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I want to prove this:

Let $u \in \mathcal{C}^4(\mathbb{R}^n).$ Then, for every $x \in \mathbb{R}^n$ and $r>0$ we get that $$\dfrac{1}{w_nr^n}\int_{B(x,r)}u(y)dy=u(x)+cr^2\Delta u(x)+O(r^4),$$ for some constant $c$.

Here, $w_n$ represents the size of $B(0,1)\subset \mathbb{R}^n, \ w_n=|B(0,1)|=\int_{|x|<1}dx,$ so then $|B(x,r)|=w_nr^n$ . I know that I need to use the Taylor's formula: $$u(x)=\sum_{|\alpha|\leq 3}\dfrac{D^\alpha u(0)}{\alpha!}x^\alpha +\sum_{|\alpha|=4}\dfrac{D^\alpha u(\xi)}{\alpha!}x^\alpha,$$ with $\xi \in [0,x].$ Here, $\alpha=(\alpha_1, \ldots, \alpha_n)\in \mathbb{N}_0^n, \ |\alpha|=\sum \alpha_k, \ x^\alpha=x_1^{\alpha_1}\cdots x_n^{\alpha_n}, \ \alpha!=\alpha_1!\cdots \alpha_n!,\ \ldots $

I guess I should show that $$\dfrac{1}{w_nr^n}\sum_{1\leq|\alpha|\leq 3}\dfrac{D^\alpha u(0)}{\alpha!}\int_{B(0,r)} y^\alpha dy=cr^2\Delta u(0)$$ and $$\sum_{|\alpha|=4}\dfrac{D^\alpha u(\xi)}{\alpha!}y^\alpha \leq Mr^{n+4},$$ for some $M>0.$ Please, any idea or hint would be really helpful.

Nicolay Avendaño
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2 Answers2

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I suspect that if you expand the taylor's formula the proof will be simpler, \begin{equation} \begin{split} u(y)&=u(0)+\sum_{i=1}^n \frac{\partial u(0)}{\partial y_i}y_i+\frac{1}{2}\sum_{i,j=1}^n \frac{\partial^2 u(0)}{\partial y_iy_j}y_iy_j\\ &+\frac{1}{6}\sum_{i,j,k=1}^n \frac{\partial^3 u(0)}{\partial y_iy_jy_k}y_iy_jy_k+o(y^4) \end{split} \end{equation} Then you can get \begin{equation} \begin{split} \int_{B_r(x)}u(y)dy&=u(0)\int_{B_r(x)}dy+\sum_{i=1}^n \frac{\partial u(0)}{\partial y_i}\int_{B_r(x)}y_idy+\frac{1}{2}\sum_{i,j=1}^n \frac{\partial^2 u(0)}{\partial y_iy_j}\int_{B_r(x)}y_iy_jdy\\ &+\frac{1}{6}\sum_{i,j,k=1}^n \frac{\partial^3 u(0)}{\partial y_iy_jy_k}\int_{B_r(x)}y_iy_jy_kdy+o(y^4) \end{split} \end{equation} So, you could use the fact \begin{equation} \int_{B_r(x)} f(x)dx=\int_{0}^r\int_{\partial B_r(x)} f(\omega r)r^{n-1}d\sigma(\omega)dr \end{equation} furthermore if $P(x)$ is a homogeneuos function then \begin{equation} \int_{\partial B_r(x)}P(x)dx=\int_{\partial B_r(x)} \Delta P(x)dx \end{equation} Then you can show that $\int_{\partial B_r(x)}r\omega_id\sigma(\omega)=0$ and $\int_{\partial B_r(x)}r^3\omega_i\omega_j\omega_k d\sigma(\omega)=0$, therefore the first and third derivatives vanish. So, you only need calculate $$\int_{\partial B_r(x)}\rho^2\omega_i\omega_jd\sigma(\omega)$$.

I hope it helps.

Daniel Meshir
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Hint: For $ \alpha = (\alpha_1, \dots, \alpha_n)$ if there is some $\alpha_i$ that is odd then $$ \int_{B_r} y^\alpha \, dy =0$$ since $y\mapsto y^\alpha$ would be an odd function with respect to the plane $\{y_i=0\}$. This will dramatrically simplify $$\sum_{1 \leqslant \vert \alpha \vert \leqslant 3} \frac{D^\alpha u(0)}{\alpha!} \int_{B_r} y^\alpha \, dy. $$

Also, you will need that $$\int_{B_r}y_i^2 \, d y = \frac1n \int_{B_r} \vert y \vert^2 \, dy $$ by symmetry.

JackT
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