I've been reading through Charles Fefferman's The Multiplier Problem for the Ball and going through lemma 1 sparked the following problem for me:
Let $V \in \mathbb{R}^2$ be a unit vector, let H be the half plane defined by $H=\{x\in \mathbb{R}^2 : x\cdot V \geq 0\}$, and let $D^{r}$ be a disc centered at $rV$ with radius $r>0$. Also, for $A\subseteq \mathbb{R}^2$, $\chi_{A}(x)=\begin{cases}1 \text{ if $x \in A$}\\ 0 \text{ if x $\notin A$}\end{cases}$.
The question I have is the following: $\forall x\in \mathbb{R}^2$ is $lim_{r\to \infty}\chi_{D^r}(x)=\chi_{H}(x)$?
'Morally' speaking, I understand that the disc of radius $r$ and center $rV$ approximates the line perpendicular to $V$ (at $V$'s origin) when $r$ is large, but I don't know how to give a rigorous proof as to why the above statement with characteristic functions is true.
Certainly, if $x\notin H$ then $\nexists r>0$ such that $d(x,rV)\leq r$, so $x\notin D^r$ so $\forall \epsilon>0$, $|\chi_{D^r}(x)-\chi_{H}(x)|=|0-0|=0<\epsilon$.
If $x\in H$ but not on on the boundary of the half plane (i.e. the angle between the vector $x$ and $V$ is strictly less than $\pi/2$), I can imagine there's a way to find $r$ large enough such that $x\in D^r$. Then, $|\chi_{D^r}(x)-\chi_{H}(x)|=|1-1|=0<\epsilon$.
The problem is here: if $x\in H$ and lies on the boundary of the half plane (i.e. the angle between the vector $x$ and $V$ is equal to $\pi/2$), then is there a way to find $r$ large enough so that $|\chi_{D^r}(x)-\chi_{H}(x)|<\epsilon$? I feel like if $x$ is not at the origin of $V$, then there is no $r$ such that the disc will contain $x$, and thus this cannot be true.
Maybe the half plane should instead be defined by $H=\{x\in \mathbb{R}^2 : x\cdot V > 0\}$ in order for the statement to be true?
I think you have the right idea.
Omitting the boundary from both, the open disk will eventually converge to the open half plane. You can prove that rigourously by giving a formula that given the coordinates of a point computes an $r$ that is large enough to represent the final value of that point, which won't change for larger $r$'s.
If you include the boundary in each case, then the points on the boundary of the half plane will always be part of the half plane but never part of the disk, so they fail the convergence of the point-wise characteristic. You might still want to speak about some form of convergence of shape, but not via convergence of the characteristic function.
Open half-pane and closed disk would mostly work, except for the origin which is part of the closed disk but not of the open half plane.