Problem
Consider the space $\mathcal D$ of continuously differentiable functions on the unit circle and the operator on $L^2$ (of the unit circle) with domain $\mathcal D$ given for $f\in\mathcal D$ by $A(f)=if'$. Prove that $A$ is symmetric. Prove that if $f$ is in the domain of $A^\dagger$ one may find a continuous function $g$ such that $f=g$ almost everywhere. Hint: this is difficult if you do not use Fourier series.
Exercise 2.5.12 in What is a Quantum Field Theory by Michel Talagrand. It's the second part I'm asking about.
Failed attempt
Suppose $f\in\mathcal D(A^\dagger)$. Then by definition we may choose a constant $C$ such that $|(f,A(g))|\le C\Vert g\Vert$ for all $x\in\mathcal D(A)=\mathcal D$.
The Fourier series of $f^*$ is $g(y)=\sum_{n\in\mathbb Z}a_n e^{inx}$ where $a_n=\frac1{2\pi}\int_0^{2\pi}f(x)^*e^{-inx}dx$.
For each $n\in\mathbb Z\setminus\{0\}$, define $h_n\in L^2$ by $h_n(x)=n^{-1}e^{-inx}$. Then $h_n\in\mathcal D(A)$, since $h_n(0)=h_n(2\pi)=n^{-1}$; and $(A(h_n))(x)=ih_n'(x)=e^{-inx}$. Note also $\Vert h_n\Vert^2=2\pi/n^2$. Hence $(f,A(h_n))=\int_0^{2\pi}f(x)^*e^{-inx}dx=2\pi a_n$, so by our choice of $C$ we have $|a_n|\le\frac C{2\pi}\Vert h_n\Vert=\frac C{\sqrt{2\pi}}|n|^{-1}$.
…
I was aiming to use the Weierstrass M-test to show that the Fourier series $g=a_0 + \sum_{n\in\mathbb Z\setminus\{0\}}a_ne^{inx}$ converges uniformly, and then show that $g$ is the continous function sought. But I fall short: we have $|a_ne^{inx}| \le M_n = \frac{C}{\sqrt{2\pi}}|n|^{-1}$, but $\sum_{n=1}^\infty n^{-1}$ does not converge.
Can this be tightened up, or do I need a different approach?
Best Answer
For simplicity, we think of the circle as the interval $[0,1]$, $L^2(\mathbb{S}^1)$ as $L^2([0,1])$, and $\mathcal{D}$ as $\{ f \in C^{1}([0,1]): f(0)=f(1), f'(0)=f'(1) \}$.
Let $(y,z) \in A^*$, i.e $y \in \text{Dom}(A^*), z \in L^2$ and $A^*y=z$. Thus, $\int_0^1 z(t) \mathrm{d}t=0$ ( * ).
Define, $\hat{y}(t)=-i\int_0^t z(s)\mathrm{d}s +\int_0^1 y(s)\mathrm{d}s.$ Because of ( * ), for all $x \in \mathcal{D}$, we have: \begin{align}\langle Ax,\hat{y} \rangle &=\int_0^1 ix'(t)\overline{\left( -i\int_0^t z(s)\mathrm{d}s +\int_0^1 y(s)\mathrm{d}s\right)} \mathrm{d}t \\ &=-\int_0^1\left(\int_0^t x'(t) \overline{z}(s)\mathrm{d}s\right)\mathrm{d}t\stackrel{\text{Fubini}}{=}\langle x,z \rangle.\end{align} Or, $\langle Ax,\hat{y} \rangle = \langle Ax,y \rangle $ for all $x \in \mathcal{D}.$ Therefore, $\hat{y}-y=\mathrm{const}$ a.e. Besides, $\int_0^1 \hat{y}(t)\mathrm{d}t =\int_0^1 y(t)\mathrm{d}t $. Hence, the desired conclusion.