# Convergence in Hardy spaces

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 ?

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

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 $$n$$.
• 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 Commented Nov 27, 2021 at 19:58
• @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 $H^\infty(\mathbb{D}) \subset H^p(\mathbb{D})$. Commented Nov 27, 2021 at 23:11
• @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 $H^p$