# Family of harmonic functions is normal if and only if family of harmonic conjugates is normal

(a) Let $$u$$ be a harmonic function in a simply connected region and $$v$$ be its conjugate. Is it true that $$u$$ is bounded if and only if $$v$$ is bounded.

(b) Assume $$\mathcal{U} = \{u_n\}$$ is a family of harmonic functions and $$\mathcal{V}$$ be family of conjugates. Is it true that $$\mathcal{U}$$ is normal if and only if $$\mathcal{V}$$ is normal.

(c) What if we impose extra condition that there is a point $$z$$ such that $$u_n(z) = v_n(z) = 0$$ for every $$n$$.

My attempt:

(a) This is false since we have an example of holomorphic $$\log(z) = \ln|z| + i\arg(z)$$ where $$\arg(z) \in (-\pi,\pi)$$ so in particular it is bounded but $$\ln|z|$$ is not bounded.

(b) This is also false. Consider $$\mathcal{U} = \{n\}$$ and $$\mathcal{V} = \{0\}$$. Then $$f_n(z) = n + i0 = n$$ is holomorphic so $$\mathcal{U}$$ and $$\mathcal{V}$$ are harmonic, but $$\mathcal{U}$$ is unbounded and is not normal, while $$\mathcal{V}$$ is uniformly bounded and is thus normal.

(c) I am not sure how to incorporate the condition that $$u_n(z) = v_n(z) = 0$$ for some $$z$$. I also wonder whether we necessarily need that $$u_n(z) = v_n(z) = 0$$ or whether this can be replaced by condition that there are $$z_1, z_2 \in \mathbb{C}$$ such that $$u_n(z_1) = a$$ and $$v_n(z_2) = b$$ for some complex numbers $$a,b$$.

I would appreciate any feedback and help with part (c).

Let $$(u_n)$$ be a sequence of harmonic functions in the simply connected region $$G$$ which converges locally uniformly to the harmonic function $$u$$. Let $$(v_n)$$ be the corresponding sequence of harmonic conjugates normalized by $$v_n(z_0) = 0$$, and $$v$$ the harmonic conjugate of $$u$$ with $$v(z_0) = 0$$.

The functions $$f_n = u_n + i v_n$$ and $$f=u+iv$$ are holomorphic in $$G$$. Let $$a \in G$$ and $$r > 0$$ such that the closed disk with center $$a$$ and radius $$2r$$ is contained in $$G$$. By the Borel–Carathéodory theorem (with $$R=2r$$) is $$|f_n(z) - f(z)| \le 2 \max \{ |u_n(w) - u(w)| : |w-a| \le 2r \} + 3 |f_n(a) - f(a)|$$ for $$|z-a| \le r$$. It follows that $$f_n \to f$$ uniformly in $$B_r(a)$$, and therefore $$v_n \to v$$ uniformly in $$B_r(a)$$.

This proves that $$v_n \to v$$ locally uniformly in $$G$$.

Therefore, if $$\cal U$$ is a normal family of harmonic functions in $$G$$ and $$\cal V$$ the family of harmonic conjugates with $$v(z_0) = 0$$ then every sequence in $$\cal V$$ has a locally uniformly convergent subsequence, i.e. $$\cal V$$ is normal.

Of course the condition $$v(z_0) = 0$$ for all $$v \in \cal V$$ can be replaced by $$v(z_0) = b$$ for any $$b \in \Bbb R$$.

If even suffices that $$\{ v(z_0) \mid v \in \cal V \}$$ is bounded for some $$z_0 \in G$$. Then the Borel–Carathéodory theorem shows that the functions $$f_n = u_n + v_n$$ are locally uniformly bounded, so that $$(f_n)$$ and consequently $$(v_n)$$ has a local uniformly convergent subsequence.

If both $$\{ u(z_0) \mid u \in \cal U \}$$ and $$\{ v(z_0) \mid v \in \cal V \}$$ are bounded for some $$z_0 \in G$$ then the above argument works in both directions, and we have that $$\cal U$$ is normal if and only if $$\cal V$$ is normal.

• Thank you for the answer. I just wanted to see whether I understand the argument. The Borel Caratheodory theorem gives us the bound on the ball, so in particular for any compact set K we can cover it with finitely many balls and take the maximum as the bound. Would that give us convergence of $v_n$ on the whole compact set. Also, I do not quite see where did we use the fact that $v_n(z_0)=0$. Commented Aug 2 at 0:24
• @SparklyCape290: Because of $v_n(z_0) = v(z_0) = 0$ we have $|f_n(a) - f(a)| \to 0$ at every point $a$. Then B-C gives $v_n \to v$ uniformly on every compact set because that can be covered with finitely many disks. Commented Aug 2 at 1:46