Prove or disprove that $f:\mathbb{R^2}\to \mathbb{R}$ with $$f(x, y) =\begin{cases} x^2+y^2-1 & \text{ if } (x,y) \in\{(z_1,z_2)\in\mathbb{R}^2:\sqrt{z_1^2+z_2^2}\leq 1\} \\ x^2+y^2 & \text{ if } (x,y) \in\{(z_1,z_2)\in\mathbb{R}^2:\sqrt{z_1^2+z_2^2}> 1\} \end{cases} $$ is lower semicontinuous on $\mathbb{R}^2$.
I think it it lower semicontinuous. $f$ is discontinuous in $z_0=||z||_{\mathbb{R}^2} =1$ and I need to show that $$lim inf_{z\to z_0}f(z)\geq f(z). $$
I don't know how to work with multidimensional $f$... what is lim inf?
I tried: for $||z||_{\mathbb{R}^2} =\sqrt{z_1^2+z_2^2}\leq 1$ it is $f(z)\leq 1-1=0$ and for $||z||>1$ it is $f(z)>1$ with $z=(z_1,z_2)$.
So $0=f(z)\leq lim_{z\uparrow z_0}f(z)\leq0<1<lim_{z\downarrow z_0}f(z)$. Is $lim_{z\uparrow z_0}f(z)=lim inf_{z\to z_0}f(z)$?
Thanks for any help!
In two dimensions you only have to consider sequences $(x_n,y_n)$ when dealing with $\lim\,,$ $\limsup\,,$ and/or $\liminf$ of $f(x,y)\,.$
The simplest way to show lower semicontinuity for your function is however this:
Take the definition that $f$ is l.s.c. when for each $c$ the set $\{c<f\}$ is open in $\mathbb R^2\,.$
When $f$ is continuous in $(x,y)$ and $c<f(x,y)$ then it is trivial that there is an open neighborhood $U$ such that $(x,y)\in U\subset\{c<f\}\,.$
Therefore we only have to consider points with $x^2+y^2=1$ at which $f$ is discontinuous. Since $f\le 1$ inside the unit disc it follows that every $c$ with $c<f(x,y)$ and $x^2+y^2=1$ must be strictly less than one: $c<1\,.$
The function $g(x,y)=x^2+y^2-1$ is continuous and agrees with $f$ in the unit disc and on its boundary. Therefore, when $x^2+y^2=1$ there is an open neighborhood $U$ such that $(x,y)\in U\subset\{c<g\}\,.$ Since $f\ge 1$ outside the unit disc and since $c<1$ it follows that $c<f(z_1,z_2)$ for all $(z_1,z_2)\in U\,.$ Therefore, $$ (x,y)\in U\subset\{c<f\}\,. $$
Here is a proof that uses $\liminf$:
Since $f\ge -1$ the $\liminf\limits_{n\to\infty}f(z_n)$ exists for every sequence $z_n\,.$
When $z_0$ is on the boundary and the sequence $z_n$ converges from outside to $z_0$ then clearly $\liminf f(z_n)=\lim f(z_n)=1\,.$
When $z_n$ converges from inside to $z_0$ (on the boundary) then $\liminf\limits_{n\to\infty} f(z_n)=\lim\limits_{n\to\infty} f(z_n)=f(z_0)$ and $f(z_0)=0\,.$
Now consider a general sequence $z_n$ that converges to $z_0$ on the boundary. Any subsequence for which $f(z_n)$ converges has itself a subsequence $z_{n_i}$ that converges either from inside or from outside to $z_0\,.$ Therefore, in both cases $f(z_{n_i})$ converges to a value that is at least $f(z_0)=0$.