So where did we leave? Ah here:

Nice! I can “see” why the fact is true. Can we do better? Can you make me “feel” why it’s true? We’ll see that in the next post!

Part 1

And the answer is…… Yes, we can! But we would have to consult the physicists. We are unlucky in one aspect: *relativity*. They (*“who?”* is also a rather interesting question to ask) have taken out all the things we see in our surroundings out of math, and made it abstract. But the fact that we can always take them back, is a plus point.

Imagine a rope enclosing a two-dimensional gas, with vacuum outside the rope. The gas will expand, pushing the rope to enclose a maximal area at equilibrium.

When the system is at equilibrium, the tension in the rope must be constant, because if there were a tension gradient at some point, there would be a non-zero net force at that point in the direction of the rope, but at equilibrium the net force must be zero in all directions.

The gas exerts a force outward on the rope, so tension must cancel this force. Take a small section of rope, so that it can be thought of as a part of some circle, called the osculating circle. The force on this rope segment due to pressure is PL where ‘P’ is pressure and ‘L’ length. The net force due to tension is 2T sin(L/2R) with ‘T’ tension and ‘R’ the radius of the osculating circle.

Because the pressure is the same everywhere, and the force from pressure must be canceled by the force from tension, the net tension force must be the same for any rope segment of the same length. That means the radius of the osculating circle is the same everywhere, so the rope must be a circle.

A big merit of the above solution is that you can actually *imagine* and *see* the gas molecules pushing the curve and making it round. Plot twist: the Greeks could see planets in the sky, they did not have to imagine them. But being believers of the geocentric theory, they wanted to explain the retrograde motion of the planets. So they found an argument using epicycles that was extremely complicated. But soon it was discovered that

“Epicycles are everything”

Fourier Analysis

So let us use the snake’s oil- **Fourier Analysis**– in our next proof. A neat proof using Green’s Theorem and Cauchy-Schwarz is given in [3] (These two proof are really the same :P).

Nah, it’s not that fancy.

And no, not that stupid. I would prefer to have a brief recap of linear algebra that we need here rather than diving right in because that would illustrate the point I want to make better.

What really are *independent elements*? The definition of linear independence that ∑λ_i v_i ↔ **λ=0, **but what does it really mean?

Linear independence is a symbol of **heterogeneity**. You can really “separate” out the different parts as you can in a heterogeneous mixture of substances. The only problem in every particular case is *how*.

Suppose we are in real numbers, and we want to filter out √2 out of 9√2+7√3 then how should we go about doing this? One way is to define an additive f:R→R using Hamel basis (you can also check out handouts of Evan Chen in case you don’t know what this term means) of over such that f(√2)≔√2 and f(h)≔0 for all other basis elements. Then f(9√2+7√3)=9f(√2)+7f(√3)=9√2 done! **But** this is not the only way we can separate the elements. There is a more “intrinsic” way of separation. And that is …

The idea at heart of Fourier Analysis is rather nice – **orthogonality**. We will need some linear algebra preliminaries (learning which is left as an exercise for the diligent reader).

Fourier Analysis wants to use the orthogonal basis {1,sin(x),sin(2x), …} and {1,cos(x),cos(2x), …} for real-valued functions and for complex-valued functions. Their respective inner products are and . Can you show that the elements are orthogonal with respect to the respective inner products?

All unspecified sums are from to . Let us assume that f(x) has the Fourier expansion . Then notice the following curious thing: and so,

Now we know that when and then and therefore integrating in that range,

The idea of bilinear Parseval’s, informally, is not very different. In that case assuming g(x) has the Fourier expansion we have

So similarly,

The above manipulations disregard convergence issues. We urge the interested reader to look it up in [5]. We remark that in case of real-valued functions, .

Get happy people! Your hard work will now pay off. Let us call the original curve **C** = (x(s),y(s)) where 0≤s≤2π and normalize L=2π. Then we can parametrize it by arc-length so that the speed always.

Let us consider their Fourier series

We clearly have

Thus, by Parseval’s identity,

By continuous shoelace formula (for geometers) By Green’s formula (for others) and bilinear Parseval’s Identity,

Now by the triangle inequality,

Thus, because |n|≤|n|^2, therefore,

A≤π ∑ = π

And yay, we are done! All these have been short and tricky proofs. There are many more that I will tell about later because you have to internalize the ideas, and I do not wish to test your patience. Anyway, here are some exercises to test whether you understood what I just said above. In case you are able to solve them, kudos! But don’t post your solutions below until enough people have seen the post (alternately, use something like spoiler tags in discord, but I don’t know about their availability here).

**Exercise 1.** How can the inner product be used for “separating” orthogonal elements?

**Exercise 2.** I said that are linearly independent and orthogonal. Considering functions that are finite sums of these, can you conclude their Fourier coefficients will be unique?

**Exercise 3.** Prove that any two elements of the set {1,sin(x),cos(x),sin(2x),cos(2x),…} are orthogonal with respect to the real inner product.

**Exercise 4.** Does the “dents to bumps” process take arbitrarily moves to terminate in general if we are not wise? Can you provide an example?

**Exercise 5.** “Separate” √2 and √3 using modular arithmetic. Can you generalize your idea?

**Exercise 6.** Justify, “Epicycles are everything”.

**Exercise 7.** Complete the hand-wavy sketches and find answers to in-text questions. Also, try reading the formal proofs and details.

**Exercise 8.** Prove the following inequality (or explain modulo formalism). It gives yet another proof of the theorem. Can you find it?

References:

[1] Jakub Konieczny https://math.stackexchange.com/users/10674/jakub-konieczny

Among all shapes with the same area, a circle has the shortest perimeter, URL (version: 2017-04-13): https://math.stackexchange.com/q/389362

[2] Mark Eichenlaub https://math.stackexchange.com/users/3305/mark-eichenlaub

Why does a circle enclose the largest area?, URL (version: 2020-07-26): https://math.stackexchange.com/q/26213

[3] Agustí Roig https://math.stackexchange.com/users/664/agust%c3%ad-roig

Why does a circle enclose the largest area?, URL (version: 2011-10-11): https://math.stackexchange.com/q/4818

[5]Fourier Analysis, Elias M. Stein and Rami Shakarchi