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Let $H^2(\mathbb{D})$ denote the complex Hardy space, this is: analytic functions defined unit disc $\mathbb{D}$ whose coefficients form a sequence in $\ell^2$. Functions in $H^2(\mathbb{D})$ have a radial limit for almost all $z\in\mathbb{T}$ the border of $\mathbb{D}$.

It is said that $q\in H^2$ is inner if $|q(z)|=1$ for almost all $z\in\mathbb{T}$.

Now, I have read the following lemma in Helson's book Lectures on Invariant Spaces:

There exists $f\in H^2$ such that $q\cdot\overline{f}\notin H^2$ for all $q\in H^2$ inner.

The proof relies on two facts:

  1. if $f\in H^2$ and $q\cdot\overline{f}\in H^2$ then the real part of $f$ must be non-zero almost everywhere in $\mathbb{T}$ (unless $f$ is identically zero),
  2. there exists $f\in H^2$ non identically zero such that its real part vanishes in a set of positive measure of $\mathbb{T}$.

I fail to see why any of these two facts hold and the book gives no further explanation either. Any ideas?

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First of all both of these facts are wrong. For the first one we have to assume that $f$ is not a constant (since $f(z) = i$ works). For the second one it is actually a classical and very important result that there are no such functions, but surely Helson meant that there is a function $f\in H^2$ not constant such that its real part vanishes on the set of positive measure, to connect it with the first statement, maybe you incorrectly copied it from the book?

For the first one assume that $\text{Re} f$ is zero on a set of positive measure. Then $f\in H^2$, so $qf\in H^2$ ($qf$ is analytic and boundary values are in $L^2$), and since $q\bar{f}\in H^2$, we get $qf+q\bar{f} = q \text{Re} f\in H^2$, so this function is zero on a set of positive measure, hence it is identically zero by a result I mentioned in the first paragraph. Since $q\neq 0$ almost everywhere it implies that $\text{Re} f = 0$ almost everywhere, so the function $f$ is purely imaginary, but since it is also analytic, it must be constant.

For the second one simply take a non-constant $L^2$ harmonic function $u$ whose boundary values are zero on a set of positive measure (e.g. by taking the Poisson extension of the characteristic function of some set which is not $\varnothing$ and not $\mathbb{T}$ up to measure zero), and then define $f(z) = u(z) + i\tilde u (z)$, where $u$ is a harmonic conjugate function.

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    $\begingroup$ (1) is stated as in the book, but it seems you're right that it is wrong. (2) is about the same that you say. $\endgroup$
    – pipenauss
    Commented Aug 11 at 21:13

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