# analysis – Whether \$Delta u=0 Rightarrow sumlimits_{i=1}^n partial_i^4 u=0\$?

$$Omegasubset mathbb R^n$$ is a bounded smooth domain. If
$$Delta u=0$$
on $$Omega$$, whether we have $$sumlimits_{i=1}^n partial_i^4 u=0$$ on $$Omega$$ ? Following is my proof, but I am not sure it is right, and I think it is complex, I feel there should be a direct proof.

$${bf What~~I ~~try:~}$$ By the mean value property, we have
$$u(y) = frac{1}{omega_n R^n}int_{B_{_R} ~(y)} u(x) dx. tag{1}$$
On the other hand, by Taylor, we have
$$u(x+y)=u(y)+u_i(y)x^i + frac{1}{2}u_{ij}(y)x^ix^j+…$$
then
$$int_{B_R~(0)} u(x+y)dx= u(y)|B_R~(0)|+ C_2Delta u(y) R^2 + C_4 (sumlimits_{i=1}^n partial_i^4 u(y))R^4+…$$
therefore,
$$frac{int_{B_R~(0)} u(x+y)dx}{|B_R~(0)|}= u(y)+ frac{C_2Delta u(y) R^2}{|B_R~(0)|} + frac{C_4 (sumlimits_{i=1}^n partial_i^4 u(y))R^4}{|B_R~(0)|} +…$$
since
$$Delta u=0$$
it become
$$frac{int_{B_R~(0)} u(x+y)dx}{|B_R~(0)|}= u(y)+ frac{C_4 (sumlimits_{i=1}^n partial_i^4 u(y))R^4}{|B_R~(0)|} +…$$
when $$R>0$$ small enough, by (1), we have $$sumlimits_{i=1}^n partial_i^4 u=0$$. By induction, we have $$sumlimits_{i=1}^n partial_i^{2k} u=0, forall kin mathbb Z^+$$.