Existence Adjoint Operator: Boundedness?

Question

Context

This would make the check on the GNS construction much more simple.

Problem

Given a Hilbert space $\mathcal{H}$.

Consider a merely linear operator $A:\mathcal{H}\to\mathcal{H}$.

Suppose it has a formal adjoint: $$\langle A^*\psi,\varphi\rangle=\langle\psi,A\varphi\rangle\quad(\varphi,\psi\in\mathcal{H})$$ (Obviously, it must be unique!)

Does it imply boundedness: $$\|A\varphi\|\leq\|A\|\cdot\|\varphi\|$$ $$\|A^*\psi\|\leq\|A^*\|\cdot\|\psi\|$$

I guess not as one has only: $$\|A\varphi\|=\sup_{\|\hat{\chi}\|=1}|\langle\hat{\chi},A\varphi\rangle|=\sup_{\|\hat{\chi}\|=1}|\langle A^*\hat{\chi},\varphi\rangle| \\\leq\sup_{\|\hat{\chi}\|=1}\|A^*\hat{\chi}\|\cdot\|\varphi\|=\|A^*\|\cdot\|\varphi\|$$ $$\|A^*\psi\|=\sup_{\|\hat{\chi}\|=1}|\langle A^*\psi,\hat{\chi},\rangle|=\sup_{\|\hat{\chi}\|=1}|\langle\psi,A\hat{\chi}\rangle| \\\leq\|\psi\|\cdot\sup_{\|\hat{\chi}\|=1}\|A\hat{\chi}\|=\|\psi\|\cdot\|A\|$$ (So they must be either both bounded or both unbounded.)

Is there an example where boundedness fails?

Besides, this is equivalent to the adjoint over Banach spaces satisfying a purely functional relation...

Greetings

Thanks to so many grate answers!! :)

Answer 1

Verify directly that $A$ is closed.

Assume $\{ x_n \}$ and $\{ Ax_{n}\}$ converge to $x$, $y$ respectively. Then, for all $z \in \mathcal{H}$, $$ (Ax_n, z) = (x_n, A^{\star}z) \implies (y,z)=(x,A^{\star}z)=(Ax,z) \implies y = Ax $$

Answer 2

You're saying that for every $\psi \in \mathcal H$, the linear functionals $\phi \to \langle \psi, A \phi \rangle$ are bounded. Use the uniform boundedness principle...