# Is an operator between Hilbert spaces bounded iff the adjoint is bounded?

Let $$X,Y$$ be Hilbert spaces. Let $$D(A)\subset X$$ be a dense subspace and let $$A:D(A)\to Y$$ be a linear operator. Define the following (not necessarily dense) subspace of $$Y$$:

$$D(A^*):=\{y\in Y|\exists x\in X\forall z\in D(A): \langle x,z\rangle = \langle y,Az\rangle\}$$

Define $$A^*:D(A^*)\to X$$ by $$\langle A^*y,z\rangle = \langle y,Az\rangle$$ for all $$y\in D(A^*),z\in D(A)$$.

If $$A$$ is bounded, it is straightforward to show that so is $$A^*$$ and that $$\|A^*\|\le \|A\|$$.

Is the converse true? I. e., if $$A^*$$ is bounded, does this imply that $$A$$ is bounded and that $$\|A\|\le \|A^*\|$$?

Pick $$f\in L^\infty(\mathbb{R})\setminus L^2(\mathbb{R})$$ and $$0\neq \psi_0\in L^2(\mathbb{R})$$ and define $$D(A) = \{ \psi \in L^2(\mathbb{R}) \ : \ \int \vert \psi(x)\vert \cdot \vert f(x)\vert dx < \infty \}$$ One checks that $$D(A)$$ is dense in $$L^2(\mathbb{R})$$ (approximate $$L^2$$ functions by their restriction to compact invervals) and we define $$A\psi = \langle f, \psi \rangle \psi_0.$$ If now $$\psi \in D(A^*)$$, then there exists $$\varphi\in L^2(\mathbb{R})$$ such that for all $$\eta \in D(A)$$ holds $$\langle \eta, \varphi \rangle = \langle A\eta, \psi \rangle = \langle \eta, \langle \psi_0, \psi\rangle f \rangle$$ As $$D(A)$$ is dense, this implies $$\varphi=\langle \psi_0, \psi\rangle f$$. However, we assumed that $$f\notin L^2(\mathbb{R})$$ and thus, $$\langle \psi_0, \psi\rangle=0$$. Hence, we get that $$D(A^*)$$ is the orthogonal complement of $$\psi_0$$ and $$A^*=0$$. Therefore, $$A^*$$ is bounded. However, $$A$$ is not bounded (as $$\psi \mapsto \langle f, \psi \rangle$$ is not bounded due to $$f\notin L^2(\mathbb{R})$$).
Added: There is a nice property that rules out such annoying behaviour. One can show that if $$A$$ is densely defined and closable, then $$D(A^*)$$ is dense. Hence, if $$A$$ is densely defined and closed we get: $$A$$ is bounded iff $$A^*$$ is bounded (because $$A^{**}=\overline{A}=A$$ is bounded). If you weaken closed to closable you can still say that $$A$$ admits a bounded extension.