For a homework exercise ($1.8$ in the book An Introduction to Operators on the Hardy-Hilbert Space) I am asked to show
Let $u$ be a real-valued function in $L^2(S^1)$. Show that there exists a real-valued $v$ in $L^2(S^1)$ such that $u + iv \in \widetilde{\mathbf H}^2$.
Here $L^2 := L^2(S^1)$ is the space of $L^2$ functions on the unit circle $S^1$ and $\widetilde{\mathbf H}^2$ (a variation of the Hardy space) is defined as $\{f \in L^2 \colon \forall n < 0\,\,\langle f, e_n \rangle = 0\}$. The inner product is given by $\langle f, g \rangle = \frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\overline{g(e^{i\theta})}d\theta$ for $f, g \in L^2$ and $\{e_n \colon n \in \mathbf Z\}$ is the orthonormal basis of $L^2$ defined by $e_n(e^{i\theta}) = e^{in\theta}$.
What I have tried
I was told that I could try to find necessary and sufficient conditions for the Fourier coefficients of $u$ such that it is real-valued. The first thing I could think of was to try to prove that the Fourier coefficients themselves are real. Suppose that $u = \sum_{n \in \mathbf Z}a_ne_n$, i.e. its Fourier coefficients are equal to $\{a_n \colon n \in \mathbf Z\}$. Evaluating $u$ at $1$ gives us $u(1) = u(e^{i0}) = \sum_{n \in \mathbf Z}a_ne^{in0} = \sum_{n \in \mathbf Z}a_n$. Similarly $u(e^{i\pi}) = \sum_{n \in \mathbf Z}(-1)^na_n$.
This result does not imply that all $a_n$ are real, since we could take $a_1 = -i$, $a_{-1} = i$ and $a_n = 0$ for all $n \in \mathbf Z \setminus \{-1, 1\}$. Then $u(e^{i\theta}) = ie^{-i\theta} - ie^{i\theta} = 2\sin(\theta)$ so $u$ is real-valued and it can easily be checked that $u$ is square-integrable, but it has complex/non-real Fourier coefficients.
How to proceed?
Right now I am stuck and I would appreciate it if somebody could point me in the right direction. Thank you in advance.
Let $\hat{f}_n$ be the Fourier coefficients of a function $f$. Then $f$ is real iff $\hat{f}_{-n} = \overline{\hat{f}_n}$
You need to choose $\hat{u}_n + i \hat{v}_n = 0 $ for $n <0$, which forces $\hat{v}_n = i \hat{u}_n$ for $n<0$. If $v$ is to be real valued, you need $\hat{v}_n = \overline{\hat{v}_{-n}}$ for $n \ge 0$.
Choose $\hat{v}_0 = 0$, $\hat{v}_n =-i \hat{u}_n$ for $n >0$. (Since $\sum_n |\hat{v}_n|^2 \le \sum_n |\hat{u}_n|^2$, we see that $v \in L^2$.)