Domain of an adjoint operator I'm studying a bit of functional analysis for quantum mechanics and I'm stuck on a definition our professor gave us.
Given an operator and its domain $(A,\mathcal{D}(A))$ densely defined in $\mathcal H$ Hilbert space, there exists the adjoint operator $(A^*,\mathcal{D}(A^*))$ which satiesfies
$$(Ax,y)=(x,A^*y),      x\in\mathcal{D}(A),y\in\mathcal{D}(A^*)$$
and it is defined in
$$\mathcal{D}(A^*)=\{y\in \mathcal H|L_y:\mathcal H\rightarrow\mathbb{C},x\mapsto (Ax,y) \text{ is a continuous linear functional}\}$$
It is the continuity part of the statement that I dont understand. I get that if $A$ is continuous that condition is satisfied for every $x,y$, but in general what does $y$ have to do with the continuity of $L_y$, since it acts on $x$? I mean, based on the way $L_y$ is defined I can't understand why it would be continuous for some $y$ but not for others.
 A: This definition of the adjoint operator is a direct application of Riesz' lemma. (I prefer another equivalent definition which does not use Riesz' lemma.)
If $H$ is a Hilbert space, a linear functional $H \ni x \to f(x)\in \mathbb{C}$ is continuous if and only if it is of the form $f(x)=\langle x, z_f\rangle$ for some $z_f\in H$ and this element turns out to be uniquely determined by $f$.
To see that the proposed  definition makes sense, suppose you already know the adjoint operator $A^*$ with its domain  $D(A^*)$ as defined in question.
Saying that  $y \in D(A^*)$ means that $f: D(A) \ni x \mapsto \langle Ax,y\rangle = \langle x, A^*y\rangle$.
This  map is trivially continuous since the inner product is continuous.
Now, exploiting Riesz' lemma, I show that the result is reversible, just applying the definition of $D(A^*)$ given in the question.
If, for some $y\in H$, $f: D(A) \ni x \mapsto \langle Ax,y\rangle$ is continuous, since $D(A)$ is dense, we can extend this functional to an everywhere defined continuous functional we can indicate with the same symbol. Hence we can apply Riesz' lemma obtaining that
$$\langle Ax, y\rangle = \langle x, z_y\rangle \quad \forall x \in D(A)\:.$$
It is not difficult to see that the vectors $y\in H$ satisfying the identity above form a linear subspace, we denote by $D(A^*)$, and that the map $D(A^*) \ni y \mapsto z_y$ is linear. Finally we can define $A^*y := z_y$.
So, regarding your last question, saying that $L_y$ is not continuous, is equivalent to saying that there is no $z_y$ such that $\langle Ax,y\rangle = \langle x, z_y\rangle$ for all $x\in D(A)$.
A: First, notice that if $L_y$ is continuous for every $y \in \mathcal H$, by the uniform boundedness principle you would conclude that $A$ is a bounded operator. So if $A$ is unbounded, there will be some $y\in\mathcal H$ such that $L_y$ is unbounded.

Now, for any $y\in \mathcal H$, we know $L_y$ is at least defined on $\mathcal D(A)$. It can be bounded, in which case it extends to a bounded linear function on the whole $\mathcal H$; but it can also be unbounded (since $A$ is unbounded).

Now, a concrete example. Consider $ \mathcal H = L^2([0,1])$ with the usual scalar product and define a derivative operator $A$ by :
\begin{align}
&\mathcal D(A) = \mathcal C^1([0,1])\\
&\forall \psi \in \mathcal D(A), A\psi =\psi'
\end{align}
Then, if $\phi \in \mathcal C^1([0,1])$ and $\phi(0) = \phi(1) =0$, an integration by part shows that for any $\psi \in \mathcal D(A)$ :
\begin{align}
\langle A\psi,\phi\rangle &= -\langle\psi,\phi'\rangle \\
|L_\phi(\psi)| \leq \|\psi\|\|\phi'\|
\end{align}
Therefore, $L_\phi$ is bounded and extends to a continuous functional $L_\phi:\mathcal H\to \mathcal C$, ie $\phi \in \mathcal D(A^*)$ and :
$$\{\phi\in\mathcal C^1([0,1])|\phi(0)=\phi(1) = 0\}\subset \mathcal D(A^*)$$
However, consider the constant function $\phi = 1$. Then, for $\psi\in\mathcal D(A)$, integration by part gives :
$$\langle A\psi,\phi\rangle = \psi(1)-\psi(0)$$
which is unbounded for the $L^2$ norm. Therefore, $\phi \notin \mathcal D(A^*)$.
