Is a family of bounded biharmonic functions defined in the unit disk an equicontinuous family of functions on compacts? A biharmonic function $u$ is a solution of the equation $\Delta^2 u =0$.
Yes. First, $\Delta u\in C^\infty$ because it is harmonic, thus $u\in C^\infty$, because $\Delta$ is elliptic. Next, a harmonic function $f$ is its own mean; for instance $$2\pi f(x)=\int_0^{2\pi}f(x+re^{i\theta})d\theta,$$ for every $r<1x$. From this we deduce that if $\phi\in C^\infty(x1,1x)$ is even, has compact support and satisfies $$2\pi\int_0^1\phi(r)dr=1,$$ then we have $$f(x)=\int_D\phi(yx)f(y)dy.$$ Applying this to $f=\Delta u$, we find $$\Delta u(x)=\int_D\phi(yx)\Delta u(y)dy=\int_D\Delta \phi(yx)u(y)dy,$$ whence $$\Delta u(x)\le 2\pi M\int_0^{1x}r\phi(r)dr,$$ where $M$ is the supremum of $u$. Finally, $\phi$ can be chosen of the form $$\phi(r)=(1x)^{2}\rho\left(\frac{r}{1x}\right)$$ for a fixed $\rho$, and this gives an upper bound $$\Delta u(x)\le C(1x)^{2}M$$ for an absolute constant $C$. This bound is uniform over every compact subset of $D$. It is not difficult to finish the argument, which looks like AscoliArzela. 


Yes. Let $x$ and $y$ be two points in the interior. $f(x)$ and $f(y)$ can be expressed as an integral of $f$ against a continuous function, the Poisson kernel on a circle around them. As $y$ converges to $x$ the continuous function converges pointwise to $0$. Obviously we have $f(x)f(y)\leq b P_xP_y_{L_1}$ where $b$ is the global bound and $P_x$ and $P_y$ the Poisson kernels, and the $L_1$ norm of the difference in poisson kernels is bounded as a function of $xy$ on every compact set. 

