# Regarding subspace generated by the polynomial multiples of outer functions

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: $$H^{p}:= \left\{ f\in Hol\left( \mathbb{D}\right) :\sup _{r < 1}\int ^{2\pi }_{0}\left| f\left( re^{i\theta}\right) \right| ^{p}d\theta < \infty \right\} \;\;\;\;(0 $$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$$. Let $$\mathscr{P}[h]$$ denote the closed subspace generated by the functions $$z^n h(z),\; n=0,1,2,....$$, ie $$\mathscr{P}[h]$$ consists of all $$H^p$$ functions that can be approximated by polynomial multiples of $$h$$. Note that $$\mathscr{P}[1]=H^p$$, since polynomials are dense in $$H^p$$.

I know that if $$h$$ is not outer, then $$\mathscr{P}[h]\neq H^p$$. Will this also imply that $$1\notin \mathscr{P}[h]$$ with dist$$(1,\mathscr{P}[h])>0.$$? Will this distance be bounded below by some +real number(irrespective of given outer function h)?

• The answer to (1) depends on what you mean by a "subspace generated by (...)". If you mean "closed subspace generated by (...)", as suggested by the remaining part of the quoted sentence, then yes, $\mathscr P[h]$ is closed. :-) – Mateusz Kwaśnicki Jul 31 '20 at 13:32
• You have basically answered your own question already (and the answer is, yes, $1\notin P[h]$ if $h$ is not outer). This follows because $P[h]$ is invariant under multiplication by $z$, so if we had $1\in P[h]$, then this space would contain all polynomials. (And then the distance is positive automatically, as $P[h]$ is closed.) – Christian Remling Jul 31 '20 at 18:25
• By the way, math.stackexchange.com is a better site to ask such questions. – Christian Remling Jul 31 '20 at 18:27
• @ChristianRemling I still have a part of the question remaining which asks if the distance is bounded below by any positive real number... – user429197 Jul 31 '20 at 18:47

The answer to your final question is no if $$p < \infty$$ and yes if $$p = \infty$$.
Let $$a > 0$$ and let $$h_a$$ be a singular inner function given by $$h_a(z) = \exp(-a(1-z)/(1+z)).$$ Note that $$|h_a(z)| \leqslant 1$$ and $$h_a(z) \to 1$$ as $$a \to 0^+$$. Therefore, by the dominated convergence theorem, $$h_a$$ converges to $$1$$ in $$H^p$$ for every $$p < \infty$$ (at least when $$p \geqslant 1$$; I bet the same is true when $$p < 1$$, but I have never worked with these spaces and so I am not entirely sure). In particular, the distance between $$\mathscr P[h_a]$$ and $$1$$ converges to zero as $$a \to 0^+$$.
On the other hand, the distance in $$H^\infty$$ between $$1$$ and any function $$f$$ which is not outer is at least $$1$$, because either $$f$$ has a zero in the unit disk or some non-tangential boundary limit of $$f$$ is equal to $$0$$. Thus, the distance between $$\mathscr{P}[h]$$ and $$1$$ is either $$0$$ or $$1$$.
• The $H^\infty$ norm is equal to the supremum norm inside the unit disk, so if $f$ takes values arbitrarily close to $0$ in the unit disk, then $\|f - 1\|_{H^\infty}$ is at least $1$. – Mateusz Kwaśnicki Jul 31 '20 at 21:36
• There's nothing special about constant $1$, I think, just approximate any $H^p$ function by a polynomial $P_n$, and then this polynomial by an inner function $h_{a_n}$ multiplied by $P_n$. – Mateusz Kwaśnicki Aug 2 '20 at 20:26