Let $X$ be a simply connected set in $\mathbb{C}$. Of course $X$ can be quite jagged and nasty. Let $\epsilon>0$ be given, and define $C(X;\epsilon)=\{z\in\mathbb{C}:\min(|z-w|:w\in X)=\epsilon\}$.
Essentially one may find $C(X;\epsilon)$ by taking a radius-$\epsilon$ ball, and rolling it around the outside of $X$, and tracing the motion of the center of the ball (I know this ignores possible extra components of $C(X;\epsilon)$, which is ok with me for the purpose of this question).
My impression is that $C(X;\epsilon)$ smooths out the boundary of $X$ quite a bit.
Can this be quantified? For example, it seems to me that $C(X;\epsilon)$ must be at least rectifiable, parameterized by a closed path $\gamma:[0,1]\to\mathbb{C}$. Are there better quantifications?
If this is quantified in a meaningful way, will $C(C(X;\epsilon);\epsilon)$ be any smoother than $C(X;\epsilon)$? My guess is no.
If one first forms $C(X;\epsilon)$ and then in some appropriate way shrinks this set by a distance of $\epsilon$ back towards the boundary of $X$, it seems like this will be a nice smoothing approximation of the boundary of $X$, getting ever better as $\epsilon\to0$. Is this method of approximating known to anyone?
EDIT: Some additional questions, after coming back to this for the first time in three years:
Assuming $X$ is bounded, as $\epsilon \to \infty$, $C(X;\epsilon)$ approaches a circle. For large $\epsilon$, let $x_\epsilon$ denote the center of the largest disc contained within the "almost circle" $C(X;\epsilon)$ (ie. $x_\epsilon$ is the "centerpoint" of $C(X;\epsilon)$). Where is $x_\epsilon$ approaching as $\epsilon\to\infty$? Can that point $x_\infty$ be deduced geometrically from $X$? It probably only depends on the convex hull of $X$, not $X$ itself.
For large $\epsilon$, $C(X;\epsilon)$ has a single component. Let $N_\epsilon$ denote the number of components of $C(X;\epsilon)$. Will ${N_\epsilon}$ reduces to a sequence? I think not necessarily. What might the function $f_X:(0,\infty)\to\mathbb{N}$ defined by $$f_X(\epsilon) = \text{Number of components of }C(X;\epsilon)$$ look like? Need it be measurable? Can any $f:(0,\infty)\to\mathbb{N}$ be realized as $f_X$ for some $X$?
Assuming $C(X;\epsilon)$ is rectifiable, let the $\epsilon$-perimeter of $X$, denoted $\mathcal{l}_\epsilon$, be the length of $C(X;\epsilon)$. Pretty clearly $\mathcal{l}_\epsilon$ need not be monotonic (just think, for example, of a very long, skinny $V$-shape). I think $l_\epsilon$ must be at least continuous? Other smoothness properties?
This is only a very partial answer to question n°6. It can be tackled using the co-area formula (see [1], chapter 2.10) and also in the more general case $X \subset \mathbb{R}^n$ compact. A particular case of the formula is stated as follows :
Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ be Lipschitz continuous, then for any measurable $E \subset \mathbb{R}^n$, we have,
\begin{equation} \int_E |\nabla f| d\mathcal{L}^n = \int_{-\infty}^{+\infty} \mathcal{H}^{n-1}(E \cap f^{-1}(t))d t \end{equation}
Let us apply the co-area formula to $f(x) = d_X(x) := \inf_{y \in X} |x-y|$ which is $1$-Lipschitz and $E = X^\epsilon = \{x \in \mathbb{R}^n | d_X(x) \leq \epsilon\}$. Note that $\partial X^\epsilon \subset C(X, \epsilon) = \{x \in \mathbb{R}^n | d_X(x) = \epsilon\} = d_X^{-1}(\epsilon)$. We then obtain,
\begin{equation} \mathcal{L}^n(X^\epsilon\setminus X) = \int_{0}^{\epsilon} \mathcal{H}^{n-1}(d_X^{-1}(t))d t \end{equation}
The left-hand side is finite by compactness of $X^\epsilon$ and $\sigma$-finiteness of the Lebesgue measure $\mathcal{L}^n$ on $\mathbb{R}^n$, therefore we obtain that the function
\begin{equation} \tag{1} t \mapsto \mathcal{H}^{n-1}(d_X^{-1}(t)) \end{equation}
is locally integrable on $\mathbb{R}$, hence for almost all $t > 0$, $C(X, t)$ has finite perimeter.
I think that $C(X, \epsilon)$ is definitely rectifiable, regardless of the shape of $X$, it might be handled via a characterisation using cones (see [1] chapter 2.8) but I was not able to show it yet.
EDIT : In order to give a more precise answer to the OP's intuition on question n°6, the perimeter map $\mathcal{P}$ given by $(1)$ is actually not continuous in general. As a counter-example take $X = \{bi \ |\ b \in [0,1]\} \cup \{1 + bi \ |\ b \in [0,1]\}$, which is the union of two unit and parallel line segments in $\mathbb{C}$ separated by a distance $1$. It seems quite intuitive to prove that $\mathcal{P}\left(\frac{1}{2}\right) - \mathcal{P}\left(\frac{1}{2}^+\right) = 1$ since the line segment $S = \{ 1/2 + bi \ | \ b \in [0,1]\}$ is a subset of $C(X,\frac{1}{2})$ and immediately "disappears" once the parameter gets past $\frac{1}{2}$.
Reference : [1] : L. Ambrosio, N. Fusco, D Pallara, Functions of bounded variations, Courier Corporation 2000, https://scholar.google.com/scholar?hl=en&as_sdt=0%2C5&q=functions+of+bounded+variation+ambrosio&btnG=