Is a family of bounded bi-harmonic functions defined in the unit disk an equicontinuous family of functions on compacts? A bi-harmonic function $u$ is a solution of the equation $\Delta^2 u =0$.



Biharmonic function
**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 $r<1-|x|$. From this we deduce that if $\phi\in C^\infty(|x|-1,1-|x|)$ is even, has compact support and satisfies
$$2\pi\int\_0^1\phi(r)dr=1,$$
then we have
$$f(x)=\int\_D\phi(|y-x|)f(y)dy.$$
Applying this to $f=\Delta u$, we find
$$\Delta u(x)=\int\_D\phi(|y-x|)\Delta u(y)dy=\int\_D\Delta \phi(|y-x|)u(y)dy,$$
whence
$$|\Delta u(x)|\le 2\pi M\int\_0^{1-|x|}r\phi(r)dr,$$
where $M$ is the supremum of $|u|$. Finally, $\phi$ can be chosen of the form
$$\phi(r)=(1-|x|)^{-2}\rho\left(\frac{r}{1-|x|}\right)$$
for a fixed $\rho$, and this gives an upper bound
$$|\Delta u(x)|\le C(1-|x|)^{-2}M$$
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 Ascoli-Arzela.


