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.
Best Answer
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 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 $$\Gamma(A^{\ast\ast}) = \overline{\Gamma(A)}.$$