Can the usual Fourier series of a function be viewed as that defined with respect to an orthonormal system of functions?

Question

This query is related to how the Fourier series of a function in defined in the book Mathematical Analysis by Apostol.

Let $I$ be an interval, and let $L^2(I)$ be the set of complex-valued functions $f$ which are measurable on $I$ such that $\int_I |f(x)|^2 dx < \infty$. In subsection 11.4 of the book, the Fourier series of a function $f$ relative to an orthonormal system of functions $S = \{\varphi_0, \varphi_1, \varphi_2, \ldots\}$ on I is defined in the following way:

Let $S = \{\varphi_0, \varphi_1, \varphi_2, \ldots\}$ be orthonormal on $I$ and assume that $f \in L^2(I)$. The notation $$ f(x) \sim \sum_{n=0}^\infty c_n \varphi_n(x) \tag{1}\label{1} $$ will mean that the numbers $c_0, c_1, c_2, \ldots$ are given by the formulas: $$ c_n = (f, \varphi_n) = \int_I f(x) \overline{\varphi_n(x)}\; dx \qquad (n = 0, 1, 2, \ldots). \tag{2}\label{2} $$ The series in $\eqref{1}$ is called the Fourier series of $f$ relative to $S$, and the numbers $c_0, c_1, c_2, \ldots$ are called the Fourier coefficients of $f$ relative to $S$.

Next, the usual Fourier series generated by $f$ is introduced in the following way. Let $S$ be the system of trigonometric functions given by $$ \varphi_0(x) = \frac{1}{\sqrt{2 \pi}}, \quad \varphi_{2n - 1}(x) = \frac{\cos nx}{\sqrt{\pi}}, \quad \varphi_{2n}(x) = \frac{\sin nx}{\sqrt{\pi}}, \qquad (n = 1, 2, \ldots). \tag{3}\label{3} $$ Clearly, $S$ is orthonormal on any interval of length $2\pi$, and hence on $I$ when $I = [0, 2\pi]$. In subsection 11.4 of the book, for this particular $S$ and $I$, the series \eqref{1} is denoted as the Fourier series generated by $f$. Then, it is stated there that we can write \eqref{1} in the following form:

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx),$$ the coefficients being given by the following formulas: $$ a_n = \frac{1}{\pi}\int_0^{2\pi} f(t) \cos nt\; dt, \qquad b_n = \frac{1}{\pi}\int_0^{2\pi} f(t) \sin nt\; dt . $$

I don't understand how it works out. Taking the $\varphi$-functions as given in \eqref{3} and $c_n$ as in \eqref{2}, the expression in \eqref{1} is not the same as $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$. Further, in the expressions of $a_n$ and $b_n$, the $\varphi$-functions do not seem to form an orthonormal set. Is this exposition then incorrect in Apostol's book?

Also, can the expression $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$ be expressed as some form of orthonormal decomposition like in \eqref{1}?

Answer 1

The two expressions are indeed the same. Recall that in a Hilbert space, the orthogonal projection of a vector $f$ along the subspace (line) spanned by another vector $\varphi$ is given by

$$P_\varphi(f)=\frac{\langle f,\varphi\rangle}{\langle \varphi,\varphi\rangle}\varphi,$$

which, in the case $||\varphi||=1$, reduces to $P_\varphi(f)=\langle f,\varphi \rangle\varphi$. If you have a complete system of orthonormal [resp. orthogonal] vectors $\{\varphi_n\}_{n\in \mathbb{N}}$ [resp] $\{\phi_n\}_{n\in \mathbb{N}}$ you can write

$$f=\sum_{n=0}^\infty \langle f,\varphi_n\rangle\varphi_n \quad [\text{resp.}] \quad f=\sum_{n=0}^\infty \frac{\langle f,\phi_n\rangle}{\langle \phi_n,\phi_n\rangle}\phi_n.$$

Your system of functions (3) is a complete system of orthonormal vectors, so you can write $f$ as the first sum above, if you multiply each of the $\varphi_j$ by their denominator (getting $\phi_j$), you can write $f$ as the second sum above. Finally observe that for even indexes we get sines, and for odd ones we get cosines, we might aswell re-indexing the sum above and write

$$f(x)=a_0/2+\sum_{n=1}^\infty a_n \cos(nx)+b_n\sin(nx).$$

Answer 2

I'll just detail the calculation to show in the end you get the usual form of the Fourier Series.

So you have the orthonormal basis S={ $\frac{1}{\sqrt{2\pi}}$,$\frac{cos(x)}{\sqrt\pi}$,$\frac{sin(x)}{\sqrt\pi}$, ... ,$\frac{cos(nx)}{\sqrt\pi}$,$\frac{sin(nx)}{\sqrt\pi}$, .... }.

If you think back to Linear Algebra, the orthogonal projection of a vector $v$ in an orthonormal basis {$a_1, ... ,a_n$} was given by $<a_1,v>a_1 + ... + <a_n,v>a_n$. If we follow this reasoning and switch the usual scalar product <,> to the product given by $<f(x),g(x)>=\int_0^{2\pi} f(x)g(x)dx $, we should have the orthogonal projection of f in S is:

<f(x),$\frac{1}{\sqrt{2\pi}}$> $\frac{1}{\sqrt{2\pi}}$+ <f,$\frac{cos(x)}{\sqrt\pi}$> $\frac{cos(x)}{\sqrt\pi}$ + <f,$\frac{sin(x)}{\sqrt\pi}$> $\frac{sin(x)}{\sqrt\pi}$ + ... + <f,$\frac{cos(nx)}{\sqrt\pi}$> $\frac{cos(nx)}{\sqrt\pi}$ + <f,$\frac{sin(nx)}{\sqrt\pi}$> $\frac{sin(nx)}{\sqrt\pi}$ + ...... =

=$(\int_0^{2\pi} \frac{f(t)}{\sqrt{2\pi}}dt)$$\cdot$$\frac{1}{\sqrt{2\pi}}$ + $\sum_{n=1}^{\infty} <f,\frac{cos(nx)}{\sqrt\pi}> \frac{cos(nx)}{\sqrt\pi} +$ $ <f,\frac{sin(nx)}{\sqrt\pi}> \frac{sin(nx)}{\sqrt\pi}=$

$= \frac{\int_0^{2\pi} {f(t)}dt}{2\pi} $+ $\sum_{n=1}^{\infty} (\int_0^{2\pi}f(t)\frac{cos(nt)} {\sqrt{\pi}} dt )\frac{cos(nx)}{\sqrt{\pi}} + (\int_0^{2\pi}f(t)\frac{sin(nt)} {\sqrt{\pi}} dt )\frac{sin(nx)}{\sqrt{\pi}}=$

$=\frac{\int_0^{2\pi}{f(t)}dt}{2\pi}+\sum_{n=1}^{\infty}\frac{1}{\pi}[\int_0^{2\pi}f(t){cos(nt)}dt]\cdot{cos(nx)} + \frac{1}{\pi}[\int_0^{2\pi}f(t){sin(nt)}dt]\cdot{sin(nx)}.$

If you call $a_n$ the terms $\frac{1}{\pi}\int_0^{2\pi}f(t){cos(nt)}dt$, for $n\geq0$, and $b_n$ the terms $\frac{1}{\pi}\int_0^{2\pi}f(t){sin(nt)}dt$, for $n\geq1$, then have the orthogonal projection of f in S is:

$ \frac{a_0}{2} $+ $\sum_{n=1}^{\infty} a_n {cos(nx)} + b_n{sin(nx)}$ and this is exactly the Fourier Series of $f$ in its usual form!!!