# Is the following solution to the isoperimetric problem correct?

I came up with a solution over here: http://keplerlounge.com/ but I am not entirely sure that it's free from error as much as I have checked it many times. If people can offer constructive feedback I would really appreciate that.

Update1: @hardmath has brought to my attention that it's expected that I reproduce the proof on the forum. If that's the only issue that remains, here's the proof:

1) A minimal condition on any closed curve C that satisfies the isoperimetric inequality is that it must be convex. A necessary property of any closed curve $C$ in order that for $A(C) = \underset{C}{\text{max}} A(C)$ is that the closed curve be convex. In consequence, this may be calculated using any interior-point with the following integral:

{expression} . This may be deduced using Steiner reflection.

Without loss of generality, O = (0,0) is chosen to be the origin and for the remainder of the proof I'll assume that we're dealing with a closed convex curve with finite perimeter that will be denoted by $\displaystyle \gamma \displaystyle$.

2) Without loss of generality, any such curve $\displaystyle \gamma \displaystyle$ may be parametrized by time and inflated.

a) By inflation I mean that all proportions are kept equal while the area enclosed by the curve increases. That is to say:

i) $\delta t > 0 \iff r(\theta, t+\delta t) > r(\theta,t)$

ii) $f(t)$ is monotonically increasing and $r(\theta,t) = f(t)r(\theta)$

iii) $f(0) = 0$

iv) $\displaystyle \lim_{t \to +\infty}\displaystyle$ $\displaystyle f(t)=\infty\displaystyle$

b) The perimeter and area enclosed by the curve may be approximated by the lower-bounds:

$P_{m}(t)=2 \pi r_{m}(t) \leq P_{\gamma}(t)$

$A_{m}(t)=\pi r_{m}^2(t) \leq A_{\gamma}(t)$

where $r_{m}(t) = \underset{\theta}{\text{min}} r(\theta,t)$

c) Clearly, while the curve $\displaystyle \gamma \displaystyle$ is being inflated $\displaystyle P_{\gamma}\displaystyle$and $\displaystyle A_{\gamma}\displaystyle$ are both monotonically increasing functions, and we have the following boundary conditions on any curve $\displaystyle \gamma \displaystyle$ that is being inflated:

i) $\displaystyle P_{\gamma}(t=0)=0\displaystyle$

ii) $\displaystyle A_{\gamma}(t=0)=0\displaystyle$

iii) $\displaystyle \lim_{t \to +\infty}\displaystyle$ $\displaystyle P_{\gamma}(t)=\infty\displaystyle$

iv) $\displaystyle \lim_{t \to +\infty}\displaystyle$ $\displaystyle A_{\gamma}(t)=\infty\displaystyle$

3) Due to the finite perimeter property of $\displaystyle \gamma \displaystyle$ we have $\displaystyle \frac{dr(\theta,t)}{dt} < \infty\displaystyle$

If we define $\displaystyle f(\theta,t) =\frac{dr(\theta,t)}{dt} \displaystyle$, it follows that for any finite time interval $I = [0,k]$ there exists an M such that $\displaystyle \underset{\theta,t}{\text{max}} f(\theta,t) \leq M \displaystyle$ on that interval.

4) Now, in order to maximize the area enclosed by $\gamma$ we need to find $r(\theta,t)$ such that the distance travelled by points on the perimeter of $\gamma$ is maximized over the time-interval $I = [0,k]$.

This is a simple optimisation problem:

max: $G(k) = r(\theta)(f(k)-f(0))$

constraint: $r(\theta)f'(t) \leq M$

The answer is straightforward. It would suffice to choose $f'(t) = M/r(\theta)$ Then we have $f(t) = Mt/r(\theta) + Cst$ and since $f(0) = 0$ this simplifies to the unique solution: $f(t)r(\theta) = Mt$.

5) From this point onwards, we may easily deduce the isoperimetric inequality:

$r(\theta,t) = f(t)r(\theta)=Mt$ is differentiable for all $t$ in $I=[0,k]$ so the perimeter and area may be calculated using:

$\displaystyle P_{\gamma}(t) = \int_0^{2\pi}\sqrt{r(\theta,t)^2 - (\frac{dr(\theta,t)}{d\theta})^2} d\theta \displaystyle$

and $\displaystyle A_{\gamma}(t) = \frac{1}{2}\int_0^{2\pi} r(\theta,t)^2 \,d\theta \displaystyle$

This simplifies to $P_{\gamma}(t) = 2\pi Mt$ and $A_{\gamma}(t) = \pi (Mt)^2$

Furthermore, we must note that on the interval $I=[0,k]$ all points on $\gamma$ are equidistant with respect to the origin so $\gamma$ is a circle.

As for the inequality:

In general, $\displaystyle A_{\gamma}(t) \leq \pi\frac{P_{\gamma}(t)}{4\pi^2}^2 \displaystyle$ so we obtain, $\displaystyle 4\pi A_{\gamma} \leq P_{\gamma}^2 \displaystyle$

Update2: The comments of @hardmath have been taken into account when editing this demonstration on 07/09/2014.

Update3: Today(29/09/2014) a professor at my university clearly explained that there were several serious holes in my proof. After his explanation I had to agree, and I must apologize to the people such as @hardmath who have patiently tried to explain my errors. I think I was too caught up in my own enthusiasm.

I'm not sure whether this question should be removed. But, I think that the discussion on this page and the way it ended could serve other members of this community.

• It would be appreciated if your solution were presented here, using the MathJax/LaTeX formatting options as appropriate. At a minimum some description of what about your solution is novel or in need of clearer exposition would better motivate a Reader to pursue the link to an external resource. – hardmath Sep 4 '14 at 12:50
• I don't see that you've proved the isoperimetric inequality. You've shown, under the assumptions you've imposed, that the area is at most $\pi(Mt)^2$. To get the isoperimetric inequality from that, you'd need the additional information that the perimeter is at least $2\pi Mt$. You note that the perimeter equals $2\pi Mt$ in the case of a circle, but you seem to have not tried to establish the necessary inequality in the general case. – Andreas Blass Sep 4 '14 at 13:55
• To the contrary, "intuitive proofs" were known to the ancient Greeks, but these are incomplete. In more modern times the proposed proof by Steiner has been critiqued as incomplete because it shows only that the maximum area for given perimeter is attained by a circle, not that this is the unique way of doing it. The proof of Peter Lax remedies this and can be given historical context. – hardmath Sep 4 '14 at 13:58
• You wrote that you think I misunderstood your demonstration. If there is a demonstration here, then indeed I did not and still do not understand it. I see nothing here that looks like a proof of the isoperimetric inequality. – Andreas Blass Sep 4 '14 at 18:11
• @AidanRocke: "Off topic" is the reason is was placed "on hold" (see title & explanatory highlighted text). Now that you've posted the proof, I cast a vote to re-open. "On hold" is a kind of grace period for the sort of editing you've done, and I'm hopeful the other re-open votes needed will appear shortly. – hardmath Sep 5 '14 at 0:54

$0$. We are missing a proposition, the isoperimetric inequality, to be proven. I want to reach the meat of the argument, but skipping preliminaries is an invitation to misunderstanding. Consider how best to state the proposition to be proven.

$1$. The first step mentions a (simple?) closed curve $C$ (in the plane), and essays a reduction to the case of convexity and a parameterization of $C$ by polar angle $\theta$:

$$r(\theta) \gt 0 \; \text{ for } \; 0 \le \theta \le 2\pi$$

where $r(0) = r(2\pi)$ is periodic. There is also introduced an alternative symbol $\gamma$ for the closed curve $C$, perhaps in anticipation of the "inflation" technique which seems to involve a second parameter, $\gamma_t$ defined by function $r(\theta,t)$.

There are several objections I would make here. No detailed argument is given about reducing to a convex region, and the statement of this is flawed: "A minimal condition on any closed curve $C$ that satisfies the isoperimetric inequality is that it must be convex." Surely it is the bounded interior of curve $C$ that is to be assumed convex, and rather than saying it's a "minimal condition" of curves that satisfy the inequality (non-convex regions with finite perimeter also satisfy it), an argument similar to Steiner's reflection is wanted. If we must preserve $\mathcal{C}^1$ smoothness of the boundary (depends on statement of proposition), some care is needed in invoking that argument.

My view is that these flaws are "fixable", and we can proceed tentatively to consider the next step.

$2$. The notion of "inflation" of the curve $C = \gamma$ is introduced with the partial explanation "all proportions are kept equal while the area enclosed by the curve increases." This doesn't seem to be a complete specification of the family of curves $\gamma_t$ generated by the new "time" parameter, but if it is, it would seem to require that $r(\theta,t) = t r(\theta)$.

At any rate such an isotropic expansion seems to meet all the criteria for "inflation" that are outlined:

(a) $\delta t > 0 \iff r(\theta, t+\delta t) > r(\theta,t)$

(b) perimeter $P_{\gamma}(t=0)=0$ and $\lim_{t \to +\infty} P_{\gamma}(t)=\infty$

(c) area $A_{\gamma}(t=0)=0$ and $\lim_{t \to +\infty} A_{\gamma}(t)=\infty$

This seems to me an unfixable obstacle, unless some criterion for "inflation" has been overlooked (by me) or left unstated (by you). A simple dilation $t$ of the curve $C$ will increase the perimeter by a factor $t$ and the area by a factor $t^2$. This only conserves the satisfaction or dissatisfaction of the isoperimetric inequality as $t$ varies. It cannot be the basis for asserting a proof of it.

$3$. Here it is asserted that $r(\theta,t)$ has a uniformly bounded (partial?) derivative with respect to time $t$ on compact intervals $[0,k]$. This would certainly be so for the case of dilation $r(\theta,t) = t r(\theta)$ discussed above, so you can arrange for it to be so. However it does not seem to follow from the little that was assumed about "inflation" (and no argument is advanced that justifies the claim).

If this is really needed for the argument, I'm willing to grant it is likely "fixable" by providing more specific definition of $r(\theta,t)$.

$4$. At "this point onwards" you declare success, "we may easily deduce the isoperimetric inequality," and move on. But the immediate "iff" statement does not seem to represent a statement of the isoperimetric inequality. The left hand side, $||r(\theta,t)|| \le Mt$ (why are we taking a norm, rather than simply absolute value, or even leave $r$ as is since it is positive?), follows from a Mean Value Thm. application of step $3$. above, but the corresponding bound on $A_\gamma$ does not give any comparison to $P_\gamma$.

In the next line we have a blatant switcheroo, where instead of the inequality $||r(\theta,t)|| \le Mt$ we suddenly have an equality $||r(\theta,t)|| = Mt$, and "$\gamma$ is a circle".

To summarize: I don't see what technique is being developed that will establish even the "easy" part of the isoperimetric inequality (the inequality itself), much less the more difficult part (that equality is attained only by a circle).

Added in response to Question Edits:

In the new step 2 a function $f(t)$ is introduced as defining the "inflation" process, i.e. $r(\theta,t) = f(t)r(\theta)$, with the apparent meaning that factor $f(t)$ will depend on time $t$ but not on angle $\theta$.

However in new step 3, "we define $f(\theta,t) = \frac{dr(\theta,t)}{dt}$". Not only does this open the door to a function that appears to depend on $\theta$ as well as $t$, it confuses the issue of partial vs. ordinary derivatives.

This confusion becomes acute in new step 4, where "it would suffice to choose $f'(t) = M/r(\theta)$". This choice is not possible unless $f(t)$ is allowed to be a function of $\theta$ as well as $t$.

Moreover to the extent that $f(t)$ varies with $\theta$, the "inflation" cannot provide that "all proportions [of the curve] are kept equal". The inconsistency is manifest in new steps 4,5 where it is claimed:

$$r(\theta,t) = f(t)r(\theta) = Mt$$

That is, the curve no longer depends on $\theta$ but only on $t$, i.e. the curve was a circle all along. The perimeter and area of the curve are then "simplified" to the usual formulas for a circle, giving the appearance that the isoperimetric inequality has been proven. However the argument is fallacious in suppressing the dependence of the family of curves on $\theta$.

• I understand the points that you've brought up but I can also anticipate that an immediate answer would probably lead to other questions. For this reason, I'll work on the proof over the weekend. But, you'll definitely hear from me by Monday. – user93511 Sep 5 '14 at 16:19
• Thank you for your observations. I have responded to the following: 1) The flawed statement: "A minimal condition on any closed curve C that satisfies the isoperimetric inequality is that it must be convex." 2) The assumption that the closed curve is smooth is dropped. 3) A more detailed explanation of what is meant by inflation(i.e. isotropic expansion). 4) A more structured arrangement of ideas so that the conclusion follows logically from the assumptions made. The "switcheroo" as you call it is removed. – user93511 Sep 7 '14 at 3:13
• @AidanRocke: The complaint is that given $r(\theta)$ depending on $\theta$, then you cannot have $r(\theta,t) = Mt$ not depending on the $\theta$ as you claim finally. If this is not clear, I have not any idea of how to better explain the fallacy in your argument. – hardmath Sep 7 '14 at 17:45
• In general, dependence on $\theta$ exists. But, in the special case where we want to maximize the area enclosed by the curve, we obtain $\gamma$ such that $r(\theta)$ is constant. Your statement is just as false as the assertion that $f(x) = Cst$ is undefined because the function doesn't depend on $x$. – user93511 Sep 7 '14 at 17:57
• The following points appear to have been misunderstood: 1) $r(θ)$ is chosen prior to inflation so "all proportions are kept equal" as stated. 2) When I take the partial derivative of $r(θ,t)$ wrt time I have a function of time because $r(θ)$ doesn't vary with time. 3) The curve was not a circle all along as you say. But, if it's required that there's an upper-bound on $f(θ,t)=dr(θ,t)/dt$ and this upper-bound is attainable for all $t$ in $I=[0,k]$ then we do obtain a circle if our intent is to maximize the area enclosed by the curve. $γ$ is a circle only in that special case. – user93511 Sep 7 '14 at 18:26