$f(x):=\ln\sqrt {(x_1-\zeta_1)^2+(x_2-\zeta_2)^2}$. $f \in C^2(\mathbb{R}^2\setminus \{0\})$?

Question

Let $\zeta \in \mathbb{R^2}$

$f:\mathbb{R}^2 \setminus \{\zeta\} \longrightarrow \mathbb{R}$,

$f(x):=\ln\|x-\zeta\|_2=\ln\sqrt {(x_1-\zeta_1)^2+(x_2-\zeta_2)^2}$

I need to find out whether $f \in C^2(\mathbb{R}^2\setminus \{\zeta\})$.

Can I simply derive $f$ wrt $x$ twice and check if the solution is continuous? Or do I write $x$ as $(x_1,x_2)$ and $\zeta$ as $(\zeta_1, \zeta_2)$ and derive wrt $x_1$ and then wrt $x_2$?

Furthermore I need to determine the gradient $\nabla f(x)$ and $\frac{\partial^2 f }{\partial x^2}+\frac{\partial^2f}{\partial y^2}$.

Thanks for your help!

Answer 1

View $f$ as a chain of smooth functions: Let $a : \mathbb{R}^2\setminus \{\zeta\}\to (0,\infty)$ be defined by $a(x_1,x_2) = (x_1-\zeta_1)^2+(x_2-\zeta_2)^2.$ This is a polynomial, so it is $C^\infty.$ Then note $b:(0,\infty)\to (0,\infty)$ defined by $b(y)=\sqrt y$ is $C^\infty.$ Finally, observe $c:(0,\infty)\to \mathbb R,$ defined by $c(u)=\ln u,$ is $C^\infty.$ It follows that the chain $f= c\circ b \circ a$ is a $C^\infty$ function from $\mathbb{R}^2\setminus \{\zeta\}$ into $\mathbb R.$