Let $T: L_2(\mathbb{R}) \to L_2(\mathbb{R})$ be the linear operator given by $$(Tx)(t) = \exp(it) x(t)$$ where $x \in L_2(\mathbb{R})$ and $t \in \mathbb{R}$.
Now I need to determine if the adjoint operator of $T$ is well-defined on the Hilbert space $L_2(\mathbb{R})$.
I am not sure what well-defined means in this context. One can easily check that $Tx \in L_2(\mathbb{R})$. I think in this context it just means that I should check if $D(T^*) = L_2(\mathbb{R})$ where
$$D(T^*) = \{y \in L_2(\mathbb{R}) | f_y: x \mapsto (Tx,y) \text{ is a bounded linear functional on } L_2(\mathbb{R})\}$$
is the domain of the adjoint operator. If that is true, we shall check if for each $y \in L_2(\mathbb{R})$ the (obviously linear) functional $f_y$ is bounded.
Let $y \in L_2(\mathbb{R})$, i.e. $\int_{\mathbb{R}} |y(t)|^2 dt < \infty$. It is $$ f_y(x) = (Tx,y) = \int_{\mathbb{R}} \exp(it) x(t) \overline{y(t)} dt. $$
Now I am stuck here and do not know what to do. I have the feeling that one has to choose a specific element $x \in L_2(\mathbb{R})$ depending on $y$ to deduce some information about the boundedness or maybe about $y$. But I have no idea how exactly this should work. The integral and the exponential function there is confusing me too.
Could someone please help me with this problem? Thanks!
You're absolutely correct, the domain of $T^\ast$ is $$D(T^\ast) := \{y \in L_2(\mathbb{R}) \mid \text{$(T(\cdot),y) : L_2(\mathbb{R}) \to \mathbb{C}$ is bounded}\}$$ by definition, and your task is to show that $D(T^\ast) = L_2(\mathbb{R})$.
Consider $$(Tx,y) = \int_{-\infty}^\infty e^{it}x(t)\overline{y(t)}\,dt = \int_{-\infty}^\infty x(t) \overline{e^{-it}y(t)} \, dt.$$ Is Cauchy–Schwarz still applicable, and if so, what does it now tell you?