# In which sense is the Fourier transform the continuous analogue of the discrete Fourier transform? A mathematical point of view.

I try to understand in which sense the Fourier transform is the continuous analogue of the discrete Fourier transform. I know, there are many books and many questions on this site concerning this topic, but almost all of them are written from an engineer's point of view. I'm a mathematician, and as such, I'm not familiar with the application of Fourier analysis in engineering. So, I would like to handle the following objects as mathematical objects without any physical interpretation.

Let me first introduce the relevant objects: Let $n\in\mathbb N$ and $\omega\in\mathbb C$ be a primitive $n$-th root of unity.

It's easy to verify that the columns of $$\tilde V_\omega:=\frac 1{\sqrt n}\underbrace{\begin{pmatrix} \omega^{0\cdot 0}&\cdots&\omega^{0\cdot (n-1)}\\\vdots&\ddots&\vdots\\\omega^{(n-1)\cdot 0}&\cdots&\omega^{(n-1)\cdot (n-1)} \end{pmatrix}}_{=:V_\omega}$$ are an orthonormal basis of $\mathbb C^n$.

Now, $$\operatorname{DFT}_\omega:\mathbb C^n\to\mathbb C^n\;,\;\;\;z\mapsto\left(f_z\left(\omega^0\right),\ldots,f_z\left(\omega^{n-1}\right)\right)^T$$ with $$f_z(w):=\sum_{k=0}^{n-1}z_kw^k\;\;\;\text{for }z=\left(z_0,\ldots,z_{n-1}\right)^T\text{ and }w\in\mathbb C$$ is called discrete Fourier transform.

Again, it's easy to verify that $V_\omega$ is the transformation matrix of $\operatorname{DFT}_\omega$, i.e. $$\operatorname{DFT}_\omega z=V_\omega z\color{blue}{\stackrel{\text{def}}=\sqrt n\tilde V_\omega z}\;\;\;\text{for all }z\in\mathbb C^n\;,$$ and is invertible with $$V_\omega^{-1}=\frac 1nV_{\omega^{-1}}$$ and hence $\operatorname{DFT}_\omega$ is an isomorphism. We can easily prove that $$\langle\operatorname{DFT}_\omega z,\operatorname{DFT}_\omega w\rangle=n\langle z,w\rangle\;\;\;\text{for all }z,w\in\mathbb C^n$$ (using that $\tilde V_\omega$ is unitary).

Now, let $a,b\in\mathbb R$ with $a<b$ and $D:=(a,b)$.

It's a well-known fact that $$\phi_k(x):=\frac 1{\sqrt{b-a}}e^{2\pi{\rm i}\frac k{b-a}x}\;\;\;\text{for }k\in\mathbb Z\text{ and }x\in D$$ is an orthonormal basis of $L^2(D;\mathbb C)$ and hence $$u=\sum_{k\in\mathbb Z}\underbrace{\langle u,\phi_k\rangle_{L^2(D;\mathbb C)}}_{=:u_k}\phi_k\;\;\;\text{for all }u\in L^2(D;\mathbb C)\;.$$

Now, $$\sqrt{b-a}\sum_{k\in\mathbb Z}\tilde u_k\phi_k$$ with $$\tilde u_k:=\frac 1{\sqrt{b-a}}u_k\;\;\;\text{for }k\in\mathbb Z$$ is called Fourier series of $u\in L^2(D;\mathbb C)$.

Let $$\mathcal F:L^2(D;\mathbb C)\to L^2(\mathbb Z;\mathbb C)\;,\;\;\;u\mapsto(\tilde u_k)_{k\in\mathbb Z}\;.$$ We can show that $$\langle\mathcal Fu,\mathcal Fv\rangle_{L^2(D;\mathbb C)}=(b-a)\langle(\tilde u_k)_{k\in\mathbb Z},(\tilde v_k)_{k\in\mathbb Z}\rangle_{L^2(\mathbb Z;\mathbb C)}\;\;\;\text{for all }u,v\in L^2(D;\mathbb C)\;.$$

Is $\mathcal F$ what we call the Fourier transform? How are $\operatorname{DFT}_\omega$ and $\mathcal F$ related?

I would like to conclude that the discrete Fourier transform is an isomorphic isometry $(\mathbb C^n,\langle\;\cdot\;,\;\cdot\;\rangle)\to(\mathbb C^n,n\langle\;\cdot\;,\;\cdot\;\rangle)$ and that an analogue statement holds true for the Fourier transform on $(L^2(D;\mathbb C),\langle\;\cdot\;,\;\cdot\;\rangle_{L^2(D;\mathbb C)})$.

Please keep this in mind, when you formulate an answer. I suppose $\mathcal F$ is not the mapping I'm looking for, cause its nature seems to be different $\operatorname{DFT}_\omega$. I guess the mapping should be $(L^2(D;\mathbb C)\to (L^2(D;\mathbb C)$ where the image is equipped with an inner product of the form $c\langle\;\cdot\;,\;\cdot\;\rangle_{L^2(D;\mathbb C)}$ for some $c\in\mathbb C$.

• It might be worth to note that $$\omega:=e^{-{\rm i}\frac{2\pi}n}$$ is somehow the unique primitive $n$-th root of unity. It's worth to note, cause the form of $\omega$ is similar to the form of the $\phi_k$. – 0xbadf00d Feb 28 '16 at 11:46
• google.com/url?sa=t&source=web&rct=j&url=http://… Here you go. A signal that is discrete in one domain is continuous in the other. – D J Sims Feb 28 '16 at 11:48
• start by showing how the Fourier series is the limiting case when $N \to \infty$ of the discrete Fourier transform on $\mathbb{C}^N$ – reuns Feb 28 '16 at 12:15
• @Mustang Your link doesn't contain anything new (something which is not part of the statements in the question), does it? I suppose you mean the following: The "continuous signal" $u\in L^2(D;\mathbb C)$ is mapped to the "discrete signal" $(\tilde u_k)_{k\in\mathbb Z}$ by $\mathcal F$. Is that what you've meant? – 0xbadf00d Feb 28 '16 at 18:06
• @user1952009 I've no idea how to start. – 0xbadf00d Feb 28 '16 at 19:10

The operator $\frac{1}{i}\frac{d}{dt}$ considered on the domain of absolutely continuous functions $f \in L^2[-\pi,\pi]$ for which $f'\in L^2$ and $f(0)=f(2\pi)$ is selfadjoint, and the map $U : L^2[-\pi,\pi]\rightarrow\ell^2(\mathbb{Z})$ defined by $$Uf = \{ \hat{f}(n) \}_{n=-\infty}^{\infty}$$ is a unitary map such that $$U\frac{1}{i}\frac{d}{dx}f = \{n\hat{f}(n)\}=M\{\hat{f}(n)\},$$ where $M$ is the multiplication operator on $\ell^2(\mathbb{Z})$ $$M \{ a_n \} = \{ n a_n \}.$$ So the discrete Fourier transform is a unitary map that turns differentiation into multiplication.
The operator $\frac{1}{i}\frac{d}{dt}$ considered on the domain consisting of all absolutely continuous $f\in L^2(\mathbb{R})$ for which $f'\in L^2(\mathbb{R})$ is selfadjoint, and the Fourier transform $\mathcal{F} : L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ is a unitary map such that $$\mathcal{F}\frac{1}{i}\frac{d}{dx}f=s(\mathcal{F}f)(s)=M\mathcal{F}f$$ where $M$ is the multiplication function on $L^2(\mathbb{R})$ given by $$(Mf)(s) = sf(s).$$ So the Fourier continuous Fourier transform is a unitary map that turns differentiation into multiplication.
Matrix diagonalization works the same way on a finite-dimensional Hilbert space $H$. If $A$ is selfadjoint with an orthonormal basis of eigenvectors $\{ e_n \}$ with corresponding eigenvalues $\{ \lambda_n \}$, then the map $$U : x \in H \mapsto \{ (x,e_n) \}_{n=1}^{N}\in\mathbb{C}^{N}$$ is a unitary map that turns $A$ into multiplication on $\ell^2(\{1,2,\cdots,N\})=\mathbb{C}^{N}$: $$UAx = \{ \lambda_n (x,e_n)\} = M\{ (x,e_n) \}$$ where $M$ is the multiplication operator on $\ell^2(\{1,2,3,\cdots,N\})$ given by $$M\{ a_n \}_{n=1}^{N} = \{ \lambda_n a_n \}_{n=1}^{N}.$$