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Let $f(z) = \sin(z),$ viewed as a function in the hardy space $H^1.$ How do I factor $f$ into inner and outer factors? I know that the formula for the outer factor is $$Q_f(z)= \exp(\dfrac{1}{2\pi} \int_{-\pi}^\pi \dfrac{e^{it}+z}{e^{it}-z} \log|\sin(e^{it})| dt)$$ for $z$ in the open unit disk.

This looks like a mess to compute. Is there a way to explicitly find the inner and outer factors of $\sin(z)?$

Obviously, we need to factor out the zeros of $\sin(z)$ in the unit disk, and there is precisely one zero in the unit disk, namely at $0.$So the inner factor $M_f$ contains $z.$ Now, I believe that since $\sin(z)/z$ is analytic in a larger disk than the unit disk, $\sin(z)/z$ is already an outer function. Does this mean that $Q_f(z) = \sin(z)/z?$

How does one in general find the outer and inner factors of a function $f \in H^1?$ What are some good examples to keep in mind?

inequalitynoob2
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    Try to to find all zeros and construct the inner factor $B$ first. Then $\sin(z)/B(z)$ is the outer one. – A.Γ. May 13 '18 at 13:24
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    @A.Γ. That is what I tried to do. Am I correct that the inner factor is $z?$ How do I find the inner factor? In general, I know that the inner factor splits into a factor that has no zeros, and a Blaschke product. The Blaschke product is not hard to find in general, but how do I go about finding the factor of the inner factor with no zeros? – inequalitynoob2 May 13 '18 at 13:43

1 Answers1

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This is a very old question, but seeing as it is unanswered I'll provide an answer.

I'll use the following very useful result, which is a simple exercise in applying the uniqueness of the inner-outer factorization:

Proposition. Let $p\in(0,\infty]$, and suppose $f\in H^p$ is such that $\frac{1}{f}\in H^p$. Then $f$ is an outer function.

We show that the inner-outer factorization of $f(z)=\sin z$ is precisely

$$f(z)=\underbrace{\frac{\sin z}{z}}_{\text{outer factor}}\cdot \underbrace{z}_{\text{inner factor}}$$

just as you claim (treating $\frac{\sin z}{z}$ as being equal to $1$ at $z=0$ for obvious reasons). By the uniqueness of the inner-outer factorization, it suffices to show that $\frac{\sin z}{z}$ is outer and $z$ is inner. That $z$ is inner is trivial. To show that $F(z)=\frac{\sin z}{z}$ is outer, we can use the proposition. Clearly $F\in H^\infty$ (it is an entire function), and $\frac{1}{F}\in H^\infty$ as the singularity at the origin is removable, and any other zero of $\sin$ occurs at an integer multiple of $\pi$, so $\frac{1}{F}$ is holomorphic on a disc strictly larger than the unit disc. Thus the proposition implies that $F$ is an outer function. The claim now follows.

Lorago
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