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 ?
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 $n$.