1 Introduction

The adjoint of an unbounded linear operator was first introduced by John von Neumann in [5] as a profound ingredient for developing a rigorous mathematical framework for quantum mechanics. By definition, the adjoint of a densely defined linear transformation S, acting between two Hilbert spaces, is an operator T with the largest possible domain such that

$$\begin{aligned} {(}Sx{\,\vert \,y}{)}={(}x{\,\vert \,Ty}{)} \end{aligned}$$
(1)

holds for every x from the domain of S. The adjoint operator, denoted by \(S^*\), is therefore “maximal” in the sense that it extends every operator T that has property (1). On the other hand, every restriction T of \(S^*\) fulfills that adjoint relation. Thus, in order to decide whether an operator T is identical with the adjoint of S it seems reasonable to restrict ourselves to investigating those operators T that have property (1). This issue was explored in detail in [16] by means of the operator matrix

$$\begin{aligned} \begin{bmatrix} I & \quad -T \\ S& \quad I\end{bmatrix}, \end{aligned}$$

cf. also [8, 11, 13, 14].

In the present paper we continue to examine the conditions under which an operator T is equal to the adjoint \(S^*\) of S. Nevertheless, as opposed to the situation treated in the cited papers, we do not assume that S and T are adjoint to each other in the sense of (1). Observe that condition (1) is equivalent to identity

$$\begin{aligned} S^*\cap T=T. \end{aligned}$$
(2)

So, still under condition (1), T is equal to the adjoint of S if and only if \(S^*\cap T=S^*\). In the present paper we are going to guarantee equality \(S^*=T\) by imposing new conditions, weaker than (1), by means of the kernel and range spaces. Roughly speaking, we only require that the intersection of the graphs of \(S^*\) and T be, in a sense, “large enough”. We also establish a criterion in terms of the norm of the resolvent of the operator matrix

$$\begin{aligned} M_{S,T}=\begin{bmatrix} 0 & \quad -T \\ S& \quad 0\end{bmatrix}. \end{aligned}$$

As an application we gain some characterizations of self-adjoint, skew-adjoint and unitary operators, thereby generalizing some analogous results by Nieminen [7] (cf. also [9]).

2 Preliminaries

Throughout this paper, \({\mathcal {H}}\) and \({\mathcal {K}}\) will denote real or complex Hilbert spaces. By an operator S between \({\mathcal {H}}\) and \({\mathcal {K}}\) we mean a linear map \(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\) whose domain \({\mathrm{dom\,}}S\) is a linear subspace of \({\mathcal {H}}\). We stress that, unless otherwise indicated, linear operators are not assumed to be densely defined. However, the adjoint of such an operator can only be interpreted as a “multivalued operator”, that is, a linear relation. Therefore we are going to collect here some basic notions and facts on linear relations.

A linear relation between two Hilbert spaces \({\mathcal {H}}\) and \({\mathcal {K}}\) is nothing but a linear subspace S of the Cartesian product \({\mathcal {H}}\times {\mathcal {K}}\), respectively, a closed linear relation is just a closed subspace of \({\mathcal {H}}\times {\mathcal {K}}\). To a linear relation S we associate the following subspaces

$$\begin{aligned} {\mathrm{dom\,}}S&={\{}{h\in {\mathcal {H}}\,:\,(h,k)\in S}{\}}&{\mathrm{ran\,}}S&={\{}{k\in {\mathcal {K}}\,:\,(h,k)\in S}{\}}\\ \ker \, S&={\{}{h\in {\mathcal {H}}\,:\,(h,0)\in S}{\}}&{{\,\mathrm{mul}\,}}S&={\{}{k\in {\mathcal {K}}\,:\,(0,k)\in S}{\}}, \end{aligned}$$

which are referred to as the domain, range, kernel and multivalued part of S, respectively. Every linear operator when identified with its graph is a linear relation with trivial multivalued part. Conversely, a linear relation whose multivalued part consists only of the vector 0 is (the graph of) an operator.

A notable advantage of linear relations, compared to operators, lies in the fact that one might define the adjoint without any further assumption on the domain. Namely, the adjoint of a linear relation S will be again a linear relation \(S^*\) between \({\mathcal {K}}\) and \({\mathcal {H}}\), given by

$$\begin{aligned} S^*{:}{=}V( S)^{\perp }. \end{aligned}$$

Here, \(V:{\mathcal {H}}\times {\mathcal {K}}\rightarrow {\mathcal {K}}\times {\mathcal {H}}\) stands for the ‘flip’ operator \(V(h,k){:}{=}(k,-h)\). It is seen immediately that \(S^*\) is automatically a closed linear relation and satisfies the useful identity

$$\begin{aligned} \overline{S}=S^{**}({=}{:}(S^*)^*). \end{aligned}$$

On the other hand, a closed linear relation S entails the following orthogonal decomposition of the product Hilbert space \({\mathcal {K}}\times {\mathcal {H}}\):

$$\begin{aligned} S^*\oplus V(S)={\mathcal {K}}\times {\mathcal {H}}. \end{aligned}$$

Note that another equivalent definition of \(S^*\) is obtained in terms of the inner product as follows:

$$\begin{aligned} S^*={\{}{(k',h')\in {\mathcal {K}}\times {\mathcal {H}}\,:\,{(}k{\,\vert \,k'}{)}={(}h{\,\vert \,h'}{)}~\text{ for } \text{ all } (h,k)\in S}{\}}. \end{aligned}$$

In other words, \((k',h')\in S^*\) holds if and only if

$$\begin{aligned} {(}k{\,\vert \,k'}{)}={(}h{\,\vert \,h'}{)}\qquad \forall (h,k)\in S. \end{aligned}$$

In particular, if S is a densely defined operator, then the relation \(S^*\) coincides with the usual adjoint operator of S. Recall also the dual identities

$$\begin{aligned} \ker \, S^*=({\mathrm{ran\,}}S)^{\perp },\qquad {{\,\mathrm{mul}\,}}S^*=({\mathrm{dom\,}}S)^{\perp }, \end{aligned}$$

where the second equality tells us that the adjoint of a densely defined linear relation is always a (single valued) operator. For further information on linear relations we refer the reader to [1, 2, 4, 10].

3 Operators which are adjoint of each other

Arens [1] characterized the equality \(S=T\) of two linear relations in terms of their kernel and range (see Corollary 2). Below we provide a similar characterization of \(S\subset T\). Observe that the intersection \(S\cap T\) of the linear relations S and T is again a linear relation, but this is not true for their union \(S\cup T\) as it is not a linear subspace in general. The linear span of \(S\cup T\) will be denoted by \(S\vee T\), which in turn is a linear relation.

Proposition 1

Let S and T be linear relations between two vector spaces. Then the following three statements are equivalent:

  1. (i)

    \(S\subset T\),

  2. (ii)

    \(\ker \,S\subset \ker \, T\)and\({\mathrm{ran\,}}S\subset {\mathrm{ran\,}}(S\cap T)\),

  3. (iii)

    \({\mathrm{ran\,}}S\subset {\mathrm{ran\,}}T\)and\(\ker \, (S\vee T)\subset \ker \, T\).

Proof

It is clear that (i) implies both (ii) and (iii). Suppose now (ii) and let \((h,k)\in S\) then there exists u with \((u,k)\in T\cap S\). Consequently, \((h-u,0)\in S\), i.e., \(h-u\in \ker \, S\subset \ker \, T\). Hence

$$\begin{aligned} (h,k)=(h-u,0)+(u,k)\in T+T\subset T, \end{aligned}$$

which yields \(S\subset T\), so (ii) implies (i). Finally, assume (iii) and take \((h,k)\in S\). Then \((u,k)\in T\) for some u and hence \((h-u,0)\in S\vee T\), i.e., \(h-u\in \ker \, T\). Consequently,

$$\begin{aligned} (h,k)=(h-u,0)+(u,k)\in T, \end{aligned}$$

which yields \(S\subset T\). \(\square\)

Corollary 1

Let S and T be two linear relations between two vector spaces. The following three statements are equivalent:

  1. (i)

    \(S=T\),

  2. (ii)

    \(\ker \, S=\ker T\)and\({\mathrm{ran\,}}S+{\mathrm{ran\,}}T\subseteq {\mathrm{ran\,}}(S\cap T)\),

  3. (iii)

    \({\mathrm{ran\,}}S={\mathrm{ran\,}}T\)and\(\ker \, (S\vee T)\subseteq \ker \, (S\cap T)\).

Corollary 2

LetSandTbe linear relations between two vector spaces such that\(S\subset T\). Then the following assertions are equivalent:

  1. (i)

    \(S=T\),

  2. (ii)

    \(\ker \, S=\ker T\)and\({\mathrm{ran\,}}S={\mathrm{ran\,}}T\).

In [17, Theorem 2.9] M. H. Stone established a simple yet effective sufficient condition for an operator to be self-adjoint: a densely defined symmetric operator S is necessarily self-adjoint provided it is surjective. In that case, it is invertible with bounded and self-adjoint inverse due to the Hellinger–Toeplitz theorem. Here, density of the domain can be dropped from the hypotheses: a surjective symmetric operator is automatically densely defined (see also [16, Corollary 6.7] and [15, Lemma 2.1]).

Below we establish a generalization of Stone’s result for a pair of operators.

Proposition 2

Let\({\mathcal {H}},{\mathcal {K}}\)be real or complex Hilbert spaces and let\(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\)and\(T:{\mathcal {K}}\rightarrow {\mathcal {H}}\)be (not necessarily densely defined or closed) linear operators such that

$$\begin{aligned} {\mathrm{ran\,}}(S\cap T^*)={\mathcal {K}}\qquad \text{ and }\qquad {\mathrm{ran\,}}(T\cap S^*)={\mathcal {H}}. \end{aligned}$$

ThenSandTare both densely defined operators such that\(S^*=T\)and\(T^*=S\).

Proof

For brevity, introduce the following notations

$$\begin{aligned} S_0{:}{=}S\cap T^*,\qquad T_0{:}{=}T\cap S^*. \end{aligned}$$

Observe that \(S_0\) and \(T_0\) are adjoint to each other in the sense that

$$\begin{aligned} {(}S_0x{\,\vert \,y}{)}={(}x{\,\vert \,T_0y}{)},\qquad x\in {\mathrm{dom\,}}S_0,y\in {\mathrm{dom\,}}T_0. \end{aligned}$$

We claim that \(S_0\) and \(T_0\) are densely defined: let \(z\in ({\mathrm{dom\,}}S_0)^{\perp }\), then by surjectivity, \(z=T_0v\) for some \(v\in {\mathrm{dom\,}}T_0\). Hence

$$\begin{aligned} 0={(}x{\,\vert \,z}{)}={(}x{\,\vert \,T_0v}{)}={(}S_0x{\,\vert \,v}{)},\qquad x\in {\mathrm{dom\,}}S_0, \end{aligned}$$

which implies \(v=0\) and also \(z=0\). The same argument shows that \(T_0\) is densely defined too. We see now that S and \(T^*\) are densely defined operators such that

$$\begin{aligned} \ker \, S\subseteq ({\mathrm{ran\,}}S^*)^\perp =\{0\}, \qquad \ker \, T^*= ({\mathrm{ran\,}}T)^\perp =\{0\}, \end{aligned}$$

and \({\mathrm{ran\,}}(S\cap T^*)={\mathcal {K}}\). Corollary 1 applied to S and \(T^*\) implies that \(S=T^*\). The same argument yields equality \(S^*=T\). \(\square\)

Corollary 3

Let\(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\)and\(T:{\mathcal {K}}\rightarrow {\mathcal {H}}\)be (not necessarily densely defined) surjective operators such that

$$\begin{aligned} {(}Sx{\,\vert \,y}{)}={(}x{\,\vert \,Ty}{)},\qquad x\in {\mathrm{dom\,}}S,y\in {\mathrm{dom\,}}T. \end{aligned}$$

ThenSandTare both densely defined operators such that\(S^*=T\)and\(T^*=S\).

From Proposition 2 we gain a sufficient condition of self-adjointness without the assumptions of being symmetric or densely defined:

Corollary 4

Let\({\mathcal {H}}\)be a Hilbert space and let\(S:{\mathcal {H}}\rightarrow {\mathcal {H}}\)be a linear operator such that\({\mathrm{ran\,}}(S\cap S^*)={\mathcal {H}}\). ThenSis densely defined and self-adjoint.

Proof

Apply Proposition 2 with \(T{:}{=}S\). \(\square\)

Clearly, if S is a symmetric operator, then \(S\cap S^*=S\). Hence we retrieve [17, Theorem 2.9] by M. H. Stone as an immediate consequence (cf. also [16, Corollary 6.7]):

Corollary 5

Every surjective symmetric operator is densely defined and self-adjoint.

In the next result we give a necessary and sufficient condition for an operator S to be identical with the adjoint of a given operator T.

Theorem 1

Let\({\mathcal {H}},{\mathcal {K}}\)be real or complex Hilbert spaces and let\(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\)and\(T:{\mathcal {K}}\rightarrow {\mathcal {H}}\)be (not necessarily densely defined or closed) linear operators. The following two statements are equivalent:

  1. (i)

    Tis densely defined and\(S=T^*\),

  2. (ii)
    1. (a)

      \(({\mathrm{ran\,}}T)^{\perp }=\ker \, S\),

    2. (b)

      \({\mathrm{ran\,}}S+{\mathrm{ran\,}}T^*\subset {\mathrm{ran\,}}(S\cap T^*)\).

Proof

It is obvious that (i) implies (ii). Assume now (ii) and for sake of brevity introduce the operator

$$\begin{aligned} S^{}_0{:}{=}S\cap T^*. \end{aligned}$$

We start by establishing that T is densely defined. Let \(g\in ({\mathrm{dom\,}}T^*)^{\perp }\), then \((0,g)\in T^*\), i.e., \(g\in {\mathrm{ran\,}}T^*\). By (ii) (a),

$$\begin{aligned} T^*g=S_0h=Sh \end{aligned}$$

for some \(h\in {\mathrm{dom\,}}S_0\). Then it follows that \((h,Sh)\in T^*\) and therefore

$$\begin{aligned} {(}Tk{\,\vert \,h}{)}={(}k{\,\vert \,Sh}{)}={(}k{\,\vert \,g}{)}=0,\qquad k\in {\mathrm{dom\,}}T, \end{aligned}$$

whence we infer that \(h\in ({\mathrm{ran\,}}T)^{\perp }\). Again by (ii) (a) we have \(h\in \ker \, S\) and thus \(g=Sh=0\). This proves that T is densely defined and as a consequence, \(T^*\) is an operator. Next we prove that

$$\begin{aligned} T^*\subset S. \end{aligned}$$
(3)

To see this consider \(g\in {\mathrm{dom\,}}T^*\). By (ii) (b),

$$\begin{aligned} T^*g=S_0h=Sh=T^*h \end{aligned}$$

for some \(h\in {\mathrm{dom\,}}S_0\). Then it follows that \(T^*(g-h)=0\), i.e.,

$$\begin{aligned} g-h\in \ker \, T^*=({\mathrm{ran\,}}T)^\perp =\ker \, S. \end{aligned}$$

Consequently, \(g=(g-h)+h\in {\mathrm{dom\,}}S\) and \(Sg=Sh=T^*g\), which proves (3). It only remains to show that the converse inclusion

$$\begin{aligned} S\subset T^* \end{aligned}$$
(4)

holds also true. For let \(g\in {\mathrm{dom\,}}S\) and choose \(h\in {\mathrm{dom\,}}S_0\) such that

$$\begin{aligned} Sg=S_0h=T^*h=Sh. \end{aligned}$$

Then \(g-h\in \ker \, S=({\mathrm{ran\,}}T)^\perp =\ker \, T^*\) whence we get \(g=(g-h)+h\in {\mathrm{dom\,}}T^*\) and \(T^*g=T^*h=Sg\), which proves (4). \(\square\)

A celebrated theorem by von Neumann [6] states that \(S^*S\) and \(SS^*\) are positive and selfadjoint operators provided that S is a densely defined and closed operator between \({\mathcal {H}}\) and \({\mathcal {K}}.\) In that case, \(I+S^*S\) and \(I+SS^*\) are both surjective. In [12] it has been proved that the converse is also true: If \(I+S^*S\) and \(I+SS^*\) are both surjective operators, then S is necessarily closed (cf. also [3]). Below, as the main result of the paper, we establish an improvement of Neumann’s theorem:

Theorem 2

Let\({\mathcal {H}},{\mathcal {K}}\)be real or complex Hilbert spaces and let\(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\)and\(T:{\mathcal {K}}\rightarrow {\mathcal {H}}\)be linear operators and introduce the operators\(S_0{:}{=}S\cap T^*\)and\(T_0{:}{=}T\cap S^*\). The following statements are equivalent:

  1. (i)

    S, Tare both densely defined and they are adjoint of each other: \(S^*=T\)and\(T^*=S\),

  2. (ii)

    \({\mathrm{ran\,}}(I+T_0S_0)={\mathcal {H}}\)and\({\mathrm{ran\,}}(I+S_0T_0)={\mathcal {K}}\).

Proof

It is clear that (i) implies (ii). To prove the converse implication observe first that

$$\begin{aligned} {(}S^{}_0u{\,\vert \,v}{)}={(}u{\,\vert \,T^{}_0v}{)},\qquad u\in {\mathrm{dom\,}}S^{}_0, v\in {\mathrm{dom\,}}T^{}_0. \end{aligned}$$

We start by showing that \(S^{}_0\) is densely defined. Take a vector \(g\in ({\mathrm{dom\,}}S_0)^{\perp }\), then there is \(u\in {\mathrm{dom\,}}S^{}_0\) such that \(g=u+T^{}_0S^{}_0u\). Consequently,

$$\begin{aligned} 0={(}u{\,\vert \,g}{)}={(}u{\,\vert \,u}{)}+{(}T^{}_0S^{}_0u{\,\vert \,u}{)}=\Vert u\Vert ^2+\Vert S^{}_0u\Vert , \end{aligned}$$

whence \(u=0\), and therefore also \(g=0\). It is proved analogously that \(T^{}_0\) is densely defined too, and therefore the adjoint relations \(S_0^{*}\) and \(T_0^{*}\) are operators such that \(S^{}_0\subset T_0^*\) and \(T^{}_0\subset S_0^*\).

We are going to prove now that \(S^{}_0\) and \(T^{}_0\) are adjoint of each other, i.e.,

$$\begin{aligned} S_0^*=T^{}_0,\qquad T_0^*=S_0^{}. \end{aligned}$$
(5)

Consider a vector \(g\in {\mathrm{dom\,}}T_0^*\) and take \(u\in {\mathrm{dom\,}}S^{}_0\) and \(v\in {\mathrm{dom\,}}T^{}_0\) such that

$$\begin{aligned} g=u+T_0S_0u\qquad \text{ and }\qquad T^{*}_0g=v+S^{}_0T^{}_0v. \end{aligned}$$

Since u is in \({\mathrm{dom\,}}T_0^*\) we infer that \(T_0S_0u\in {\mathrm{dom\,}}T_0^*\) and hence

$$\begin{aligned} T_0^*g=T_0^*u+T_0^*T^{}_0S^{}_0u. \end{aligned}$$

It follows then that

$$\begin{aligned} 0&=v-T_0^*u+S^{}_0T_0^{}v-T^{*}_0T^{}_0S^{}_0u=(I+T^{*}_0T^{}_0)(v-S^{}_0u) \end{aligned}$$

which yields \(v=S_0u\in {\mathrm{dom\,}}T_0.\) As a consequence we obtain that

$$\begin{aligned} g=u+T^{}_0S^{}_0u=v+T^{}_0v, \end{aligned}$$

and therefore that \(g\in {\mathrm{dom\,}}S_0\). This proves the first equality of (5). The second one is proved in a similar way.

Now we can complete the proof easily: since \(S_0\subset T^*\) and \(T_0\subset T\) it follows that

$$\begin{aligned} T_0^*=S^{}_0\subset T^*\subset T_0^*, \end{aligned}$$

whence \(T^*=T_0^*=S^{}_0\), and therefore \(T^*\subset S\). On the other hand, \(T_0\subset S^*\) implies

$$\begin{aligned} S\subset S^{**}\subset T^{*}_0=T^*, \end{aligned}$$

whence we conclude that \(S=T^*\). It can be proved in a similar way that \(T=S^*\). \(\square\)

As an immediate consequence we conclude the following result:

Corollary 6

Let \({\mathcal {H}}\) and \({\mathcal {K}}\) be real or complex Hilbert spaces and let \(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\) be a densely defined operator. The following statements are equivalent:

  1. (i)

    Sis closed,

  2. (ii)

    \(S^*S\)and\(SS^*\)are self-adjoint operators,

  3. (iii)

    \({\mathrm{ran\,}}(I+S^*S)={\mathcal {H}}\)and\({\mathrm{ran\,}}(I+SS^*)={\mathcal {K}}\).

Proof

Apply Theorem 2 with \(T{:}{=}S^*\). \(\square\)

In the ensuing theorem we provide a renge-kernel characterization of operators T that are identical with the adjoint \(S^*\) of a densely defined symmetric operator S. We stress that no condition on the closedness of the operator or density of the domain is imposed. On the contrary: we get those properties from the other conditions.

Theorem 3

Let\({\mathcal {H}}\)be a real or complex Hilbert space and let\(T:{\mathcal {H}}\rightarrow {\mathcal {H}}\)be a (not necessarily densely defined or closed)linear operator and let\(T_0{:}{=}T\cap T^*\). The following two statements are equivalent:

  1. (i)

    there exists a densely defined symmetric operatorSsuch that\(S^*=T\),

  2. (ii)
    1. (a)

      \(\ker \, T=({\mathrm{ran\,}}T^*)^{\perp }\),

    2. (b)

      \({\mathrm{ran\,}}T_0={\mathrm{ran\,}}T^{**}={\mathrm{ran\,}}T^*\).

In particular, if any of the equivalent conditions (i), (ii) is satisfied, thenTis a densely defined and closed operator such that\(T^*\subset T\).

Proof

It is straightforward that (i) implies (ii) so we only prove the converse. We start by proving that T is densely defined. Take \(g\in ({\mathrm{dom\,}}T)^{\perp }\), then \(g\in {{\,\mathrm{mul}\,}}T^*\subseteq {\mathrm{ran\,}}T^*\). By (ii) (b), there exists \(h\in {\mathrm{dom\,}}T_0\) such that \(g=T_0h=Th\). Consequently, \((h,g)\in T^*\) and for every \(f\in {\mathrm{dom\,}}T\),

$$\begin{aligned} {(}Tf{\,\vert \,h}{)}={(}f{\,\vert \,g}{)}=0, \end{aligned}$$

which yields \(h\in ({\mathrm{ran\,}}T)^{\perp }\). Observe that (ii) (a) and (b) together imply that

$$\begin{aligned} ({\mathrm{ran\,}}T)^\perp =\ker \, T, \end{aligned}$$
(6)

whence we infer that \(h\in \ker \, T\) and therefore that \(g=Th=0\). This means that \(T^*\) is a (single valued) operator. Our next claim is to show that

$$\begin{aligned} T^*\subset T. \end{aligned}$$
(7)

To this end, let \(g\in {\mathrm{dom\,}}T^*\), then \(T^*g=T_0h\) for some \(h_0\in {\mathrm{dom\,}}T_0\). From inclusion \(T_0\subset T^*\) we conclude that \(g-h\in \ker \, T^*=({\mathrm{ran\,}}T)^\perp\), thus \(g=(g-h)+h\in {\mathrm{dom\,}}T\) and

$$\begin{aligned} Tg=Th=T_0h=T^*g, \end{aligned}$$

which proves (7). Next we show that \(T^*\) is densely defined too, i.e., T is closable. To this end consider a vector \(g\in ({\mathrm{dom\,}}T^*)^\perp ={{\,\mathrm{mul}\,}}T^{**}\). Since \({{\,\mathrm{mul}\,}}T^{**}\subseteq {\mathrm{ran\,}}T^{**}\), we can find a vector \(h\in {\mathrm{dom\,}}T_0\) such that \(g=T_0h\). For every \(k\in {\mathrm{dom\,}}T^*\),

$$\begin{aligned} {(}h{\,\vert \,T^*k}{)}={(}Th{\,\vert \,k}{)}={(}g{\,\vert \,k}{)}=0, \end{aligned}$$

thus \(h\in ({\mathrm{ran\,}}T^*)^{\perp }\). By (ii) (a) we infer that \(h\in \ker \, T\) and hence \(g=Th=0\) and thus \(({\mathrm{dom\,}}T^*)^\perp =\{0\}\), as it is claimed. Finally we show that T is closed. Take \(g\in {\mathrm{dom\,}}T^{**}\), then \(T^{**}g=Th\) for some \(h\in {\mathrm{dom\,}}T\), according to assumption (ii) (b). Hence \(g-h\in \ker \, T^{**}=({\mathrm{ran\,}}T^*)^{\perp }\), thus \(g-h\in \ker \, T\) because of (ii) (a). Consequently, \(g=(g-h)+h\in {\mathrm{dom\,}}T\) which proves identity \(T=T^{**}\). Summing up, \(S{:}{=}T^*\) is a densely defined operator such that \(S\subset T=S^*\). In other words, T is identical with the adjoint \(S^*\) of the symmetric operator S. \(\square\)

4 Characterizations involving resolvent norm estimations

Let \({\mathcal {H}}\) and \({\mathcal {K}}\) be real or complex Hilbert spaces. For given two linear operators \(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\) and \(T:{\mathcal {K}}\rightarrow {\mathcal {H}}\), let us consider the operator matrix

$$\begin{aligned} M_{S,T}{:}{=}\begin{bmatrix} 0 &{} \quad -T \\ S&{} \quad 0\end{bmatrix}, \end{aligned}$$

acting on the product Hilbert space \({\mathcal {H}}\times {\mathcal {K}}\). More precisely, \(M_{S,T}\) is an operator acting on its domain \({\mathrm{dom\,}}M_{S,T}{:}{=}{\mathrm{dom\,}}S\times {\mathrm{dom\,}}T\) by

$$\begin{aligned} M_{S,T}(h,k){:}{=}(-Tk,Sh) \qquad (h\in {\mathrm{dom\,}}S, k\in {\mathrm{dom\,}}T). \end{aligned}$$

Assume that a real or complex number \(\lambda \in {\mathbb {K}}\) belongs to the resolvent set \(\rho (M_{S,T})\), which means that

$$\begin{aligned} M_{S,T}-\lambda =\begin{bmatrix} -\lambda &{} \quad -T \\ S&{} \quad -\lambda \end{bmatrix} \end{aligned}$$

has an everywhere defined bounded inverse. In that case, for brevity’s sake, we introduce the notation

$$\begin{aligned} R_{S,T}(\lambda ){:}{=}(M_{S,T}-\lambda )^{-1} \end{aligned}$$

for the corresponding resolvent operator.

In the present section, we are going to establish some criteria, by means of norms of the resolvent operator \(R_{S,T}(\lambda )\), under which the operators S and T are adjoint of each other. Our approach is motivated by the classical paper of T. Nieminen [7] (cf. also [9]). We emphasize that our framework is more general than that of [7] for many ways: we do not assume that the operators under consideration are densely defined or closed, and also the underlying space may be real or complex.

Theorem 4

Let\(S:{\mathcal {H}}\rightarrow {\mathcal {K}}\)and\(T:{\mathcal {K}}\rightarrow {\mathcal {H}}\)be linear operators between the real or complex Hilbert spaces\({\mathcal {H}}\)and\({\mathcal {K}}\). The following assertions are equivalent:

  1. (i)

    SandTare densely defined such that\(S^*=T\)and\(T^*=S\),

  2. (ii)

    every non-zero real numbertbelongs to the resolvent set of\(M_{S,T}\)and

    $$\begin{aligned} \Vert R_{S,T}(t) \Vert \le \frac{1}{{|}t{|}},\qquad \forall t\in {\mathbb {R}}, t\ne 0. \end{aligned}$$
    (8)

Proof

Let us start by proving that (i) implies (ii). Assume therefore that S is densely defined and closed and that \(T=S^*\). Consider a non-zero real number t and a pair of vectors \(h\in {\mathrm{dom\,}}S\) and \(k\in {\mathrm{dom\,}}S^*\), then we have

$$\begin{aligned} \left\| \begin{bmatrix} t &{} \quad -S^* \\ S&{} \quad t\end{bmatrix}\begin{bmatrix}h \\ k \end{bmatrix}\right\| ^2&=\Vert th-S^*k\Vert ^2+\Vert Sh+tk\Vert ^2\\&=t^2[\Vert h\Vert ^2+\Vert k\Vert ^2]+\Vert Sh\Vert ^2+\Vert S^*k\Vert ^2\\&\ge t^2\left\| \begin{bmatrix}h \\ k \end{bmatrix}\right\| ^2 \end{aligned}$$

which implies that \(M_{S,T}+t\) is bounded from below and the norm of its inverse \(R_{S,T}(-t)\) satisfies (8). However it is not yet clear that \(R_{S,T}(-t)\) is everywhere defined. Since we have

$$\begin{aligned} \begin{bmatrix} t &{} -S^* \\ S&{} t\end{bmatrix}\begin{bmatrix}h \\ k \end{bmatrix}=t\left( \begin{bmatrix}h \\ \tfrac{1}{t}Sh \end{bmatrix} +\begin{bmatrix}-\tfrac{1}{t}S^*k \\ k \end{bmatrix}\right) , \end{aligned}$$

it follows that

$$\begin{aligned} {\mathrm{ran\,}}(M_{S,T}+t)=\tfrac{1}{t}S\oplus W \left( \tfrac{1}{t}S^*\right) , \end{aligned}$$
(9)

where W is the ‘flip’ operator \(W(k,h){:}{=}(-h,k)\). Since S is densely defined and closed according to our hypotheses, the subspace on the right hand side of (9) is equal to \({\mathcal {H}}\times {\mathcal {K}}\). This proves statement (ii).

For the converse direction, observe that (8) implies

$$\begin{aligned} \left\| \begin{bmatrix} t &{} \quad -T \\ S&{} \quad t\end{bmatrix}\begin{bmatrix}h \\ k \end{bmatrix}\right\| ^2\ge t^2\left\| \begin{bmatrix}h \\ k \end{bmatrix}\right\| ^2,\qquad h\in {\mathrm{dom\,}}S, k\in {\mathrm{dom\,}}T. \end{aligned}$$

Hence from (8) we conclude that

$$\begin{aligned} 0&\ge \Vert Sx\Vert ^2+\Vert Ty\Vert ^2+t\{{(}Sx{\,\vert \,y}{)}-{(}x{\,\vert \,Ty}{)}+{(}y{\,\vert \,Sx}{)}-{(}Ty{\,\vert \,x}{)}\}\\&=\Vert Sx\Vert ^2+\Vert Ty\Vert ^2+2t \, \hbox {Re}\{{(}Sx{\,\vert \,y}{)}-{(}x{\,\vert \,Ty}{)}\} \end{aligned}$$

for every \(t\in {\mathbb {R}}\). Consequently,

$$\begin{aligned} \hbox {Re} {(}Sx{\,\vert \,y}{)}=\hbox {Re} {(}x{\,\vert \,Ty}{)},\qquad x\in {\mathrm{dom\,}}S, y\in {\mathrm{dom\,}}T. \end{aligned}$$

In the real Hilbert space case it is straightforward that S and T are adjoint to each other. In the complex case, replace x by ix to get

$$\begin{aligned} \hbox {Im} {(}Sx{\,\vert \,y}{)}=\hbox {Im} {(}x{\,\vert \,Ty}{)},\qquad x\in {\mathrm{dom\,}}S, y\in {\mathrm{dom\,}}T. \end{aligned}$$

So, in both real and complex cases, we obtain that \(S\subset T^*\) and \(T\subset S^*\). With notation of Theorem 2 this means that \(S_0=S\) and \(T_0=T\). Since we have

$$\begin{aligned} \begin{bmatrix} I &{} \quad -T \\ S&{} \quad I\end{bmatrix}= M^{}_{S,T}+1 ,\qquad \begin{bmatrix} I &{} \quad T \\ -S&{} \quad I\end{bmatrix}= 1-M^{}_{S,T}, \end{aligned}$$

we conclude that

$$\begin{aligned} \begin{bmatrix} I+TS &{} \quad 0 \\ 0&{} \quad I+ST\end{bmatrix}=[M^{}_{S,T}+1][ 1- M^{}_{S,T}] \end{aligned}$$

is a surjective operator onto \({\mathcal {H}}\times {\mathcal {K}}\), which entails \({\mathrm{ran\,}}(I+TS)={\mathcal {H}}\) and \({\mathrm{ran\,}}(I+ST)= {\mathcal {K}}\). An immediate application of Theorem 2 completes the proof. \(\square\)

As an immediate consequence of Theorem 4 we can establish the following characterizations of self-adjoint, skew-adjoint and unitary operators.

Corollary 7

Let \({\mathcal {H}}\) be a real or complex Hilbert space. For a linear operator \(S:{\mathcal {H}}\rightarrow {\mathcal {H}}\) the following assertions are equivalent:

  1. (i)

    Sis densely defined and self-adjoint,

  2. (ii)

    Every non-zero real number t is in the resolvent set of \(M_{S,-S}\) and

    $$\begin{aligned} \Vert R_{S,-S}(t) \Vert \le \frac{1}{{|}t{|}},\qquad \forall t\in {\mathbb {R}}, t\ne 0. \end{aligned}$$
    (10)

Proof

Apply Theorem 4 with \(T{:}{=}S\) to conclude the desired equivalence. \(\square\)

Corollary 8

Let \({\mathcal {H}}\) be a real or complex Hilbert space. For a linear operator \(S:{\mathcal {H}}\rightarrow {\mathcal {H}}\) the following assertions are equivalent:

  1. (i)

    Sis densely defined and skew-adjoint,

  2. (ii)

    every non-zero real number t is in the resolvent set of \(M_{S,S}\) and

    $$\begin{aligned} \Vert R_{S,S}(t) \Vert \le \frac{1}{{|}t{|}},\qquad \forall t\in {\mathbb {R}}, t\ne 0. \end{aligned}$$
    (11)

Proof

Apply Theorem 4 with \(T{:}{=}-S\). \(\square\)

Corollary 9

Let\({\mathcal {H}}\)and\({\mathcal {K}}\)be a real or complex Hilbert spaces. For a linear operator\(U:{\mathcal {H}}\rightarrow {\mathcal {K}}\)the following assertions are equivalent:

  1. (i)

    Uis a unitary operator,

  2. (ii)

    \(\ker U=\{0\}\), every non-zero real numbertis in the resolvent set of\(M_{U,U^{-1}}\)and

    $$\begin{aligned} \Vert R_{U,U^{-1}}(t) \Vert \le \frac{1}{{|}t{|}},\qquad \forall t\in {\mathbb {R}}, t\ne 0. \end{aligned}$$
    (12)

Proof

An application of Theorem 4 with \(S{:}{=}U\) and \(T{:}{=}U^{-1}\) shows that U is densely defined and closed such that \(U^*=U^{-1}\). Hence, \({\mathrm{ran\,}}U^*\subseteq {\mathrm{dom\,}}U\). Since we have \({\mathrm{ran\,}}U^*+{\mathrm{dom\,}}U={\mathcal {H}}\) for every densely defined closed operator U, we infer that \({\mathrm{dom\,}}U={\mathcal {H}}\) and therefore U is a unitary operator. \(\square\)