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Let $\mathbb{D}$ and $\mathbb{T}$ denote the open unit disk and unit circle in $\mathbb{C}$ respectively. We write $Hol(\mathbb{D})$ for the space of all holomorphic functions on $\mathbb{D}.$ The Hardy spaces on $\mathbb{D}$ are defined as: $$H^{p}:= \left\{ f\in Hol\left( \mathbb{D}\right) :\sup _{r < 1}\int ^{2\pi }_{0}\left| f\left( re^{i\theta}\right) \right| ^{p}d\theta < \infty \right\} \;\;\;\;(0<p<\infty), $$ $$H^{\infty }:= \left\{ f\in Hol\left( \mathbb{D}\right) :\sup_{z\in D}\left| f\left( z\right) \right| < \infty \right\}.$$ A function $g\in H^p(\mathbb{D})$ is outer if there exists a function $G:\mathbb{T}\longrightarrow [0,\infty)$ with $G\in L^1(\mathbb{T})$ such that $$g\left( z\right) =\alpha \text{exp}\left( \int ^{2\pi }_{0}\dfrac {e^{i\theta }+z}{e^{i\theta }-z}G\left( e^{i\theta }\right) \dfrac {d\theta }{2\pi }\right) \qquad(z\in \mathbb{D})$$ and $|\alpha|=1$.

Is it true that any outer function can be written as the quotient of two bounded outer functions? If yes, then how? Also if $f\in H^p$ is outer, then is $\frac{1}{f}$ outer?

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  • $\begingroup$ I dont think reciprocal of a outer function in $H^p$ is outer, because the reciprocal of $\alpha e^{\int_0^{2\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}}G(ie^{i\theta})\frac{d\theta}{2\pi}$ may not be bounded $\endgroup$ – vidyarthi Aug 19 '19 at 11:08
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Your defintion is not quite correct as an outer function in $H^p$ is:

$f(z)= e^{i\theta}\exp\left( \int ^{2\pi }_{0}\dfrac {e^{i\theta }+z}{e^{i\theta }-z}\log G\left( e^{i\theta }\right) \dfrac {d\theta }{2\pi }\right) \qquad(z\in \mathbb{D})$, where $G \ge 0, G \in L^p, \log G \in L^1$

Then both questions are easily shown to be true:

For the first let $\log G = \log^+ G - \log^{-}G$ the standard decomposition of $\log G$ in positive and negative parts; then obviously both:

$f_{\pm}(z)=\exp(-\left( \int ^{2\pi }_{0}\dfrac {e^{i\theta }+z}{e^{i\theta }-z}\log^{\pm} G\left( e^{i\theta }\right) \dfrac {d\theta }{2\pi }\right) \qquad(z\in \mathbb{D}))$ are bounded by $1$

and $f= \frac{f_{-}}{f_+}$, while trivially $ \frac{1}{f}= \frac{f_{+}}{f_{-}}$ corresponding to $\frac{1}{G}$ which is outer in the restricted Nevalinna space (it may not be in $H^q$ for any $q >0$ though)

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  • $\begingroup$ Can you tell how $f_+$ and $f_-$ are bounded by 1? $\endgroup$ – user31459 Aug 20 '19 at 1:44
  • $\begingroup$ corrected the typos (the minus obviously has to be in the exponent), so now everything should be clear since the real parts of the exponent in both $f_{\pm}$ are negative as the real parts of the integrands are $\ge 0$ since the Poisson Kernel (the real part of $\frac{e^{i\theta} +z}{e^{i\theta} -z}$) and $\log^{\pm} G$ are non-negative $\endgroup$ – Conrad Aug 20 '19 at 10:27
  • $\begingroup$ @Conrad Yes. Thank you! $\endgroup$ – user510271 Aug 20 '19 at 10:34

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