Are there closed and simple plane curves (Jordan curves) of finite length that are not piecewise $C^1$ curves? Are there closed and simple plane curves (Jordan curves) of finite length that are not piecewise $C^1$ curves (or $C^1$ curves for parts, this is, continuous curves that are made up by a finite number of $C^1$  arcs)?
I'm studying some versions of the inequality isoperimetric  and I've arrived at the following version:
Theorem (inequality isoperimetric for piecewise $C^1$ curves): Let $\alpha$ a closed and simple plane curve of class $C^1$ for parts of length $L$ and that delimits a region of area A. Then,
$$L^2 - 4 \pi A \geq 0.$$
Moreover, the equality is valid iff $\alpha$ is a circumference.
I have found some proofs using plane geometry that apparently generalize this fact.But thinking about it, I couldn't find an example of a curve that doesn't satisfy these conditions. The statement I found is as follows:
Theorem (inequality isoperimetric): Let $\alpha$ a closed plane of length $L$ and that delimits a region of area A. Then,
$$L^2 - 4 \pi A \geq 0.$$
Moreover, the equality is valid iff $\alpha$ is a circumference.
Does the first result imply the second?
Thank you for your help.
 A: If "$\gamma$ is piecewise $C^{1}$ on $[a, b]$" means the interval $[a, b]$ may be subdivided into finitely many subintervals $[t_{i-1}, t_{i}]$ on which $\gamma$ is continuously-differentiable (including one-sided derivatives at endpoints), then a simple closed curve of finite length need not be piecewise $C^{1}$.
At least three problems can occur:

*

*The path can fail to be differentiable on any interval.

*The path may "formally" piecewise $C^{1}$ if the domain is broken into infinitely many subintervals.

*The derivative may exist everywhere, but fail to have one-sided limits.

Here are functions whose graphs have these behaviors. Simple closed paths of finite length are easily constructed: Pick a portion of the graph over a closed, bounded interval with "bad behavior" and close off with vertical and horizontal line segments.

*

*Let $(a_{k})_{k=0}^{\infty}$ be a bijective enumeration of the rationals, let $S$ denote the unit step function
$$
S(x) = \begin{cases}
  0 & x < 0, \\
  1 & 0 \leq x,
\end{cases}
$$
and define $f(x) = \sum\limits_{k=0}^{\infty} 2^{-k} S(x - a_{k})$. The function $f$ is bounded and strictly increasing (hence integrable on every real interval), and has a jump discontinuity of size $2^{-k}$ at $a_{k}$. The definite integral
$$
F(x) = \int_{-\infty}^{x} f(t)\, dt
$$
is continuous and increasing, so its graph has finite length, but $F'(a_{k})$ does not exist for each $k$, i.e., $F$ is not differentiable at each rational.

*Use the same construction as in the preceding item, but take $a_{k} = \frac{1}{k+1}$. The resulting function $F$ is linear on $[\frac{1}{k+2}, \frac{1}{k+1}]$ for every $k \geq 0$, so is piecewise linear on each interval $[a, b] \subset (0, \infty)$, but is not piecewise $C^{1}$ on any interval containing $[0, 1]$.

*The function $f(x) = x^{2}\sin(\frac{1}{x})$ (and $f(0) = 0$) has bounded derivative, so its graph has finite length over every bounded interval, but the one-sided limits of $f'$ do not exist at $0$. "Worse" behavior can be produced with Volterra-type constructions.

