Understanding Hilbert-adjoint for (possibly unbounded) operators

Question

I want to understand how one defines the Hilbert-adjoint of a (possibly) unbounded operator.

Here is what I know for the bounded case:

For a bounded linear operator $T$, the Hilbert-adjoint is defined by the requirement that $\langle Tx,y \rangle=\langle x,T^*y \rangle \text{, } \forall x,y \in H$.

This is done via the Riesz-Representation Theorem. We have that $h(x,y)=\langle Tx,y \rangle$ is sesquilinear and $\lVert h \rVert = \lVert T \rVert$. Using Riesz Representation there exists an operator $S$ sucht that $h(x,y)=\langle x,Sy \rangle$. Denoting $S$ as $T^*$ justifies the definition.

The unbounded case:

Let $\mathcal{D}(T)$ denote the domain of an operator $T$.

Definition: Let $T$ be a densely defined operator on the Hilbert space $H$. On the subspace $\mathcal{D}(T^*):= \{y \in H : x \mapsto \langle Tx,y \rangle \text{ is continuous } \mathcal{D}(T) \rightarrow \mathbb{C}\}$, we define the Hilbert-adjoint $T^*$ of $T$ as:

For $y \in \mathcal{D}(T^*)$, the map $x \mapsto \langle Tx,y \rangle$ has a unique continuous extension to a linear functional on $H$. Using the Riesz-Representation Theorem, this functional can be represented in the form $x \mapsto \langle x,z \rangle$ where $z=:T^*y$.

The main difference in the definitions seems to be that we suppose that $T$ is densely defined (i.e. $\mathcal{D}(T)$ is dense in $H$). What is the reason we assume $T$ to be a densely defined operator? The other difference is that we define the domain of $T^*$. Why do we define $\mathcal{D}(T^*)$ as done above?

Answer 1

You essentially have an answer phrased in terms of operators in the comments by @Manuel. Here, I would like to introduce (and more ambitiously, convince) you to the idea that studying these guys in terms of their graphs is more intuitive/natural. In fact, set theoretically, a function is often defined via the graph, so this shouldn’t be a huge surprise. When discussing this via graphs, the basic definitions/concepts surrounding bounded/unbounded operators is almost identical, so things won’t seem so ad-hoc.

The main observation if that for a function $T:V\to W$ between vector spaces, we have that $T$ is linear if and only if its graph is a linear subspace. This allows us to go back and forth between linear maps and their graphs, and it turns out in this part of functional analysis, a lot can be phrased conveniently in terms of graphs (or linear relations more generally).

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Easy Preliminaries

Throughout, we let $X,Y$ be sets, we let $V,W$ vector spaces over a field $\Bbb{F}$, we let $E,F$ be Banach spaces (over the same field $\Bbb{R}$ or $\Bbb{C}$), and we let $H$ (or $H_1,H_2$) be Hilbert spaces.

Definition (Relations Between Sets).

A relation from $X$ into $Y$ is simply a subset $\Gamma$ of $X\times Y$. We define the domain and range/image of the relation in the obvious manner:

• $\mathrm{dom}(\Gamma):=\{x\in X\,:\, \text{there is a $y\in Y$ such that $(x,y)\in \Gamma$}\}$
• $\mathrm{ran}(\Gamma):=\{y\in Y\,:\, \text{there is an $x\in X$ such that $(x,y)\in \Gamma$}\}$.

We say that $\Gamma$ is everywhere-defined if $\mathrm{dom}(\Gamma)=X$.

A relation $\Gamma$ is said to be the graph of a function if it satisfies the “vertical line test”: for each $x\in X$ and $y_1,y_2\in Y$, if $(x,y_1),(x,y_2)\in\Gamma$, then $y_1=y_2$.

Often times people may call such a $\Gamma$ a partial relation (and the associated function a partial function) to emphasize the fact that the domain is not necessarily all of $X$. However, when dealing with unbounded operators, this will be the most common situation, so it is just more convenient to simplify terminology and omit the “partial” to allow this possibility in the definition and explicitly mention cases when things are everywhere-defined.

Definition (Linear relations)

A linear relation from $V$ into $W$ simply means a vector subspace $\Gamma$ of the product $V\times W$. From above, we have the concepts of the domain and range of $\Gamma$. We now define in addition, two more sets

• $\ker(\Gamma):=\{v\in V\,:\, (v,0)\in \Gamma\}=\text{pr}_V\left(\Gamma \cap (V\times \{0\})\right)$, the kernel of $\Gamma$.
• $\mathfrak{M}(\Gamma):=\{w\in W\,:\, (0,w)\in\Gamma\}=\text{pr}_W\left(\Gamma\cap (\{0\}\times W)\right)$, the multivalued part of $\Gamma$.

In this linear context, if $\Gamma$ is the graph of a function, we often rather say it is the graph of an operator. Also, since $\Gamma$ is a vector subspace, the induced function is actually a linear map in the usual sense $\mathrm{dom}(\Gamma)\subset V\to W$).

It is a trivial exercise to verify that for a linear relation $\Gamma$ from $V$ into $W$, the sets $\mathrm{dom}(\Gamma), \mathrm{ran}(\Gamma), \ker(\Gamma),\mathfrak{M}(\Gamma)$ are actually vector subspaces of $V,W,V,W$ respectively; also $\Gamma$ is the graph of an operator if and only if the multivalued part is trivial, i.e $\mathfrak{M}(\Gamma)=\{0_W\}$.

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Dual/Adjoint relations.

Now, for any linear relation, we can define an dual relation/adjoint relation between the dual spaces:

Definition (Dual relation)

Let $\Gamma$ be a linear relation from $V$ into $W$. We define the dual relation of $\Gamma$ to be the relation from $W^*$ into $V^*$ defined as \begin{align} \Gamma’:=\{(\omega,\nu)\in W^*\times V^*\,:\, \text{for all $(v,w)\in \Gamma$, $\omega(w)=\nu(v)$}\}. \end{align}

It is easily verified that $\Gamma’$ is also a linear relation. Also, notice carefully that this goes from $W^*$ to $V^*$ even though originally we went from $V$ to $W$. This may seem ad-hoc but it’s not. Recall that if $T:V\to W$ is a linear map, then the dual map is $T’:W^*\to V^*$ defined as $T’(\omega):=\omega\circ T$. Then, we can easily see that $[\text{graph}(T)]’=\text{graph}(T’)$ as subspaces of $W^*\times V^*$. Also, writing things this way highlights a nice “symmetry” in the definitions I think. For example, it is very obvious that the inverse of the dual relation is the dual of the inverse relation (the inverse relation being defined in the obvious manner even at the set-theoretic level: $\Gamma^{-1}=\{(y,x)\,:\, (x,y)\in\Gamma\}$).

People often call this the adjoint and denote it as $\Gamma^*$, but we shall reserve this term for the Hilbert space case.

• The definition above makes perfect sense in the Banach space context as well, we simply have to remember that $E^*,F^*$ now mean the topological dual spaces (the space of continuous linear functionals).
• If $H_1,H_2$ are Hilbert spaces and $\Gamma$ a linear relation from $H_1$ into $H_2$, then $\Gamma’$ is a linear relation from $H_2^*$ into $H_1^*$, i.e $\Gamma’\subset H_2^*\times H_1^*$. We can now use the Riesz (conjugate-linear) isomorphism $H_i\cong H_i^*$ to get a subset $\Gamma^*\subset H_2\times H_1$. This will also be a subspace, called the adjoint relation of $\Gamma$. In other words, (say we choose inner products to be conjugate linear in the first slot and linear in the second slot) \begin{align} \Gamma^*=\{(\eta,\xi)\in H_2\times H_1\,:\, \text{for all $(x,y)\in\Gamma$, $\langle \xi,x\rangle_{H_1}=\langle \eta,y\rangle_{H_2}$}\}. \end{align}

From now on, let us stick to the Banach space case (the Hilbert space case is nearly identical and just one Riesz isomorphism away). The main “theorem” (it’s simply a trivial unwinding of definitions) which will answer your questions is the following.

Theorem.

Let $\Gamma$ be a linear relation from a Banach space $E$ into a Banach space $F$. Then, the multivalued part of the dual relation $\Gamma’$ is the annihilator of the domain of $\Gamma$ in $E$, i.e \begin{align} \mathfrak{M}(\Gamma’)= [\mathrm{dom}(\Gamma)]^0:=\{\epsilon\in E^*\,:\, \text{for all $x\in \mathrm{dom}(\Gamma)$, $\epsilon(x)=0$}\}. \end{align} In particular, $\Gamma’$ is the graph of an operator if and only if $\mathrm{dom}(\Gamma)$ is dense in $E$.

To prove the first statement, we simply unwind definitions of everything: \begin{align} \mathfrak{M}(\Gamma’)&:=\{\epsilon\in E^*\,:\, (0_{F^*},\epsilon)\in \Gamma’\} \tag{def of multivalued part}\\ &:=\{\epsilon\in E^*\,:\, \text{for all $(x,y)\in \Gamma$, $\epsilon(x)=0_{F^*}(y)$}\} \tag{def of dual relation}\\ &=\{\epsilon\in E^*\,:\, \text{for all $(x,y)\in\Gamma$, $\epsilon(x)=0$}\}\\ &=\{\epsilon\in E^*\,:\, \text{for all $x\in \mathrm{dom}(\Gamma)$, $\epsilon(x)=0$}\}\\ &:=[\mathrm{dom}(\Gamma)]^0\tag{def of annihilator}. \end{align} This completes the proof of the first statement. For the second statement, recall that $\Gamma’$ is the graph of an operator if and only if its multivalued part is trivial, i.e if and only if $[\mathrm{dom}(\Gamma)]^0=\{0\}$. But the annihilator of a subspace is trivial if and only if it is dense (proof below), hence completing the proof.

For the sake of completeness here’s a proof of the final assertion about annihilators.

Lemma (Annihilators and density)

Let $E$ be a Banach space and $S\subset E$ a subspace. Then, the annihilator $S^0$ is trivial if and only if $S$ is dense in $E$.

To prove $(\impliedby)$, consider any $\epsilon\in S^0$. Then by definition, $\epsilon$ is an element of $E^*$ such that $\epsilon|_{S}=0$, and so by continuity of $\epsilon$ and density of $S$ it follows $\epsilon=0$. Hence, $S^0$ is trivial.

The implication $(\implies)$ follows by the Hahn-Banach theorem (in the Hilbert case, you can use the Riesz theorem instead): contrapositively, suppose that $S$ is not dense. Then, there is an $x\in E\setminus\overline{S}$, and so by Hahn-Banach, there is an $\epsilon\in E^*$ such that $\epsilon|_{\overline{S}}=0$ but $\epsilon(x)=1$ say. In particular, $\epsilon\in S^0$ but $\epsilon\neq 0$. (In the Hilbert space case, you could take $\epsilon=\frac{1}{\|x\|^2}\left\langle x,\cdot\right\rangle$).

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Closing Remarks.

If you now consider any operator $T:\mathrm{dom}(T)\subset E\to F$ between Banach spaces, you could instead look at its graph $\Gamma(T)\subset E\times F$, and consider the dual relation of this, $[\Gamma(T)]’\subset F^*\times E^*$. By the above remarks, this will then be the graph of an operator (called $T’$) if and only if $\mathrm{dom}(\Gamma(T))=\mathrm{dom}(T)$ is dense in $E$.

Btw, the concepts of extensions, restrictions, closability, closed, closure apply very intuitively at the level of graphs, whereas they are sometimes phrased in ad-hoc manners when discussing things only at the level of operators. This device of phrasing matters in terms of graphs is very useful in this realm of functional analysis (and a huge chunk of the utility of course has to do with the closed-graph theorem).

Answer 2

Definition : Suppose that $H$, $K$ are Hilbert spaces. An unbounded operator $T:H\to K$ is a subspace $G_T$ of $H\times K$ such that if $(0,k)\in G_T$ then, $k=0$. If $\pi_1:\times K\to H$ and $\pi_2:H\times K\to K$ are projections then, $\pi_1(G_T)$ is called the domain of $T$ and is denoted by, $\mathcal{D}(T)$ and $\pi_2(G_T)$ is called the range of $T$ and is denoted by, $\mathcal{R}(T)$.

Definition : An unbounded operator $T:H\to K$ between Hilbert spaces $H$ and $K$ is said to be densely defined if $\mathcal{D}(T)$ is a dense subspace of $H$.

Definition : An unbounded operator $T:H\to K$ between Hilbert spaces $H$ and $K$ is said to be closable if $\overline{G_T}$ has the property : whenever $(0,k)\in\overline{G_T}$ we have, $k=0$.

Definition : Suppose that $T:H\to K$ is an unbounded operator between Hilbert spaces $H$ and $K$ with domain $\mathcal{D}(T)$ and range $\mathcal{R}(T)$. We defined the adjoint domain for $T$ (denoted by, $\mathcal{D}(T^\star)$ by : $$\mathcal{D}(T^\star)=\{y\in K\,:\,\exists\, c_y>0\,\text{satisfying}\,||Tx||\leq c_y||x||\,\forall\,x\in\mathcal{D}(T)\}$$

Motivation behind the last definition : Recall the existence proof of adjoint of a bounded operator between Hilbert spaces. We use the Riesz representation theorem. But to prove it we have to use boundedness of a sesquilinear form. Now look at the adjoint domain. The set is defined in such a way that the "underlying sesquilinear form" becomes continuous/bounded.

Theorem : Suppose that $H$, $K$ are Hilbert spaces and $T:H\to K$ is a densely defined unbounded operators. Then, there exists a unique unbounded operator $T^\star:K\to H$ with domain $\mathcal{D}(T^\star)$ such that, $\langle Tx,y\rangle=\langle x,T^\star y\rangle$ for all $x\in\mathcal{D}(T)$, for all $y\in\mathcal{D}(T^\star)$.

Proof : We first prove the existence. Let, $y\in\mathcal{D}(T^\star)$. Now define, $F:\mathcal{D}(T)\to\mathbb{C}$ by, $Fx=\langle Tx,y\rangle$. Then, $|Fx|=|\langle Tx,y\rangle|\leq ||Tx||.||y||\leq c_y||y||.||x||$. So, $F$ is a bounded linear functional on $\mathcal{D}(T)$. Since, $T$ is densely defined, $F$ extends to a (necessarily unique) bounded linear functional $F:H\to\mathbb{C}$. Then by Riesz's representation theorem, there exists a unique $h_y\in H$ such that, $Fx=\langle x,h_y\rangle$ for all $x\in H$. From this argument, we define, $T^\star:\mathcal{D}(T^\star)\to H$ by, $T^\star y=h_y$. One can then show that $T^\star$ is a linear map and hence it's graph (or you can identity the graph with $T^\star$ itself) is an unbounded operator $K\to H$. This proves the existence part. We will see that for the uniqueness part one uses the densely defined fact !

Suppose that there is another $S:\mathcal{D}(T^\star)\to H$ such that $\langle x,Sy\rangle=\langle Tx,y\rangle$ for all $x\in\mathcal{D}(T)$ and $y\in\mathcal{D}(T^\star)$. Then one has, $\langle x,T^\star y\rangle=\langle x,Sy\rangle$ for all $x\in\mathcal{D}(T)$ and $y\in\mathcal{D}(T^\star)$. Now to show that $T^\star y=Sy$ you need density of $\mathcal{D}(T)$ in $H$.