Lipschitz continuity of conformal maps I have a basic question on Lipschitz continuity of maps. 
Let $\mathbb{D}$ be the unit disk and $T$ be an equilateral triangle. We have a conformal map $\phi :\mathbb{D} \to T$, which is extended to a homeomorphism from $\overline{\mathbb{D}}$ to $\overline{T}$. In fact, $\phi$ can also be given specifically by the Schwarz--Christoffel formula. Let $p$ denote a vertex of $T$. 
Then, it follows from the formula that there exist $C>0$ and $R>0$ such that 
\begin{align*}
|\phi^{-1}(z)-\phi^{-1}(p)| \le C|z-p|^3
\end{align*}
 for any $z \in B(p,R)$. Here, we denote by $B(p,R)$ the ball centered at $p$ with radius $R>0$. In particular, $\phi^{-1}$ is Lipschitz continuous at vertices of $T$. We also see that $\phi^{-1}$ is smooth at any points other than vertices. 
Then, can we show that $\phi^{-1}$ is Lipschitz continuous on $\overline{T}$?
Now, $\phi^{-1}$ is smooth function on $\overline{T}$. Hence, we conclude that $|(\phi^{-1})'|$ is bounded on $\overline{T}$. Therefore, we can conclude that $\phi^{-1}$ is Lipschitz continuous. Is this proof correct?

What you wrote is almost correct but your logic is muddled: You did not prove that $\phi^{-1}$ is $C^1$ on $\bar{T}$. 
What one observes is that there is a constant $L$ such that $\phi^{-1}$ is $L$-Lipschitz on   $T$. (Use the inverse function theorem to prove that $\phi^{-1}$ has uniformly bounded derivative.) 
From this, it follows that $\phi^{-1}$ is $L$-Lipschitz  on $\bar{T}$. (This is very general: If $A$ is a subset of a metric space $M$ and $f: M\to M'$ is $K$-Lipschitz, then $f$ admits a unique continuous extension to $\bar{A}$, which is  again $K$-Lipschitz.)  Hence, $\phi^{-1}$ is Lipschitz-continuous on $\bar{A}$. 
