A discrete approach to Wirtinger's inequality

TL;DR

This paper develops a discrete version of Wirtinger's inequality for piecewise functions, offering a new elementary proof and insights into the equality case, while connecting to Fourier series development.

Contribution

It introduces a novel discrete approach to Wirtinger's inequality and provides a new elementary proof, also exploring the equality case and Fourier series implications.

Findings

Discrete version of Wirtinger's inequality derived

Elementary proof of the inequality established

Connection to Fourier series development of periodic functions

Abstract

Considering Wirtinger's inequality for piece-wise equipartite functions we find a discrete version of this classical inequality. The main tool we use is the theorem of classification of isometries. Our approach provides a new elementary proof of Wirtinger's inequality that also allows to study the case of equality. Moreover it leads in a natural way to the Fourier series development of $2\pi$-periodic functions.

Equations

$${\int }_0^{2\pi }{f}^2(t)dt\le {\int }_0^{2\pi }{f}^{\prime 2}(t)dt,$$

$$\sum _{j=1}^n{x}_j{x}_{j-1}\le \frac{3n^2-4{\pi }^2}{3n^2+2{\pi }^2},$$

$$\sum _{i=1}^n{x}_i{x}_{i-1}\le \cos (\frac{2\pi }{n}),$$

$$x_i=a\cos (\frac{2\pi }{n}i)+b\sin (\frac{2\pi }{n}i),i=1,\dots ,n,$$

$$T(x_1,x_2,\dots ,x_n)=(x_n,x_1,x_2,\dots ,x_{n-1}),$$

$${\int }_0^{2\pi }{f}^2(t)dt=\frac{2\pi }{3n}\sum _{j=1}^n(2x_j^2+x_j{x}_{j-1})$$

$${\int }_0^{2\pi }{f}^{\prime 2}(t)dt=\frac{n}{\pi }\sum _{j=1}^n(x_j^2-x_j{x}_{j-1}).$$

$${\int }_0^{2\pi }{f}^2(t)dt\le {\int }_0^{2\pi }{f}^{\prime 2}(t)dt$$

$$\sum _{j=1}^n{x}_j{x}_{j-1}\le \frac{3n^2-4{\pi }^2}{3n^2+2{\pi }^2}\sum _{j=1}^n{x}_j^2.$$

$$\sum _{j=1}^n{x}_j{x}_{j-1}\le \frac{3n^2-4{\pi }^2}{3n^2+2{\pi }^2},\text{mbox}with\sum _{j=1}^n{x}_j=0,\sum _{j=1}^n{x}_j^2=1.$$

$$T(x_1,x_2,\dots ,x_n)=(x_n,x_1,x_2,\dots ,x_{n-1}),$$

$$\sum _{j=1}^n{x}_j{x}_{j-1}=⟨X,T(X)⟩,$$

$$T(e_1)$$

$$T(e_2)$$

$$T(e_{2k+1})$$

$$T(e_{2k+2})$$

$$T(e_1)$$

$$T(e_{2k})$$

$$T(e_{2k+1})$$

$$e_1$$

$$e_2$$

$$e_{2k+1}$$

$$e_{2k+2}$$

$$e_1$$

$$e_{2k}$$

$$e_{2k+1}$$

$$⟨X,T(X)⟩$$

$$⟨X,T(X)⟩=y_1^2-y_2^2+\sum _{k=1}^{(n-2)/2}(y_{2k+1}^2+y_{2k+2}^2)\cos {\alpha }_k,$$

$$⟨X,T(X)⟩=y_1^2+\sum _{k=1}^{(n-1)/2}(y_{2k}^2+y_{2k+1}^2)\cos {\alpha }_k.$$

$$\sum _{i=1}^n{x}_i{x}_{i-1}\le \cos (\frac{2\pi }{n}),$$

$$x_i=a\cos (\frac{2\pi }{n}i)+b\sin (\frac{2\pi }{n}i),i=1,\dots ,n,$$

$$⟨X,T(X)⟩\le \cos (\frac{2\pi }{n}).$$

$$⟨X,T(X)⟩\le \cos (\frac{2\pi }{n})\sum _{k=1}^{(n-2)/2}(y_{2k+1}^2+y_{2k+2}^2)\le \cos (\frac{2\pi }{n})$$

$$⟨X,T(X)⟩\le \cos (\frac{2\pi }{n})\sum _{k=1}^{(n-1)/2}(y_{2k}^2+y_{2k+1}^2)=\cos (\frac{2\pi }{n})$$

$$\sum _{i=1}^n{x}_i{x}_{i-1}<\frac{3n^2-4{\pi }^2}{3n^2+2{\pi }^2},$$

Full text

1 Introduction

The classical Wirtinger inequality states that for a $2\pi$-periodic ${\cal C}^{1}$ function $f(t)$ with $\int_{0}^{2\pi}f(t)\,dt=0$ one has

[TABLE]

with equality if and only if $f(t)=a\sin(t)+b\cos(t)$ for some $a,b\in\mathbb{R}$.

The goal of this note is to give a discrete inequality that will imply the above result, including the case of equality. At the same time our approach leads in a natural way to the Fourier series development of a $2\pi$-periodic function.

Wirtinger did not publish his result, but he communicated it by letter to W. Blascke who included it in [1]. The original proof is based on the theory of Fourier series. Discrete approximations to Wirtinger's inequality have been given by several authors; see for instance [2], [4].

As a motivation for a discrete inequality we consider Wirtinger's inequality for piece-wise equipartite linear functions, that is for continuous functions $f:[0,2\pi]\longrightarrow\mathbb{R}$, linear on each interval $[\frac{2\pi}{n}(j-1),\frac{2\pi}{n}j]$, $j=1,\dots,n$ and such that $f(0)=f(2\pi)$. Denoting $f(\frac{2\pi}{n}j)$ by $x_{j}$, $j=1,\dots,n$, and taking $x_{0}=x_{n}$, Wirtinger's inequality for this class of functions is equivalent to the discrete inequality

[TABLE]

for $x_{j}\in\mathbb{R}$, $j=1,\dots,n$, $x_{0}=x_{n}$, $\sum_{j=1}^{n}x_{j}=0$ and $\sum_{j=1}^{n}x_{j}^{2}=1$.

Wirtinger's inequality can then be obtained from the above inequality by a limiting process.

We shall obtain (2) as a consequence of the following

Theorem 2.1. Let $x_{1},\dots,x_{n}\in\mathbb{R}$, for $n\geq 4$, with $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=1$. Then

[TABLE]

with $x_{0}=x_{n}$. Equality holds if and only if

[TABLE]

for $a,b\in\mathbb{R}$ satisfying $a^{2}+b^{2}=2/n$.

This result that can be considered as the Wirtinger discrete inequality was obtained by Fan, Taussky and Todd in [2] where it is used to obtain classical Wirtinger inequality (1) but, as the authors say, without the equality clause. Other proofs of Theorem 2.1 have been published later, see for instance [4].

For completeness we provide here a simple different proof of the above result based on the theorem of classification of isometries applied to the cyclic isometry $T$ given by

[TABLE]

since the left hand-side of (3) can be written as $\langle X,T(X)\rangle$, where $X=(x_{1},x_{2},\dots,x_{n})$.

As we have said our approach, based on inequality (2), leads to inequality (1), and allows to caracterize functions for which equality holds. This characterization is somewhat delicate but the argument used has a surprising consequence: the Fourier series development of a $2\pi$-periodic function.

2 Discrete Wirtinger's inequality

In order to find a discrete version of the Wirtinger inequality we consider this inequality for piece-wise equipartite linear functions.

For $n\in\mathbb{N}$, $n\geq 4$, let $f:[0,2\pi]\longrightarrow\mathbb{R}$ be a continuous function, linear on each interval $[\frac{2\pi}{n}(j-1),\frac{2\pi}{n}j]$, $j=1,\dots,n$ and such that $f(0)=f(2\pi)$. Denoting $f(\frac{2\pi}{n}j)$ by $x_{j}$, $j=1,\dots,n$, and taking $x_{0}=x_{n}$, a computation shows that

[TABLE]

and

[TABLE]

So the inequality

[TABLE]

is equivalent to

[TABLE]

Assuming now $\int_{0}^{2\pi}f(t)\,dt=0$, that means $\sum_{i=1}^{n}x_{j}=0$, it follows that Wirtinger's inequality for piece-wise linear functions is equivalent to (6) with this additional hypothesis or, normalizing,

[TABLE]

This is a problem of maximizing a given quadratic form under some restrictions. It can be solved by different methods such as Lagrange multipliers or by the determination of the least characteristic value of a Hermitian matrix, as done in [2]. As said our approach is based on the theorem of classification of isometries.

The canonical expression of the quadratic form.

The left-hand side of (7) leads in a natural way to consider the cyclic isometry

[TABLE]

since

[TABLE]

where $X=(x_{1},\dots,x_{n})$ and $\langle\,,\,\rangle$ is the standard scalar product. Hence, in order to prove (7) we start by analyzing the structure of the isometry $T$. This will allow us to find the canonical expression of the quadratic form $\langle X,T(X)\rangle$.

The theorem of classification of isometries (see [3]) applied to $T$ asserts that there is an orthonormal basis $(e_{1},\dots,e_{n})$ such that, denoting $\alpha_{k}=\frac{2\pi}{n}k$, one has for $n$ even

[TABLE]

and for $n$ odd

[TABLE]

In fact, it can be seen by using elementary trigonometric formulas that for $n$ even, this basis is given by

[TABLE]

and for $n$ odd by

[TABLE]

Since $\langle e_{i},T(e_{j})\rangle+\langle e_{j},T(e_{i})\rangle=0,\mbox{ for }i\neq j,$ we get for every vector $X=\sum_{i=1}^{n}y_{i}e_{i}$,

[TABLE]

Hence the canonical expression of the quadratic form $\langle X,T(X)\rangle$ is for even $n$

[TABLE]

and for odd $n$

[TABLE]

The discrete inequality

The maximum of the quadratic form $\langle X,T(X)\rangle$ is given by the following

Theorem 2.1** (Discrete Wirtinger's inequality).**

Let $x_{1},\dots,x_{n}\in\mathbb{R}$, for $n\geq 4$, with $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=1$. Then

[TABLE]

with $x_{0}=x_{n}$. Equality holds if and only if

[TABLE]

for $a,b\in\mathbb{R}$ satisfying $a^{2}+b^{2}=2/n$.

Proof. With the previous notation we must prove

[TABLE]

Since $\langle X,e_{1}\rangle=0$, $||X||=1$ it is $X=\sum_{j=2}^{n}y_{j}e_{j}$, $\sum_{j=2}^{n}y_{j}^{2}=1$ and we get from (10)

[TABLE]

for $n$ even, and from (11)

[TABLE]

for $n$ odd. This proves the first part of the Lemma.

Equality holds when $X=y_{3}e_{3}+y_{4}e_{4}$ for $n$ even and $X=y_{2}e_{2}+y_{3}e_{3}$ for $n$ odd. Substituting $e_{2},e_{3},e_{4}$ by the expressions in (2) and (2) the Lemma follows. $\square$

As a consequence of this result we obtain inequality (7).

Proposition 2.2**.**

Let $x_{1},\dots,x_{n}\in\mathbb{R}$, for $n\geq 4$, with $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=1$. Then

[TABLE]

with $x_{0}=x_{n}$.

Proof. By Theorem 2.1 in order to prove (12) it is enough to show that

[TABLE]

Denoting $2\pi/n$ by $\alpha$ the above inequality is equivalent to

[TABLE]

which using that $\cos\alpha<1-\alpha^{2}/2+\alpha^{4}/24$ is easily verified. ∎

We remark that equality in (7) never holds.

Corollary 2.3**.**

For $n\in\mathbb{N}$, $n\geq 4$, let $f:[0,2\pi]\longrightarrow\mathbb{R}$ be a continuous function, linear on each interval $[\frac{2\pi}{n}(j-1),\frac{2\pi}{n}j]$, $j=1,\dots,n$ and such that $f(0)=f(2\pi)$. Assume that $\int_{0}^{2\pi}f(t)dt=0$. Then

[TABLE]

Proof. As said, inequality (13) with hypothesis $\int_{0}^{2\pi}f(t)\,dt=0$ is equivalent to (7). So the Corollary is a direct consequence of Proposition 2.2. $\square$

3 Wirtinger's inequality

Now we can obtain, by a limiting process, the classical Wirtinger's inequality.

Theorem 3.1** (Wirtinger's inequality).**

Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ be a $2\pi$-periodic ${\cal C}^{1}$ function such that $\int_{0}^{2\pi}f(t)\,dt=0.$ Then

[TABLE]

Equality holds if and only if $f(t)=a\cos(t)+b\sin(t)$ for some $a,b\in\mathbb{R}$.

Proof. For each $n\in\mathbb{N}$, $n\geq 4$, let $\phi_{n}(t)$ be the function linear on each interval $[\frac{2\pi}{n}(j-1),\frac{2\pi}{n}j]$, with $\phi_{n}(\frac{2\pi}{n}j)=f(\frac{2\pi}{n}j)$, $j=1,\dots,n$.

Set $x_{j,n}=f(\frac{2\pi}{n}j)$, $m_{n}=\dfrac{1}{n}\sum_{j=1}^{n}x_{j}$, and $\tilde{x}_{j,n}=x_{j,n}-m_{n}$. Let $\tilde{\phi}_{n}(t)$ be the function linear on each interval $[\frac{2\pi}{n}(j-1),\frac{2\pi}{n}j]$, with $\tilde{\phi}_{n}(\frac{2\pi}{n}j)=\tilde{x}_{j,n}$, $j=1,\dots,n$. Equivalently, $\tilde{\phi}_{n}(t)=\phi_{n}(t)-m_{n}$.

Since $\int_{0}^{2\pi}\tilde{\phi}(t)\,dt=0$ it follows, by Corollary 2.3, that

[TABLE]

Moreover since $f$ is a ${\cal C}^{1}$ function we have

[TABLE]

and

[TABLE]

Finally the hypothesis $\int_{0}^{2\pi}f(t)\,dt=0$ yields $\lim_{n\to\infty}m_{n}=0$ and so

[TABLE]

and inequality (14) follows.

It remains to analize when equality holds in (14).

From now on we will assume that $n$ is an odd integer; the case $n$ even is dealt similarly. Let $H_{k}=\langle e_{2k},e_{2k+1}\rangle$ denote the subspace of $\mathbb{R}^{n}$ generated by $e_{2k}$ and $e_{2k+1}$, the vectors introduced in Section 2, for $k=1,\dots,(n-1)/2$. Let $P_{k}$ be the orthogonal projection from $\mathbb{R}^{n}$ on $H_{k}$, and let $P_{0}$ be the orthogonal projection on $H_{0}=\langle e_{1}\rangle$.

Lemma 3.2**.**

Let $f$ be a function satisfying the hypotheses of Theorem 3.1 and such that equality holds in (14). For each $n\geq 4$ let $X_{n}$ be the vector of components $x_{j,n}=f(\frac{2\pi}{n}j)$, $j=1,\dots,n$. Then

[TABLE]

Proof. By the definition of Riemann's integral we have

[TABLE]

where $p=(n-1)/2$.

From (5), (11) and (15) it follows that

[TABLE]

As a consequence of equality in (14) we get

[TABLE]

The Lemma follows from the inequality

[TABLE]

which is true for $k\geq 2$ (which implies $n\geq 5$) using that $\cos(x)\leq 1-x^{2}/2+x^{4}/24$. $\square$

To continue the proof of Theorem 3.1, for each vector $X=(x_{1},\dots,x_{n})$ let $L_{X}$ be the function that is linear on each interval $[\frac{2\pi}{n}(j-1),\frac{2\pi}{n}j]$ with $L_{X}(\frac{2\pi}{n}j)=x_{j}$, $j=1,\dots,n$, ($x_{0}=x_{n}$).

When $X_{n}$ is the vector of components $x_{j,n}=f(\frac{2\pi}{n}j)$, $j=1,\dots,n$, $L_{X_{n}}$ is the function $\phi_{n}$ defined at the begining of this proof. So we can assume that $\sum_{j=1}^{n}x_{j,n}=0$ and we know that $\lim_{n\to\infty}L_{X_{n}}=f$.

Writing $X_{n}=y_{2}e_{2}+y_{3}e_{3}+\sum_{k=2}^{p}(y_{2k}e_{2k}+y_{2k+1}e_{2k+1})$ we have

[TABLE]

To finish the proof we need to show that

[TABLE]

Formula (4) can be writen as

[TABLE]

which gives, by using the identity of polarization,

[TABLE]

for two vectors $X,Y$.

In particular one gets $\langle L_{e_{i}},L_{e_{j}}\rangle=0$, $i\neq j$, and hence

[TABLE]

and

[TABLE]

Writing $\tilde{e}_{2}=\sqrt{\frac{n}{2}}\,e_{2}$, $\tilde{e}_{3}=\sqrt{\frac{n}{2}}\,e_{3}$ we have

[TABLE]

Thus $\lim_{n\to\infty}\alpha_{n}=a\cos t+b\sin t$, as wanted, where $a,b$ are the first Fourier coefficients of $f$.

As concerning $\lim_{n\to\infty}\beta_{n}$ we have

[TABLE]

and the proof finishes by applying Lema 3.2. $\square$

Remark. Let $f$ be a $2\pi$-periodic ${\cal C}^{1}$ function such that $\int_{0}^{2\pi}f(t)\,dt=0$. The same argument used to calculate $\lim_{n\to\infty}\alpha_{n}$ in the above proof, applied also to $\beta_{n}$ shows that $f$ can be written as

[TABLE]

with

[TABLE]

If we drop the assumption $\int_{0}^{2\pi}f(t)\,dt=0$ we need to add in the above expression of $f$ the term $\frac{1}{2\pi}\int_{0}^{2\pi}f(t)\,dt$. So, the discrete approach we have developped here leads, in a natural way, to the well known Fourier series development of a $2\pi$-periodic function.