widetilde{H_2}^{-1} \text { on } R(H_2), \end{aligned}$$

so

$$\begin{aligned} QQ^*=\widetilde{H_2}BB^*\widetilde{H_2}^{-1}\text { on } R(H_2), \quad Q^*Q=\widetilde{H_1}B^*B\widetilde{H_1}^{-1}\text { on } R(H_1), \end{aligned}$$

and

$$\begin{aligned} BB^*=\widetilde{H_2}^{-1}QQ^*\widetilde{H_2} \text { on } \widetilde{{\mathcal {D}}_2},\quad B^*B=\widetilde{H_1}^{-1}Q^*Q\widetilde{H_1}\text { on } \widetilde{{\mathcal {D}}_1}. \end{aligned}$$

Thus, by Lemma 2.3 and taking into account that $R(H_1)$, $R(H_2)$ are dense in their corresponding closures, $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ if and only if Q is unitary. $\square$

Example

Let us consider ${\mathcal {H}}={\mathcal {D}}=L^2({\mathbb {R}})$ and a bounded measurable function $r:{\mathbb {R}}\rightarrow {\mathbb {C}}$. We write $N=\{x\in {\mathbb {R}}:r(x)=0\}$. Let

$$\begin{aligned} {\mathfrak {t}}(f,g)=\int \limits _{\mathbb {R}}r(x)f(x)g(x)dx, \quad \forall f,g\in {\mathcal {H}}. \end{aligned}$$

(3.2)

The sesquilinear form ${\mathfrak {t}}$ is bounded, and it is non-negative if and only if r(x) is non-negative for a.e. $x\in {\mathbb {R}}$. For any measurable function $p:{\mathbb {R}}\rightarrow [0,+\infty )$, with $c\le p(x)\le d$ for some $c,d>0$ and every $x\in {\mathbb {R}}$, we define

$$\begin{aligned} {\mathfrak {s}}_1(f,g)=\int \limits _{\mathbb {R}}|r(x)|p(x)f(x)g(x)dx, \quad \forall f,g\in {\mathcal {H}}\end{aligned}$$

and

$$\begin{aligned} {\mathfrak {s}}_2(f,g)=\int \limits _{\mathbb {R}}|r(x)|p(x)^{-1}f(x)g(x)dx, \quad \forall f,g\in {\mathcal {H}}. \end{aligned}$$

The sesquilinear forms ${\mathfrak {s}}_1$ and ${\mathfrak {s}}_2$ are non-negative (trivially closed by boundedness) and, by the Cauchy-Schwarz inequality, $({\mathfrak {s}}_1,{\mathfrak {s}}_2)\in {\mathcal {M}}({\mathfrak {t}})$. Moreover, again independently by p, we have $\ker ({\mathfrak {s}}_1)=\ker ({\mathfrak {s}}_2)=\{f\in {\mathcal {H}}: f(x)=0\text { for } \hbox {a.e.}\; x\in {\mathbb {R}}\backslash N\}\equiv L^2(N)$, and so $\widetilde{{\mathcal {D}}_1}=\widetilde{{\mathcal {D}}_2}=\{f\in {\mathcal {H}}: f(x)=0\text { for } \hbox {a.e.}\; x\in N\}\equiv L^2({\mathbb {R}}\backslash N)$. The pair $({\mathfrak {s}}_1,{\mathfrak {s}}_2)$ is minimal in ${\mathcal {M}}({\mathfrak {t}})$ for any choice of p by Proposition 3.1. Indeed, one easily checks that the operators $\widetilde{H_1}$ and $\widetilde{H_2}$ are the multiplication operators by $|r(x)|^\frac{1}{2}p(x)^\frac{1}{2}$ and $|r(x)|^\frac{1}{2}p(x)^{-\frac{1}{2}}$ on the domain $L^2({\mathbb {R}}\backslash N)$, respectively. Hence, by comparing (3.2) and (3.1), the operator Q is the multiplication operator by $\frac{r}{|r|}$ on the domain $L^2({\mathbb {R}}\backslash N)$ and it is in particular unitary.