Suppose that a sequence of analytic functions is locally Lipschitz in an open connected set $D$ of the complex plane. (That is, for all $z\in D$, there is a constant $L$ such that $|f_n(z_1)-f_n(z_2)|\leq L|z_1-z_2|$ if $z_1,z_2$ are in a neighborhood of $z$.)
I want to show that $f(z)$ must also be analytic, and moreover $f_n'(z)\to f'(z)$, where $f(z)$ is the pointwise limit of $f_n$.
I think I can show that $f$ is continuous using a $\epsilon/3$ type argument (ie split up $|f_n(z)-f(z)|$ using the triangle inequality), but I'm not sure how this shows analyticity of the function $f$. I would really like to say that we in fact have uniform convergence, but I'm not sure if the condition given is strong enough to prove that.
Any help is appreciated!
Recall Montel's theorem. It says if $U\subseteq \mathbb{C}$ is a domain and $(f_n)_n$ is a family of complex-analytic functions $f_n : U \rightarrow \mathbb{C}$ that is locally uniformly bounded (i.e for any compact $D\subseteq U$ there exists a constant $C_D$ such that $\sup_{n} \Vert f_n \Vert_{L^\infty(D)}<C_D)$ we have that there exists a subsequence $(f_{n_k})_k$ that converges uniformly on compact subsets of $U$.
Now we show that $(f_n)_n$ and $(f_n')_n$ are both locally uniformly bounded if $(f_n)_n$ is locally equi-lipschitz (this means locally we can find an upper bound for the lipschitz constant of all functions in our family).
It is clear that $\vert f_n'\vert$ on $D$ is bounded by the Lipschitz constant of $f_n$ on $D$. Thus, locally equi-lipschitz implies that $(f_n')_n$ is locally uniformly bounded. On the other hand, if we fix some compact $D\subseteq U$, and some point $z_0\in D$, then we get $$ \vert f_n(z) \vert \leq \vert f_n(z_0) \vert + \vert f_n(z)-f_n(z_0) \vert \leq \sup_{n} \vert f_n(z_0) \vert + \text{diam}(D) \Vert f_n'\Vert_{L^\infty(D)}.$$ As $(f_n(z_0))_n$ converges and $(f_n')_n$ is locally uniformly bounded, we get that $(f_n)_n$ is locally uniformly bounded too.
In particular, we can now use Montel's theorem and pick a subsequence $(f_{n_k})_k$ and some functions $h,g$ such that $(f_{n_k})_k$ converges locally uniformly to $h$ and $(f_{n_k}')_k$ converges locally uniformly to $g$. However, as $(f_n)_n$ converges pointwise to $f$, we get that $h=f$. Also, as $f$ is locally the uniform limit of complex-analytic functions, we get that $f$ is complex-analytic.
As $(f_{n_k})_k$ converges locally uniformly to $f$ and $(f_{n_k}')_k$ converges locally uniformly to $g$ one can use the standard argument to show that $g=f'$. Namely, fix $z_0, z\in U$ and path $\gamma$ that connects $z_0, z$. As the path is compact, we have that $(f_{n_k})_k$,$(f_{n_k}')_k$ converges uniformly on $\gamma$ and hence, we obtain $$ f(z) = \lim_{k\rightarrow \infty} f_{n_k}(z) = \lim_{k\rightarrow \infty} \left( f_{n_k}(z_0)+ \int_\gamma f_{n_k}'(\tau) d\tau \right) = f(z_0) + \int_\gamma g(\tau) d\tau. $$ Taking a derivative in $z$ yields $f'(z) = g(z)$ (by the fundamental theorem of calculus).
Finally, we can now start with any subsequence $(f_{n_m})_m$ and find a subsequence $(f_{n_{m_\ell}})_\ell$ such that $(f_{n_{m_\ell}})_\ell$ converges locally uniformly to $f$ and $(f_{n_{m_\ell}}')_\ell$ converges locally uniformly to $f'$. This means, that if you fix with some $z\in U$, then there exists for every subsequence $(f_{n_m}'(z))_m$ a subsequence $(f_{n_{m_\ell}}')_\ell$ that converges to $f'(z)$. However, this implies that $(f'_n(z))_n$ converges to $f'(z)$, which is what we wanted to show.