It is well known that a complex harmonic function $f$ on a simply connected domain $D$ has a canonical decomposition of the form $$f=g+\bar{h},$$ where $g$ and $h$ are analytic functions on $D.$ In fact, this decomposition is unique if we assume $g(z\_0)=0,$ for some $z\_0\in D.$


Let $\{f\_n\}$ be a sequence of non constant complex valued harmonic functions defined on a simply connected domain $D$ such that $\{f\_n\}$ converges locally uniformly to some non constant harmonic function $f=g+\bar{h}.$


Assuming $\{f\_n\}$ has a canonical decomposition $f\_n=g\_n + \overline{h\_n}.$


Since $\{f\_n\}\longrightarrow f$ locally uniformly, can we say that $\{g\_n\}\longrightarrow g$ and $\{h\_n\}\longrightarrow h$ locally uniformly?



Component wise convergence of a sequence of complex harmonic functions
The answer is positive. Consider a small disk $D(a,r)$ in the domain of
uniform convergence (the center is at $a$, radius $r$), and expand $f\_n(a+te^{i\theta})$ into a harmonic Fourier series:
$$f\_k(a+te^{i\theta})=\sum\_{n=-\infty}^\infty c\_{k,n}t^ne^{in\theta},\quad t<r.$$
Then we have $c\_{k,n}\to c\_n$, as $k\to\infty$ for each $n$, in view of
uniform convergence. Therefore, the analytic parts
$$g\_k(a+te^{i\theta})=\sum\_{n=0}^\infty c\_{k,n}te^{in\theta}$$
converge uniformly on each smaller disk. Then $h\_k=\overline{f\_k-g\_k}$
also converge.


Remark. The Hilbert space $L^2$ of functions on the circle $\partial D(a,r)$ is a direct sum $L^2=H^2\_++H^2\_-$, where $H^2\_+$ and $H^2\_-$
are "analytic" and "anti-analytic" parts. So the map $f\mapsto g$
is just a projection onto $H^2\_+$. This gives an explicit estimate of norms of $h\_n$ in terms of norms of $f$.


