# Biharmonic function

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 $\Delta^2 u =0$.

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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.

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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

$|f(x)-f(y)|\leq b ||P_x-P_y||_{L_1}$

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 $|x-y|$ on every compact set.

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