Parseval's theorem derivation using Bessel's inequality

Question

Could someone reiterate how to derive Parseval's theorem from Bessel's inequality?

I'm just a bit confused from the text in the textbook and would appreciate some clarification!

Thanks in advance mate.

Answer 1

Normally mathmaticians prove Parseval's equation with Lebesgue's integral in $L^2[0,\pi]$. However, the proof with Bessel's inequalities is far more natural for the following reason:

For $r,u\in \mathbb R$, $r<u$, there exist $2$ numbers $s\in\mathbb Q$ and $t\notin\mathbb Q$ with $r<s<t<u$. Indeed, with the Gaussian bracket $\lfloor\cdot\rfloor$ (or the floor $\lfloor\cdot\rfloor$) we calculate

$$d=10\cdot \left\lfloor max\left\{1, \dfrac{1}{u-r}\right\}\right\rfloor\in\mathbb N.$$

Then we can take the numbers

$$s=\dfrac{1}{d}\cdot\left\lfloor\dfrac{r+u}{2}d\right\rfloor\in\mathbb Q$$

and

$$t=s+\dfrac{\sqrt{2}}{d}\notin \mathbb Q.$$

Now we introduce Dirichlet's function $\chi(x)$ to

$$\chi(x):=\begin{cases} 0 & x\notin\mathbb Q\\ 1 & x\in \mathbb Q\\\end{cases}.$$

In order to integrate this function with Lebesgue's integral, for example over the interval $[0,1]$, we need a non-decreasing series of (simple) functions $(\chi_k(x))$, that is $\chi_k(x)\le\chi_{k+1}(x)$, with

$$\lim_{k\to\infty}\chi_k(x)\to\chi(x).$$

If $\chi_k(x_0)>0$ the function cannot have an area around its functional value because for any extension $[x_0-\delta_1,x_0+\delta_2]$ (parallel to the x-axis) there exists a point $r_0\notin\mathbb Q$ in this interval $[x_0-\delta_1,x_0+\delta_2]$ according to the remarks above and we have $\chi_k(r_0)\le\chi(r_0)=0$. Therefore for $\chi_k(x)>0$ we only integrate over points or over soaring lines without breadth.

The integration of the function $\psi(x):=1-\chi(x)$ is similar. If $(\psi_k(k))$ is our non-decreasing series of functions with $\lim\psi_k(x)\to\psi(x)$ and if $\psi_k(x_1)>0$ the function can not have an area around its functional value because for any extension $[x_1-\delta_2,x_1+\delta_3]$ there exists a point $q_1\in\mathbb Q$ in this interval according to the remarks above and we have $\psi_k(q_1)\le\psi(q_1)=0$.

But the functions are considered to be Lebeque-integrable with

$$\tag{1}L-\int_0^1\chi(x)dx=0$$

and

$$\tag{2}L-\int_0^1\psi(x)dx=1.$$

'$L-$' is the notation that the Lebesgue integral is to be taken.

The idea of Lebesgue's integral is that the set $(\mathbb R\setminus \mathbb Q)\cap[0,1]$ has measure $1$ and the set $\mathbb Q\cap[0,1]$ has measure $0$. The integral $(1)$ has the value $1$ for the measure $0$ and the value $0$ for the measure $1$ and this gives the value $0$. The integral $(2)$ has the value $0$ for the measure $0$ and the value $1$ for the measure $1$ and this gives the value $1$. Clearly, the Lebesgue integral detaches itself from the notion of area. Like it or not.

The natural approach to integration is Darboux' integral that we have learnt (in a less strict manner) at school. For the partition of the interval

$$a=x_0<x_1<\dots<x_n=b$$

the upper Darboux sum of the function $f(x)$ is

$$U(f,(x_k)):=\sum_{k=0}^{n-1}(x_{k+1}-x_k)\cdot\sup_{x_k\le x\le x_{k+1}}{f(x)}$$

and the lower sum is

$$L(f,(x_k)):=\sum_{k=0}^{n-1}(x_{k+1}-x_k)\cdot\inf_{x_k\le x\le x_{k+1}}{f(x)}.$$

If $L(f,(x_k))$ and $U(f,(x_k))$ converge to a common value for

$$\max_k \vert x_{k+1}-x_k\vert\to 0$$

this value is the Darboux integral

$$D-\int_a^bf(x)dx.$$

Riemann's integral takes an arbitrary value $f(\eta_k)$, $x_k\le\eta_k\le x_{k+1}$, instead of the supremum or infimum and therefore the sum lies in between the lower and upper sum of Darboux' integral. Both the Darboux intgral and Riemann integral converge to the same value if the function $f(x)$ is continuous on the interval $[a,b]$:

$$D-\int_a^bf(x)dx=R-\int_a^bf(x)dx$$

(theorem $79.1$ and $83.1$, Lehrbuch der Analysis, Teil $1$, Harro Heuser, [H$1$], [edit an English reference here and below]). If the function $f(x)$ has a finite number of discontinuities and if the function is integrable in between these discontinuities we can extend the Darboux and Riemann integral to the sum of the integrals over the (not necessarily closed) intervals where the function is continuous. However our purpose is the deduction of Parseval's equation with Riemann's integral and therefore we cannot extend the definition to an infinite number of discontinuities because then there exists at least one point of accumulation of discontinuities and at this point the function is similar to Dirichlet's function $\chi(x)$ above. We will take Riemann's integral, though we could take Darboux' integral just as well, because Riemann developed this integral in order to tackle the problems of Fourier analysis.

Let $f(x)$ be a $2\pi$-periodic function with $f(x+2\pi)=f(x)$. If $f(x)$ has no discontinuities or only a finite number of discontinuities and if $f(x)$ is square R-integrable, or if

$$R-\int_0^{2\pi}f^2(x)dx$$

exists, then Parseval's equation

$$\tag{3}\dfrac{1}{\pi}\int_0^{2\pi}f^2(x)dx=\dfrac{1}{2}a_0^2+\sum_{k=1}^\infty (a_k^2+b_k^2)$$

with

$$\tag{4}f(x)\sim\dfrac{1}{2}a_0+\sum_{k=1}^\infty (a_k\cos kx + b_k\sin kx)$$

holds.

We define

$$\tag{5}s_n(x)=\dfrac{1}{2}+\sum_{k=1}^n (a_k\cos kx + b_k\sin kx)$$

and the Cesaro sum or C-sum

$$\tag{6}c_n(x)=\dfrac{s_0+s_1+\ldots+s_n}{n+1}.$$

First we prove the theorem for a function $f(x)$ that is continuous on $\mathbb R$. The C-sum of a continuous function $f(x)$ converges uniformly to the function $f(x)$ (theorem $139.5$, Lehrbuch der Analysis, Teil $2$, Harro Heuser, [H$2$]) or for any $\varepsilon>0$ there exists a $N\in\mathbb N$ so that

$$\vert f(x)-c_n(x) \vert < \sqrt{\dfrac{\varepsilon}{2\pi}}$$

for $n\ge N$. We immediately obtain

$$\int_0^{2\pi}\{f(x)-c_n(x)\}^2dx\le\int_0^{2\pi}\dfrac{\varepsilon}{2\pi}dx=\varepsilon$$

for $n\ge N$. Bessel's inequality (theorem $134.2$, [H$2$]) states that for any sum

$$\varphi_n(x):=\alpha_0+\sum_{k=1}^n(\alpha_k\cos kx+\beta_k\sin kx)$$

the sum $s_n(x)$ is the better approximation with respect to the norm $\Vert\cdot\Vert_2$

$$\int_0^{2\pi}\{f(x)-s_n(x)\}^2dx\le\int_0^{2\pi}\{f(x)-\varphi_n(x)\}^2dx.$$

Mind the same $n$ in this inequality. Therefore we obtain

$$0\le\pi\sum_{k=n+1}^\infty (a_k^2+b_k^2)=\int_0^{2\pi}\{f(x)-s_n(x)\}^2dx\le\int_0^{2\pi}\{f(x)-c_n(x)\}^2dx\le\varepsilon$$

and the theorem is proven for a continuous function $f(x)$.

Now we prove the theorem if the function $f(x)$ has one jump discontinuity at the point $x_0$ in some interval of length $2\pi$. Because the function is bounded there exists a $B>0$ with $\vert f(x)\vert\le B$. For any $\varepsilon>0$ there exists a $0<\delta<\pi$ so that $(2\delta)\cdot (2B)^2<\dfrac{\varepsilon}{4}$. We define the function $f_1(x):=f(x)$ on the intervals $x\in [x_0-\pi,x_0-\delta]\cup[x_0+\delta,x_0+\pi]$. On the interval $[x_0-\delta,x_0+\delta]$ the function $f_1(x)$ is the line from the point $(x_0-\delta, f(x_0-\delta))$ to the point $(x_0+\delta, f(x_0+\delta))$. Finally we extend this function to a $2\pi$-periodic function $f_1(x+2\pi)=f_1(x)$. By construction the function $f_1(x)$ is continuous and we obtain

$$\tag{7}{\int_0^{2\pi}\{f_1(x)-f(x)\}^2dx}= {\int_{x_0-\pi}^{x_0+\pi}\{f_1(x)-f(x)\}^2dx}= {\int_{x_0-\delta}^{x_0+\delta}\{f_1(x)-f(x)\}^2dx}\le {\int_{x_0-\delta}^{x_0+\delta}(2B)^2dx}< \dfrac{\varepsilon}{4}.$$

Additionally we determine a $N\in\mathbb N$ so that

$$\int_0^{2\pi}\{f_1(x)-c_{1,n}(x)\}^2dx\le\dfrac{\varepsilon}{4}$$

for $n\ge N$ with the C-sum $c_{1,n}(x)$ of the function $f_1(x)$; mind that the function $f_1(x)$ is continuous. With Bessel's inequality and Minkowski's inequality for integrals (theorem $85.3$, [H$1$]) we get

$$0\le {\sqrt{\int_0^{2\pi}\{f(x)-s_n(x)\}^2dx}}\le {\sqrt{\int_0^{2\pi}\{f(x)-c_{1,n}(x)\}^2dx}}= {\sqrt{\int_0^{2\pi}\{[f(x)-f_1(x)]+[f_1(x)-c_{1,n}(x)]\}^2dx}}\le {\sqrt{\int_0^{2\pi}\{f(x)-f_1(x)\}^2dx}+\sqrt{\int_0^{2\pi}\{f_1(x)-c_{1,n}(x)\}^2dx}}\le {\sqrt{\dfrac{\varepsilon}{4}}+\sqrt{\dfrac{\varepsilon}{4}}}=\sqrt{\varepsilon}.$$

We square these inequalities and Parseval's equality is proven for a function with only one jump discontinuity.

We follow these ideas and prove the theorem if the function $f(x)$ is unbounded at a point $x_0$. Because the function $f(x)$ is square-integrable for any $\varepsilon>0$ there exists a $0<\delta<\pi$ such that

$$\tag{8}\int_{x_0-\delta}^{x_0}f^2(x)dx<\dfrac{\varepsilon}{64}.$$

Moreover, there exists a $x_0-\delta\le\xi_1<x_0$ with

$$(x_0-\xi_1)\cdot f^2(\xi_1)<\dfrac{\varepsilon}{64}.$$

Indeed, otherwise we get

$$\int_{0}^{\delta}f^2(x_0-t)dt\ge\int_{0}^{\delta}\dfrac{\varepsilon}{64t}dt\ge\dfrac{\varepsilon}{64},$$

contrary to inequality (8). On the interval $[x_0-\pi,\xi_1]$ we define the function $f_1(x):=f(x)$, and on the interval $[\xi_1,x_0]$ the function $f_1(x)$ is the line from the point $(\xi_1, f(\xi_1))$ to the point $(x_0,0)$. We obtain with Minkowski's inequality

$$0\le {\sqrt{\int_{x_0-\pi}^{x_0}\{f(x)-f_1(x)\}^2dx}}= {\sqrt{\int_{\xi_1}^{x_0}\{f(x)-f_1(x)\}^2dx}}\le {\sqrt{\int_{\xi_1}^{x_0}f^2(x)dx}+\sqrt{\int_{\xi_1}^{x_0}f_1^2(x)dx}}\le {\sqrt{\int_{x_0-\delta}^{x_0}f^2(x)dx}+\sqrt{\dfrac{\varepsilon}{64}}}\le {\sqrt{\dfrac{\varepsilon}{64}}+\sqrt{\dfrac{\varepsilon}{64}}}= \sqrt{\dfrac{\varepsilon}{16}}.$$

Likewise there exists a point $x_0<\xi_2<x_0+\pi$ with

$$\int_{x_0}^{\xi_2}f^2(x)dx\le\dfrac{\varepsilon}{64}$$

and

$$(\xi_2-x_0)\cdot f^2(\xi_2)<\dfrac{\varepsilon}{64}.$$

We extend the function $f_1(x)$ from above. On the interval $[x_0,\xi_2]$ it is the line from the point $(x_0,0)$ to the point $(\xi_2,f(\xi_2))$. On the interval $[\xi_2,x_0+\pi]$ we define $f_1(x):=f(x)$. Finally we extend this function $f_1(x)$ to a $2\pi$-periodic function with $f_1(x+2\pi)=f_1(x)$ and obtain

$$\int_0^{2\pi}\{f^2(x)-f_1^2(x)\}dx\le\dfrac{\varepsilon}{4}.$$

By construction the function $f_1(x)$ is continuous. Therefore we only have to follow the steps after the inequality (7) in order to prove Parseval's equation for a function $f(x)$ that is unbounded at one point.

The last steps for an unbounded discontinuity could also be applied if the function $f(x)$ has a jump discontinuity or if the function is only unbounded for $f(x_0-)$ or $f(x_0+)$ though the procedure can be understood better in dealing with the case of a jump discontinuity separately. It is easy now to extend the procedure for a finite number of discontinuities and Parseval's equation has been proven.