$Omegasubset mathbb R^n$ 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 $sumlimits_{i=1}^n partial_i^{2k} u=0, forall kin mathbb Z^+$.