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I consider for $p \in [1, \infty)$ the Hardy Space

$$ H^p(\mathbb{D}) = \\ \left\lbrace f:\mathbb{D} \rightarrow \mathbb{C} \, : f \, \text{is holomorphic and} \, \sup_{0\leq r < 1} \left(\frac{1}{2 \pi} \int_{0}^{2 \pi} |f(r \exp(i \theta))|^p d\theta \right)^{1/p} < \infty \right\rbrace $$

the norm of this space is given by

$\|f\|_p = \lim_{r \rightarrow 1 } \left(\frac{1}{2 \pi} \int_{0}^{2 \pi} |f(r \exp(i \theta))|^p d\theta \right)^{1/p} $

I know that convergence in Hardy spaces in the disk implies uniform convergence on compacts: be a $K \subset \mathbb{D}$ compact, you have inequality

$$ \sup_{z \in K} |f(z)|\leq \frac{\|f\|_p}{1-r} .$$

for some $r \in (0,1)$.

Is the converse true? the uniform convergence on compacts implies convergence in norm $\mid \mid \cdot \mid \mid_p$ ? some counterexample ?

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    $\begingroup$ This is very vague - one should spell out clearly hypothesis and conclusion $\endgroup$
    – Conrad
    Commented Nov 27, 2021 at 16:02
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    $\begingroup$ sorry, I just edited my question; I hope it's clearer $\endgroup$ Commented Nov 27, 2021 at 16:30

1 Answer 1

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The converse is false: every holomorphic function on $\mathbb{D}$ it the limit (uniformly on compact subsets) of a sequence of polynomials. On the other hand, not every function on $\mathbb{D}$ is the limit of a sequence of polynomial in $\|\cdot\|_p$, simply because not every holomorphic function is in $H^p(\mathbb{D})$, e.g. $\exp\left(\frac{z+1}{1-z}\right)$.

One could also ask whether $f_n\overset{\text{unif. comp.}}{\to} f$, with $f_n,f\in H^p$, implies $\|f_n-f\|_p\to 0$. This is easy to solve for $p\in [1,\infty]$: just take $f_n=z^n$. Then $f_n\to 0$ uniformly on compact subsets, but $\|f_n\|_p=1$ for every $n$.

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  • $\begingroup$ That is of course true but doesn't quite solve the problem in the case we know apriori that $f_n, f$ are in $H^p$ and $f_n$ converges normally to $f$, does it then follow that it converges in norm too? I suspect the answer is still no $\endgroup$
    – Conrad
    Commented Nov 27, 2021 at 19:58
  • $\begingroup$ @Pelota why function $\exp(\frac{z+1}{z-1})$ is not in $H^p(\mathbb{D}) $? This function is a singular interior function and so is in $ H^\infty(\mathbb{D}) \subset H^p(\mathbb{D})$. $\endgroup$ Commented Nov 27, 2021 at 23:11
  • $\begingroup$ @Caesar The function you wrote is indeed in every $H^p$. The one I wrote, however, is the inverse of that and it is not hard to see that it cannot be contained in any $H^p$ $\endgroup$
    – user840639
    Commented Nov 28, 2021 at 1:15
  • $\begingroup$ @Conrad Added a counterexample for that case $\endgroup$
    – user840639
    Commented Nov 28, 2021 at 14:05

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