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Let $K$ be a compact convex set in $\mathbb{R}^2$. In the proof of a proposition in a paper I am reading, they are concerned with parameterizing $\partial K$ in the following way:

If $K$ is strictly convex, then given $\hat{\mathbf{n}} \in \{x \in \mathbb{R}^2\,:\, |x| = 1\} = S^1$ from the unit circle, there is a unique point $\sigma(\hat{\mathbf{n}}) \in \partial K$, where $K$ is supported by a half-plane with outward unit normal vector $\hat{\mathbf{n}}$. The curve $\sigma$ is continuous on $S^1$.

I believe I follow the above, but I don't understand their following comment:

Should $K$ fail to be strictly convex, then strict convexity of $K$ can fail in at most countably many directions $\hat{\mathbf{n}}_i \in S^1$. By inserting an interval of length $2^{-i}$ into the unit circle at $\hat{\mathbf{n}}_i$, and extending $\sigma$ linearly to $I_i$, we parametrize $\partial K$ continuously.

What is the idea here ? How does the argument work for a simple case, e.g. $K$ is the unit square?

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The idea can be illustrated on the real line.

If a function $f:\mathbb{R} \to \mathbb{R}$ has a discontinuity at a point $a$, but the one-sided limits $f(a-)$ and $f(a+)$ exist, then we can define $$ g(x) = \begin{cases}f(x),\quad & x<a \\ f(a-)+\frac{ x-a}{\epsilon}( f(a+)-f(a-)),\quad & a\le x\le a+\epsilon \\ f(x-\epsilon),\quad &x> a+\epsilon \end{cases} $$ which is continuous on $[a,a+\epsilon]$. A jump has been replaced by a ramp.

The authors apply the same to a function $f:S^1\to \partial K$. It may be easier to think of $S^1$ as an interval $[0,1]$ with endpoints glued. Inserting ramps as above, we end up with a continuous function on a larger interval $[0,b]$. This interval is still $S^1$ when the endpoints are glued.

For the square, the following happens. The initial map $\sigma$ is constant on each of the subarcs $k\pi/2<\theta<(k+1)\pi/2 $, sending this subarc to one of vertices. It jumps at $k\pi/2$ from one vertex to another. Inserting segments at these points, we get a parametrization of the square that sits at a vertex for $\pi/2$ radians, then moves linearly along a side to get to the next vertex, etc.


For general $K$, one has to prove that one-sided limits exist; one approach is to show that $\sigma$ is of bounded variation.

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