I am pretty sure other questions on this site can answer this problem, but I'm really interested in knowing if this particular solution is valid. Thanks.
Question: Let $U$ consists of the set of real-valued functions that are harmonic on the open unit disc $D$, continuous on $\bar D$, and uniformly bounded (in absolute value) by some constant $C$. Show that the $k^{th}$ derivatives are uniformly bounded on compact subsets of $D$.
Attempt:
For every $u\in U$, there exists a harmonic conjugate $v$, since $D$ is simply connected.
Let $F$ be the family of analytic functions of the form $f=u+iv$, where $u\in U$.
Let $G$ be the family of analytic functions of the form $g=e^f$, where $f\in F$.
Since $|g|=|e^f|=|e^u|$, and $U$ is a uniformly bounded family, $G$ is a uniformly bounded family.
It follows that $G$ is a normal family, i.e on compact subsets of $D$ every sequence has a uniformly convergent subsequence.
Since the partial derivative of a harmonic function are harmonic, if we can show that $u_x$ and $u_y$ are uniformly bounded on compact subsets the result follows by induction.
Wlog suppose $U_x$ the family of functions $u_x$, $u\in U$ is not uniformly bounded on the compact set $K$. Thus there exists a sequence $\{(u_n)_x \}$ such that $|(u_n)_x| \rightarrow \infty$.
Now since $G$ is normal, the sequence $\{ e^{f_n} \}= \{ e^{u_n +i v_n} \}$ has a uniformly convergent subsequence $\{ e^{f_{n_k}} \}$ converging to some analytic function $g$.
By the Cauchy estimates, the derivatives of this subsequence converge uniformly to the derivatives of $g$. Thus, $$\{ e^{f_{n_k}} f'_{n_k} \} \rightarrow g'.$$
This implies $|e^{f_{n_k}}|$ $|f'_{n_k}| \rightarrow |g'|$, and since $|g'|$ and $|e^{f_{n_k}}|$ are bounded we must have $|f'_{n_k}|$ bounded.
Thus, $|(u_{n_k})_x|<|f'_{n_k}|$, is also bounded, which is contradiction.