Prove that a function $u: u= \ln\|x\|{_{2}}$ has $\Delta u = 0$. I had a similar case some time ago and following the advices there I tried to solve this one too.
I tried to find its first partial derivative and I got:
$\frac{\partial}{\partial x_{i}}=\frac{1}{2\cdot \|x\|_{2}^{1/2}}$
Now i have to find the second derivative of this and I got $\frac{1+4\cdot (\sum x_{i}^{2})^{5}}{8\cdot(\sum x_{i}^{2})^{5/4} }$. 
And now I am stuck,I have no idea if my calculations are okay or how can I continue to solve this.
I would be incredibly thankful for some help.
Annalisa
 A: You have, for $x\in\mathbb{R}^d$, 
$$
u(x) = \ln \lVert x\rVert_2 = \ln \sqrt{\sum_{i=1}^d x_i^2} = \frac{1}{2}\ln \sum_{i=1}^d x_i^2
$$
from which, for $x\in\mathbb{R}^d$ and $1\leq i\leq d$,
$$
\frac{\partial}{\partial x_i}u(x) = \frac{1}{2}\frac{\partial}{\partial x_i}\ln \sum_{i=1}^d x_i^2= \frac{1}{2}\cdot\frac{2x_i}{\sum_{i=1}^d x_i^2}
= \frac{x_i}{\lVert x\rVert_2^2}\,.
$$
From there,
$$
\frac{\partial^2}{\partial x_i^2}u(x) = \frac{\partial}{\partial x_i}\frac{x_i}{\lVert x\rVert_2^2} = \frac{1\cdot\lVert x\rVert_2^2-x_i\cdot 2x_i}{\lVert x\rVert_2^4}= \frac{\lVert x\rVert_2^2-2x_i^2}{\lVert x\rVert_2^4}
$$
so that
$$
\Delta u(x) =\sum_{i=1}^d\frac{\partial^2}{\partial x_i^2}u(x) = \frac{d\lVert x\rVert_2^2-2\lVert x\rVert_2^2}{\lVert x\rVert_2^4}= \frac{d - 2}{\lVert x\rVert_2^4}
$$
This is only 0 if $d=2$.
A: Hint 
Let $\|\mathbf{x}\|=\sqrt{x^2+y^2}$. Then 
\begin{align*}
u & =\ln \|\mathbf{x}\|\\
\frac{\partial u}{\partial x}&=\frac{1}{\|\mathbf{x}\|}\frac{\partial}{\partial x}\sqrt{x^2+y^2}\\
&=\frac{1}{\|\mathbf{x}\|}\frac{x}{\sqrt{x^2+y^2}}\\
&=\frac{x}{\|\mathbf{x}\|^2}
\end{align*}
Likewise 
$$\frac{\partial u}{\partial y}=\frac{y}{\|\mathbf{x}\|^2}.$$
Now
\begin{align*}
\frac{\partial ^2u}{\partial x^2}&=\frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right)\\
&=\frac{(x^2+y^2)\frac{\partial x}{\partial x}-x\frac{\partial (x^2+y^2)}{\partial x}}{(x^2+y^2)^2}\\
&=\frac{\|\mathbf{x}\|^2-2x^2}{\|\mathbf{x}\|^4}.
\end{align*}
Likewise
$$\frac{\partial ^2u}{\partial \color{red}{y}^2}=\frac{\|\mathbf{x}\|^2-2\color{red}{y}^2}{\|\mathbf{x}\|^4}.$$
Hopefully you can take it from here and generalize it to higher dimensions.
