# Isoperimetric inequality implies Wirtinger’s inequality

Let C:x=x(t),y=y(t),a≤t≤bC: x=x(t), y=y(t), a\le t\le b be a C1C^1 closed curve (not necessarily simple).The isoperimetric inequality says that
A≤ℓ24π, A\le \frac{\ell^2}{4\pi}, where A=|∫Cy(t)x′(t)dt|A=\left|\int_C y(t)x'(t) dt\right| is the area enclosed by CC, and
ℓ=∫ba√(x′(t))2+(y′(t))2dt\ell=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2} dt is the arc length of CC.
My question is how to use this theorem to prove Wirtinger theorem: If f(t)f(t) is a TT-periodic C1C^1 real-valued function such that
∫T0f(t)dt=0,\int_0^T f(t) dt=0, then ∫T0|f(t)|2dt≤T24π2∫T0|f′(t)|2dt.\int_0^T |f(t)|^2 dt\le \frac{T^2}{4\pi^2}\int_0^T |f'(t)|^2 dt.

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First thing to note that if we re-parametrize t=kst = ks, we have, by the chain rule, ddsf=k(ddtf)∘t\frac{d}{ds}f = k (\frac{d}{dt}f)\circ t. Choose k=T/(2π)k = T/(2\pi), then the change of variable shows that it is sufficient to prove the claim for the period being 2π2\pi. That is, if we can show for any 2π2\pi periodic function with mean 0

∫2π0f2ds≤∫2π0|f′|2ds \int_0^{2\pi} f^2 ds \leq \int_0^{2\pi} |f’|^2 ds

we’ll be done by a re-scaling argument.

Since ff has mean 0, you can write f=F′f = F’ for FF another 2π2\pi periodic function. So the isoperimetric inequality implies

|∫2π0fF′ds|≤14π(∫2π0√(F′)2+(f′)2ds)2 \left|\int_0^{2\pi} f F’ ds \right| \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{ (F’)^2 + (f’)^2 } ds \right)^2

or

∫2π0f2ds≤14π(∫2π0√f2+(f′)2ds)2 \int_0^{2\pi} f^2 ds \leq \frac{1}{4\pi} \left( \int_0^{2\pi} \sqrt{f^2 + (f’)^2} ds \right)^2

Now use Holder’s inequality on the finite interval [0,2π][0,2\pi], we get

∫2π0√f2+(f′)2ds≤√2π(∫f2+(f′)2ds) \int_0^{2\pi} \sqrt{f^2 + (f’)^2} ds \leq \sqrt{2\pi} \left( \int f^2 + (f’)^2 ds \right)

So we get

∫2π0f2ds≤12∫2π0f2+(f′)2ds \int_0^{2\pi} f^2 ds \leq \frac{1}{2}\int_0^{2\pi} f^2 + (f’)^2 ds

which, subtracting 12∫f2\frac12 \int f^2 from both sides yields the desired inequality.

Now, a short remark on why it is necessary to first use the scaling argument. The principle is the following: Wirtinger’s inequality, as discussed in the first paragraph above, is scale invariant: changing the scale of parameter tt changes the terms f2f^2 and T2(f′)2T^2 (f’)^2 equally.

The isoperimetric inequality, however, is not scale invariant in the same way: using the ansatz where xx corresponds to FF and yy corresponds to F′=fF’ = f, you see that a change of parametrization t→kst \to ks will leave xx the same while changing yy by a multiple factor.

In particular, this means that the depending on scale, the inequality may not be sharp. In other words, the changing of scale corresponds, morally speaking, to changing the xx and yy directions in different proportions. So if we start out with a circle, which is a maximizer of the isoperimetric inequality, after this funny reparametrisation which scales xx and yy differently, we end up with an ellipse, which no longer is a maximizer.

The change of scale in the first paragraph allows us to “use a version of the isoperimetric inequality as close to the circle version as possible”. In other words, by isolating the scale you can most efficiently use the isoperimetric inequality, which then allows for a simple proof of Wirtinger’s inequality.

To clarify a bit on the paragraphs after the rule: you can still prove, by brute-force, Wirtinger from Isoperimetric without the normalisation of T=2πT = 2\pi. But in the step where I applied Holder inequality (or Cauchy Schwarz if you prefer) to √f2+(f′)2=√f2+(f′)2⋅1\sqrt{f^2 + (f’)^2} = \sqrt{f^2 + (f’)^2}\cdot 1, it will be necessary to use a different, likely much more complicated splitting to properly optimize the final result.
– Willie Wong
Dec 21 ’10 at 3:32

can you please elaborate how did you got the second inequality? The trick is to define x(t)x(t) and y(t)y(t), but obvious choice x(t)=tx(t)=t, y(t)=f(t)y(t)=f(t) clearly does not work.
– mpiktas
Dec 21 ’10 at 8:07

1

@mpiktas. Define x(t)=∫t0f(s)ds,y(t)=f(t),0≤t≤2πx(t)=\int_0^t f(s) ds, y(t)=f(t), 0\le t\le 2\pi. The condition ∫2π0f(t)dt=0\int_0^{2\pi}f(t) dt=0 implies that the curve is closed.
– TCL
Dec 21 ’10 at 14:28

@mpiktas: right, @TCL already explained it. I thought it was obvious that the introduction of the FF means I am setting x=Fx = F. I suppose it is not. Let me add a line to clarify. Also, note that setting x(t)=tx(t) = t contradicts the assumption that you are describing a closed curve, since x(2π)=2π≠x(0)=0x(2\pi) = 2\pi \neq x(0) = 0, and so in fact is not an obvious choice.
– Willie Wong
Dec 21 ’10 at 15:31

@TCL, @Willie Wong, thanks for explanation.
– mpiktas
Dec 21 ’10 at 21:17
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