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Here is a standard theorem about bounded operators: Let $H$ be a Hilbert space. For any bounded linear operator $A:H\to H$ there is a unique bounded operator $A^*$ s.t $\langle Au,v\rangle=\langle u,A^*v\rangle$ for all $u,v\in H$.

The proof that I know of uses Riesz Representation Theorem, which states that a continuous linear functional $T:H\to \mathbb{C}$ on a Hilbert space is of the form $T(u)=\langle u,w\rangle$ for a unique $w\in H$. I have two questions:

1) It's natural to expect that there are linear operators on a pre-Hilbert space that are bounded but have no adjoint. What is an example?

2) When does an unbounded operator on a Hilbert space have an adjoint? I know there are self-adjoint unbounded operators. so I am thinking if there is a characterization for when the adjoint exists?

the first question is more important for me right now. I'm not sure if the second question has a good answer.

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Why is the first claim "natural"? Look at the explicit action of $T^*$: if $Tv = a u$, with $v$, $u$ unit vectors and $a$ scalar, then $T^*u = \bar{a} v$. Seems to me such example cannot exist. –  Michael Jan 25 at 6:15
    
it's natural in the sense that Reisz's theorem uses the Hilbert space structure. And I am not sure if I understand your construction. you're defining an operator $T^*$ whose domain is the range of $T$. This does not make for an adjoint pair. –  adrido Jan 25 at 6:41

2 Answers 2

up vote 1 down vote accepted

It's natural to expect that there are linear operators on a pre-Hilbert space that are bounded but have no adjoint. What is an example?

Any bounded operator $T$ on a pre-Hilbert space $H$ has a unique extension to a bounded operator $\tilde{T}$ on the completion $\tilde{H}$ of $H$. Then $\tilde{T}$ has an adjoint, and if $T$ has an adjoint, i.e. there exists a bounded operator $T^\ast\colon H\to H$ with $\langle Tx,y\rangle_H = \langle x, T^\ast y\rangle_H$ for all $x,y\in H$, that must be the restriction of $(\tilde{T})^\ast$ to $H$, since we have $\langle Tx,y\rangle_H = \langle x, \tilde{T}^\ast y\rangle_{\tilde{H}}$.

So $T\colon H\to H$ has an adjoint if and only if $\tilde{T}^\ast(H) \subset H$.

That criterion makes it easy to construct an example. Let $H \subset \ell^2(\mathbb{N})$ the subspace of sequences with only finitely many nonzero terms. For any $\xi \in \ell^2(\mathbb{N})$, consider $T_\xi \colon x \mapsto \langle x,\xi\rangle_{\ell^2}\cdot e_1$. Since $$\langle T_\xi x, y\rangle_{\ell^2} = \langle \langle x,\xi\rangle_{\ell^2}\cdot e_1,y\rangle_{\ell^2} =\langle x,\xi\rangle_{\ell^2}\cdot \overline{y_1} = \langle x, y_1\cdot\xi\rangle_{\ell^2} = \langle x ,\langle y,e_1\rangle_{\ell^2}\cdot \xi\rangle_{\ell^2},$$

we have $\tilde{T}_\xi^\ast \colon y \mapsto \langle y,e_1\rangle_{\ell^2}\cdot \xi$, in particular $\tilde{T}_\xi^\ast(e_1) = \xi$. If $\xi \notin H$, the restriction of $T_\xi$ to $H$ is a bounded operator on $H$ without adjoint.

Generally, if $H$ is an incomplete pre-Hilbert space, many operators on $H$ will have the property $\tilde{T}^\ast(H) \not\subset H$, so working with adjoints on incomplete pre-Hilbert spaces doesn't look too fruitful (working with incomplete normed spaces generally is rarely done, one usually immediately considers their completion).

When does an unbounded operator on a Hilbert space have an adjoint?

When it is densely defined.

The characterisation of the adjoint by

$$\bigl(\forall x\in D(A)\bigr)\bigl(\forall y\in D(A^\ast)\bigr)\bigl(\langle Ax,y\rangle = \langle x, A^\ast y\rangle\bigr)\tag{1}$$

only determines $A^\ast y$ up to elements of $D(A)^\perp$, and since $(1)$ shall uniquely determine $A^\ast y$ for $y\in D(A^\ast)$, we need $D(A)^\perp = \{0\}$, i.e. $\overline{D(A)} = \{0\}^\perp = H$.

If for some $y\in H$ there exists a $z\in H$ such that for all $x\in D(A)$

$$\langle Ax,y\rangle = \langle x, z\rangle,\tag{2}$$

then $x \mapsto \langle Ax,y\rangle$ is a continuous linear form on $D(A)$ (with the topology induced by $H$).

Conversely, if $x\mapsto\langle Ax,y\rangle$ is continuous on $D(A)$, it extends to a continuous linear form on $H$, and by Riesz' theorem there is a $z\in H$ with $(2)$. That $z$ is uniquely determined by $(2)$ if and only if $D(A)$ is dense in $H$. In any case, the projection of such a $z$ in $\overline{D(A)}$ is uniquely determined by $(2)$, so viewing $A$ as an operator from a dense subspace of $\overline{D(A)}$ to $H$, its adjoint exists.

For an arbitrary densely defined operator, the adjoint is however often not useful. You can have

$$D(A^\ast) = \left\lbrace y \in H : (x\mapsto \langle Ax,y\rangle) \text{ is continuous}\right\rbrace = \{0\}.$$

The adjoint is more useful if it is itself densely defined. So when is $D(A^\ast)$ dense in $H$?

Before answering that, let's take a look how the graphs $\Gamma(A)$ of $A$ and $\Gamma(A^\ast)$ of $A^\ast$ are related. They are both linear subspaces of $H\times H$, which we endow with the inner product

$$\langle (x,y), (v,w)\rangle_{H\times H} =\langle x,v\rangle_H + \langle y,w\rangle_H$$

to obtain a Hilbert space structure. Then

$$\begin{align} (y,z) \in \Gamma(A^\ast) &\iff \bigl(\forall x\in D(A)\bigr)\bigl(\langle Ax,y\rangle_H - \langle x, z\rangle_H = 0\bigr)\\ &\iff \bigl(\forall x\in D(A)\bigr)\bigl(\langle (x,-Ax),(z,y)\rangle_{H\times H} = 0\bigr)\\ &\iff (z,y) \in \Gamma(-A)^\perp. \end{align}$$

So, with the coordinate swapping $S\colon H\times H \to H\times H; \; (x,y) \mapsto (y,x)$, we have

$$\Gamma(A^\ast) = S(\Gamma(-A)^\perp) = S(\Gamma(-A))^\perp.$$

If $A^\ast$ is densely defined, so $A^{\ast\ast}$ exists, then a short manipulation of the above shows that

$$\Gamma(A^{\ast\ast}) = \Gamma(A)^{\perp\perp} = \overline{\Gamma(A)},$$

so the closure of the graph of $A$ must be the graph of a linear operator - $A$ must be closeable.

Conversely, for any linear operator $B\colon D(B) \to H$, where $D(B) \subset H$, we can look at the subspace $R(B^\ast) := S(\Gamma(-B))^\perp$ of $H\times H$. This is the set of $(y,z) \in H\times H$ with

$$\bigl(\forall x \in D(B)\bigr)\bigl(\langle Bx, y\rangle_H = \langle x,z\rangle_H\bigr).$$

By the above, $R(B^\ast)$ is the graph of a linear operator (namely $B^\ast$) if and only if $B$ is densely defined.

So we have: Let $A\colon D(A) \to H$ be a densely defined linear operator. $A^\ast$ is densely defined if and only if $A$ is closeable, i.e. if and only if $\overline{\Gamma(A)}$ is the graph of a linear operator, then $$\Gamma(A^{\ast\ast}) = \overline{\Gamma(A)}.$$

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Thank you! +1. Can you please give some examples of unbounded closable operators? and perhaps a reference where I can learn more about them? –  adrido Jan 25 at 23:32
1  
I'm more at home with continuous operators, so I can only give a somewhat trivial example (or class of examples). Consider $H = L^2(I)$ for a compact interval $I$, and the differentiation operator $A(f) = f'$ defined on the subspace of continuously differentiable functions on $I$. Then $A$ is closeable, but not closed. The closure of $A$ would be the differentiation operator on the Sobolev space $H^1(I) \subset L^2(I)$. Usually, one immediately considers the closure of a closeable operator, since that has nicer properties. Unbounded operators play a big role in partial differential equations, –  Daniel Fischer Jan 25 at 23:44
    
so a source on the theory of PDEs or a PDE-leaning book on Functional Analysis should treat unbounded operators in some depth. Rudin treats them in the last chapter, not too extensively, but the most important things are of course done. –  Daniel Fischer Jan 25 at 23:48
    
I will make sure to read that section. thank you. –  adrido Jan 26 at 7:24

Let me start with your second question. Let $H$ be a Hilbert space. First of all, we recall that

Definition. Let $H$ be a Hilbert space. An unbounded operator on $H$ is a couple $(\mathcal D(A), A)$, where $\mathcal D(A)$ is linear subspace of $H$ and $A: \mathcal D(A) \to H$ is a linear map.

Definition. $(\mathcal D(A), A)$ is closed iff its graph $G_A = \{ (x, Ax), x \in \mathcal D(A)\}$, $G_A < H \times X$ is closed.

Definition. Let $A: \mathcal D(A) \to H$ be closed and suppose $\mathcal D(A)$ is dense in $H$. If $y\in H$ satisfies $\sup_{x \in B_H} | (Ax, y) | < \infty$, then there exists a unique $A^*y \in H$ s.t. $(Ax, y) = (x, A^*y)$ for all $x \in \mathcal D(A)$. Then $y \in \mathcal D(A^*)$. The couple $(\mathcal D(A^*), A^*)$ is the adjoint of $A$.

Why do we require $\mathcal D(A)$ be dense? Because this way we can extend by density the scalar product on $H$ and so apply the Riesz' representation theorem on $( \cdot, y )$ in order to get existence and uniqueness of $A^* y$. Closedness hypotesis ensures continuity of the scalar products $(Ax, y)$ and $(x, A^* y)$.

These notions are quite standard and you can see References for a more complete exposition.

A pre-Hilbert space $K$ is a linear space on a field on which a sesquilinear form (i.e. a scalar product) is defined. The scalar product always induces a norm on $K$, so we can always see $K$ as a normed linear space. Hence we can always define the Banach adjoint, more often called the transpose, of a bounded linear map $T$ on $K$. Recall that

Definition. Let $T : (V, \|\cdot\|_V) \to (W, \|\cdot\|_W)$ be a bounded linear map. The transpose of $T$ is the map $T^* : W' \to V'$ defined by $T^*L = LT$.

Remark. This definition and that of (Hilbert) adjoint coincide when we identify $H$ and $H'$ via Riesz' theorem.

In a pre-Hilbert space, usual definition of the Hilbert adjoint makes no sense, since the definition is based upon completeness of $H$, that ensure existence and uniqueness of projections on closed convex sets (since $H$ Hilbert is uniformly convex and the last concept makes sense only on Banach, i.e. complete, spaces), that is the necessary property to prove Riesz' representation theorem, which is needed to ensure uniqueness of the adjoint.

References.

  • Rudin, Functional analysis
  • Brezis, Functional analysis, Sobolev spaces and partial differential equations
  • Reed, Simon, Methods of modern mathematical phyisics, Functional analysis
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Thank you! I understand your argument for the second question (+1). But in regard to the first question, I would not say that the definition is based on the completeness of $H$ (the definition makes reference only to inner products). It's rather the proof (of existence+uniqueness) that depends on the completeness. –  adrido Jan 25 at 22:37
    
Well, one usually defines the adjoint as the unique operator satisfying $( Ax, y ) = ( x, A^* y) \forall x, y \in H$. In this sense, completeness is necessary for the definition, as we can't claim $A^*$ is unique by definition. However, it's likely a matter of the definition; one can also states $A^*$ is the adjoint of $A$ if satisfies the previous relation and then show it exists and it is unique. Anyway, Daniel's answer is, as usual, more complete and satisfactory than mine. –  Federico Jan 26 at 10:45

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