This is an exercise (13.8) in Rudin's Functional Analysis.

Let an operator $T$ in $L^{2}(\mathbb R)$ be defined as follows:

$\mathcal{D}(T)=\{f \textrm{ absolutely continuous}\in L^{2}(\mathbb{R}): f'\in L^{2}(\mathbb{R})\}; TF=if'$.

The exercise is to show that T is self adjoint. Here is what I have done so far:

Let $f,g \in \mathcal{D}(T)$.

Then $\int_{\mathbb{R}}Tf \bar{g}d\mu=\int_{\mathbb{R}}if'\bar{g}d\mu=-i\int_{\mathbb{R}}f\bar{g'}d\mu$ (using integration by parts and the fact that $f(t)\rightarrow 0$ as $t\rightarrow \pm\infty$).

So we have $<Tf,g>=<f,Tg> \forall f,g \in \mathcal{D}(T)$.

Hence $T$ is symmetric, i.e. $T\subset T^{*}$.

It remains to show that $T^{*}\subset T$.

I don't know if the rest of what follows is correct:

Let $f\in \mathcal{D}(T), g\in \mathcal{D}(T^{*})$ and let $\phi=T^{*}g$.

Let $\Phi(x)=\int_{-\infty}^{x}\phi(t)dt$ (Does this make sense? Is $\phi$ absolutely continuous?)

Then $\Phi'(x)=\phi(x)$

$\int_{-\infty}^{\infty}if'\bar{g}=<Tf,g>=<f,\phi>=\int_{-\infty}^{\infty}f\bar{\phi}d\mu=-\int_{-\infty}^{\infty}f'\bar{\Phi}d\mu$

Hence $\int_{-\infty}^{\infty}f'\overline{(\Phi-ig)}d\mu=0$.

Since this is true $\forall f \in \mathcal{D}(T)$, it follows that $ig-\Phi\in R(T)^{\perp}$.

Is it true that $R(T)=L^{2}$? In that case, $g$ is a scalar multiple of $\Phi$ and so $g\in \mathcal{D}(T)$.

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