I am trying to understand this example from Reed-Simon volume 1:
"Suppose that $f$ is a bounded measurable function, but that $f\notin L^2(\mathbb R)$. Let $D(T) = \{\psi \in L^2(\mathbb R) : \int_{\mathbb R} |f(x)\psi(x)| dx < \infty\}$. $D(T)$ certainly contains all the $L^2$ functions with compact support so $D(T)$ is dense in $L^2(\mathbb R)$. Let $\psi_0$ be some fixed vector in $L^2(\mathbb R)$ and define $T\psi = (f, \psi)\psi_0$ for $\psi\in D(T)$. Suppose that $\varphi\in D(T^*)$, then
$(\psi, T^*\varphi) = (T\psi, \varphi) = ((f(\psi)\psi_0,\varphi) = \overline{(f,\psi)}(\psi_0,\varphi)=(\psi,(\psi_0,\varphi)f)$
for all $\psi\in D(T)$. Thus $T^*\varphi = (\psi_0,\varphi)f$. Since $f\notin L^2(\mathbb R)$, $(\psi_0,\varphi)=0$. Thus any $\varphi\in D(T^*)$ is orthogonal to $\psi_0$, so $D(T^*)$ is not dense. In fact, $D(T^*)$ is just the vectors perpendicular to $\psi_0$, and on that domain $T^*$ is the zero operator."
Why does $f\notin L^2(\mathbb R)$ imply that $(\psi_0,\varphi)=0$? The text doesn't give any justification so there must be a simple reason why but I'm just not seeing it.
For every $\varphi \in D(T^\ast)$, we have $T^\ast\varphi \in L^2(\mathbb{R})$, by definition of the adjoint. Now, we saw that
$$T^\ast\varphi = (\psi_0,\varphi)\cdot f$$
is a constant multiple of $f$ - this part, "Thus $T^\ast\varphi = (\psi_0,\varphi)f$", needs a little justification, I'll do that below. But a constant multiple of $f$ is in $L^2(\mathbb{R})$ only if it is $0$.
The justification:
We know that there is a $g \in L^2(\mathbb{R})$ such that for all $\psi \in D(T)$ we have
$$(T\psi,\varphi) = (\psi,T^\ast\varphi) = \int_\mathbb{R} \overline{\psi(x)} g(x)\,dx.\tag{1}$$
On the other hand, we have
$$(T\psi,\varphi) = \int_\mathbb{R} \overline{\psi(x)}(\psi_0,\varphi)f(x)\,dx\tag{2}$$
as computed in the quoted part. Since $D(T)$ contains all characteristic functions of bounded measurable sets, it follows that
$$(\psi_0,\varphi)f = g$$
almost everywhere.