Poisson Transform and convergence of bounded harmonic functions

Question

Denote by $\mathscr{P}:L^\infty(\mathbb{P}^1(\mathbb{C})) \rightarrow \mathcal{H}^\infty(\mathbb{H}^3)$ the Poisson transform which is defined as

$$ \mathscr{P}(f)(x):=\int_{\mathbb{P}^1(\mathbb{C})}f(\theta)d\mu_x(\theta)=\int_{\mathbb{P}^1(\mathbb{C})}f(\theta)P(x,\theta)d\theta $$ where $f \in L^\infty(\mathbb{P}^1)$, the measure $d\theta$ is the standard $SO(3,\mathbb{R})$-invariant measure on $\mathbb{P}^1(\mathbb{C})$ and $P(x,\theta)$ is the Poisson kernel. Suppose to have a sequence $(f_k)_{k \in \mathbb{N}}$ of functions in $L^\infty(\mathbb{P}^1(\mathbb{C}))$ such that $||f_k||_\infty < C$ for every $k$. Define $\psi_k:=\mathscr{P}(f_k)$. Since $\mathscr{P}$ respects the $L^\infty$-norm, we know that the sequence $(\psi_k)_{k \in \mathbb{N}}$ is equibounded and hence it admits a subsequence which converges to a bounded harmonic function $\psi$ in the smooth topology. Being $\psi$ bounded and harmonic, there exists $f \in L^\infty((\mathbb{P}^1(\mathbb{C}))$ such that $\mathscr{P}(f)=\psi$. With an abuse of notation, knowing that $\psi_k \to \psi$ in the smooth topology what does imply in terms of convergence of $f_k$ to $f$? Does the corresponding subsequence of $f_k$ converges in some way to $f$?

Answer 1

What's below may be an answer to your question, or it may be an answer to the analogous question in a different context, which can probably be adapted to answer your question. I don't know which because I don't understand your notation.

The context in which I'm going to give an answer: $\Bbb T=\{z\in\Bbb C:|z|=1\}=\partial \Bbb D$, where $\Bbb D=\{z\in\Bbb C:|z|<1\}$. If $f\in L^\infty(\Bbb T)$ then $u=P[f]$ is the Poison integral of $f$, a bounded harmonic function in $\Bbb D$ which has boundary values $f$ almost everywhere.

My confusion over notation: First, I imagine $\Bbb P^1$ denotes one-dimensional projective space. Except if so I would have thought that $\Bbb P^1(\Bbb C)$ consists of just one point, while $\Bbb P^1(\Bbb R^2)\approx \Bbb T$. I really have no idea what you mean by $\Bbb H^3$, hence no idea whether $\Bbb H^3$ is equivalent to $\Bbb D$.

The answer, for Poisson integrals in the context I'm familiar with:

Theorem. Suppose $f_n\in L^\infty(\Bbb T)$, $||f_n||_\infty$ is bounded and $u_n=P[f_n]$. If $f_n\to f$ weak* in $L^\infty(\Bbb T)=(L^1(\Bbb T))^*$ then $u_n\to u=P[f]$ uniformly on compact subsets of $\Bbb D$. Conversely, if $u_n\to u$ uniformly on compact sets and $u=P[f]$ then $f_n\to f$ weak*.

Proof: Suppose $f_n\to f$ weak*. Then the sequence $u_n$ converges pointwise to $v=P[f]$. But since $u_n$ is uniformly bounded some subsequence converges to some harmonic function $u$ uniformly on compact sets. Hence $v=u$, hence $u_n\to u$ uniformly on compact sets.

(Or more directly, using a little bit of functional analysis: Suppose that $f_n\to f$ weak*. Then $\int gf_n\to\int gf$ _uniformly for $g$ in any compact subset of $L^1$, hence $P[f_n]\to P[f]$ uniformly on compact subsets of $\Bbb D$.)

Conversely, suppose $u_n\to u=P[f]$ uniformly on compact sets. Then $\int gf_n\to \int gf$ if $g$ is a linear combination of Poisson kernels; since these are dense in $L^1(\Bbb T)$ and $||f_n||_\infty$ is bounded this shows that $f_n\to f$ weak*.