Why is Steiner's proof of isoperimetric inquality incorrect?

Question

I've just read an exposition of an attempted proof of Isoperimetric Inequality, by Andreas Hehl, with the argument attributed to Steiner.

The statement of the theorem is as follows:

Let $c(t) = (x(t), y(t))$ be a simple, closed, positively oriented and regular parametrised $C^1$ curve with $t \in [a, b]$. Denote the area enclosed in the above defined curve $c(t)$ with A. For a given length $l$ of $c(t) = (x(t), y(t))$, we have

$$A \leq \frac{l^2}{4\pi}$$

with equality iff $c(t)$ is a circle.

The proof proceeds by a clever symmetrisation argument & is perhaps too long and (presumably) well known to give here.

The author of the document doesn't find the proof convincing:

While this construction is intuitively accessible, it cannot be seen as a complete proof of the Isoperimetric Inequality. It merely gives one way to construct the biggest area enclosed in a curve with fixed length $l$, namely the circle. Yet, we have not shown the existence of a (unique) solution, so we have to see this ’proof’ as incomplete.

Contrary to the author of the document, I find the proof entirely convincing:

1. If $c(t)$ is concave, it will enclose area smaller than a corresponding convex curve (see the pdf for exact meaning behind "corresponding convex curve").
2. If $c(t)$ is convex, but is not a circle with perimeter $l$, it will have area smaller than the circle.
3. Circle of perimeter $l$ has an area of $l^2/(r\pi)$, which can be easily shown using limits and the inscribed/outscribed polygons trick going back to Archimedes'.

We have then that if c(t) is not a circle, it has area less than the area of a circle with perimeter $l$, which is $l^2/(r\pi)$ so we get a strict inequality. If $c(t)$ is a circle we get equality. What is the catch?

Answer 1

No, Steiner's argument establishes the following new claims:

1. If $c(t)$ encloses a concave area, then there is some other curve of the same length but enclosing a larger area.
2. If $c(t)$ encloses a convex area but is not a circle, then there is some other curve of the same length but enclosing a larger area.

The third claim you list is correctly stated, but it is not new. It is also irrelevant to the hole in the proof.

Whether the other curve in the above claims encloses a convex area or not is irrelevant. The important point is that the other curves Steiner constructs are not circles. So, Steiner's argument proves just the following implication: IF $c$ is a closed curve such that for every closed curve $c'$ of the same length, it holds that the area enclosed by $c$ is larger than or equal to the area enclosed by $c'$, THEN $c$ is a circle. A separate argument is needed to show that such an extremal curve exists.

The hole in Steiner's attempted proof is perhaps easier to see by considering the following analogous argument, which obviously does not show that 1 is the smallest positive real number.

Let $x$ be a positive real number such that $x\neq 1$. Then $x^2-x=x(x-1)\neq 0$. If $x^2<x$, then $x^2$ is a positive real number less than $x$. On the other hand, if $x^2>x$, then $x(x-1)>0$, so $x-1>0$, and consequently 1 is a positive real number less than $x$. Hence, for every positive real number $x$ such that $x\neq 1$, there is another positive real number $y$ such that $y<x$.