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 $|\alpha|=1$. 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 wanted to ask:

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)?

  • 1
    $\begingroup$ 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. :-) $\endgroup$ Jul 31, 2020 at 13:32
  • $\begingroup$ 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.) $\endgroup$ Jul 31, 2020 at 18:25
  • $\begingroup$ By the way, math.stackexchange.com is a better site to ask such questions. $\endgroup$ Jul 31, 2020 at 18:27
  • $\begingroup$ @ChristianRemling I still have a part of the question remaining which asks if the distance is bounded below by any positive real number... $\endgroup$
    – user429197
    Jul 31, 2020 at 18:47

1 Answer 1


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 $1$.

  • $\begingroup$ 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 $1$. $\endgroup$ Jul 31, 2020 at 21:36
  • $\begingroup$ Huh, your comment just disappeared... $\endgroup$ Jul 31, 2020 at 21:36
  • $\begingroup$ Will there be any outer function for which this distance will be some k>0? $\endgroup$
    – user429197
    Aug 2, 2020 at 7:31
  • $\begingroup$ 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 $P_n$. $\endgroup$ Aug 2, 2020 at 20:26

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