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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 $\|A\|\le \|A^*\|$?

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The answer is no. For the counterexample I took inspiration by the example in B. Schleins script on mathematical methods in quantum mechanics (p.25)

https://www.math.uzh.ch/typo3conf/ext/qfq/Source/api/download.php?s=5c976ed6a7a55

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.

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