Does band-limited imply continuous?

Question

I was hoping someone could point me to an article or text which explores the connection between the continuity of a signal in the time domain and it being band-limited in frequency domain.

(Update)

• Band limited is defined as: There exists a finite frequency, $\omega_{0}$, such that $\mathcal{F}\left( \omega \right) = 0$ $\forall\omega$ where $\left| \omega \right| > \omega_{0}$.
• There is no assumptions regarding support for $\mathcal{F}(\omega)$ within the frequency domain or assumptions regarding membership of $f(t)$ in spaces such as $L^p(\mathbb{R})$, $S(\mathbb{R})$, or $S^{'}(\mathbb{R})$
• For the context of this question: signal is any function or any distribution which admits a Fourier transform.

(end update)

For example there are two papers entitled: Band-Limited Processes in Discrete and Continuous Time:

http://www.le.ac.uk/ec/research/RePEc/lec/leecon/dp11-11.pdf (24 pages), and http://www.le.ac.uk/users/dsgp1/SIGNALS/LODZ.pdf (54 pages)

Weaving throughout the first is the assumption that signals which are band-limited in the frequency domain are continuous in the time domain and the longer paper is even bolder by stating on page 2 that:

A band-limited function is analytic. It possesses derivatives of all orders.

While I agree that sharpness" of anyedges" within a band-limited function must be related to the highest frequency present in the signal, I can't find any discussion the in the signal analysis literature which describes when and if

A function/signal band-limited in the frequency domain is continuous in time domain.

I arrived at this question via the examination of almost periodic functions (a.p.) which are band-limited to the rationals in the interval $[-1,1]$ and (more generally) almost periodic functions which have a fintie basis set where each base is band limited; i.e. where frequencies with a particular base are connected by rationals found within a finite interval. Example of such functions would be:

$f_{1a}(t) = \sum\limits_{n = 1}^{\infty} \frac{\left( -1 \right)^n}{n} \cos\left( 2 \pi \frac{t}{n} \right) $

$ f_{1b}(t) = \sum\limits_{n = 1}^{\infty} \frac{1}{n} \cos\left( 2 \pi \frac{t}{n} \right) $

$ f_{2a}(t) = \sum\limits_{q = 2}^{\infty} \frac{\left( -1 \right)^q}{2 \phi\left( q \right) \; q} \sum\limits_{\substack{k = 1 \\ \left( k, q\right)=1}}^{q} \cos\left( 2 \pi \frac{k}{q} t \right) $

$ f_{2b}(t) = \sum\limits_{q = 2}^{\infty} \frac{1}{2 \phi\left( q \right) \; q} \sum\limits_{\substack{k = 1 \\ \left( k, q\right)=1}}^{q} \cos\left( 2 \pi \frac{k}{q} t \right) $

$ f_{3}(t) = \sum\limits_{q = 2}^{\infty} \sum\limits_{\substack{k = 1 \\ \left( k, q\right)=1}}^{q} a\left(k, q\right) \cos\left( 2 \pi \frac{k}{q} t \right) $

$ f_{4}(t) = \sum\limits_{n=1}^{N} \sum\limits_{\substack{r_{i} \in \mathbb{Q} \\ \left| r_{i} \right| \le B_{n}}} a\left(n, r_{i} \right) e^{2 \pi i \left( \beta_{n} r_{i} \right) t } $

where:

$\left( k, q\right)$ is the least common divisor of $k$ and $q$

$\phi\left( q \right)$ is the Euler Totient function.

$N$ is a finite integer $1 \le N$

$r_n$ is the set of all rational numbers within the closed interval: $\left| r_{i} \right| \le B_{n}$

The set $\left\{ \beta_{1}, \beta_{2}, \beta_{2}, \cdots, \beta_{N-2}, \beta_{N-1}, \beta_{N} \right\}$ is the finite basis of the almost periodic function function defined by $f_4$.

The definition of the basis set of an a.p. function is found on page 34 of Almost Periodic Functions by A. S. Besicovitch

$B_{n}$ is a positive, finite limit, $0 < B_{n} < \infty$, on the frequencies, $\lambda_{i,n} = r_{i} \beta_{n}$, of the almost periodic function defined by $f_4(t)$

All of the example signals have point-wise convergence (except 1b and 2b at $t=0$)

The Fourier transform of each of the signals above is a set of Dirac-$\delta$ distributions which are band-limited to the interval, $[-B,B]$; where $B = 1$ or $B = \text{max}\left( B_{n}\right)$. Such a Fourier transform form a band-limited, ``bed--of--nails".

The questions before me are:

1. Are signals with a typical, band-limited Fourier transform (i.e. not a ``bed--of--nails") continuous (analytic)? Under what conditions?
2. Are signals with a band-limited, ``bed--of--nails" Fourier transform continuous (analytic)?
3. If some such signals are continuous (analytic) and others are not, then under what conditions?
4. What are the necessary (sufficient) conditions for the signal to be continuous (analytic)?

I was hoping someone could pointing me to an article or text that explores the connection (if any) between continuity and band-limited.

Is there theorem similar to the Paley-Wiener theorem, but for functions which do not belong to $L^2$?

Is this question a variation of the Gabor limit? since a ''Bed of Nails" Fourier transform has no local support, then the signal must have local on the whole of the real line?

Answer 1

Not at all. Since band-limited means "compactly supported Fourier transform." All you have to do is find a function which has compact support and is in $L^1$ and take it's inverse Fourier transform and modify it a bit. Such an example is

$$f(x)=e^{-x^2}\cdot 1_{[-1,1]}$$

Then this will be a continuous function, but you can modify this choice by doing

$$\mathcal{F}^{-1}(f)+1_{\Bbb Q}$$

which has the same Fourier transform, but is clearly discontinuous everywhere.

Answer 2

Perhaps a better statement, given Adam Hughes example, would be that if $f$ is band limited, then it's equal almost everywhere to some continuous function $g$. I believe that this is true -- it's certainly true for functions on $S^1$, where the FT lies in $L^2(\mathbb Z)$, and "band limited" means "FT is zero outside of some finite interval", for then the function is almost everywhere equal to the IFT of the FT, and this IFT is just a finite sum of trigonmetric functions.

I'm pretty sure the corresponding statement for $L^2[0, 1]$ is also true, but I'm no confident that I could prove it. :(

Since the OP won't say what space the function $f$ lies in :), I'll give a kind of hedged answer:

If the multiplication-convolution theorem holds for your function, along with a few other assumptions, then band-limited does indeed imply continuous (in the sense of "equal to a continuous function a.e."). Here's why:

1. $\widehat{\mathrm{sinc}}$ is a box function $b$ (i.e., the indicator function for some interval). Let's suppose (by rescaling as needed) that the support of $\hat{f}$ is contained in that interval.

2. Then $\widehat{\mathrm{sinc}}* \hat{f} = \hat{f}$, so $\widehat{\widehat{\mathrm{sinc}}} \star \hat{\hat{f}} = \hat{\hat{f}}$.

3. Assuming that $\hat{\hat{f}} = f $ a.e., this means that $\mathrm{sinc} \star {f} = f$ (ae).

4. Since $\mathrm{sinc} \star {f}$ is continuous (because $\mathrm{sinc}$ is), we have that $f$ is continuous ae.

I'm making lots of assumptions here -- that $f$ is nice enough that convolution with sinc raises continuity from $C^{-1}$ to $C^0$, that the IFT of the FT of $f$ is still $f$ under whatever FT you're using, that the convolution-multiplication theorem holds, etc.

But you can also use this proof to show that $f$ is differentiable (by convolving with sinc again), twice differentiable, etc., so perhaps this is what those authors were thinking of when they made those claims. (I haven't read the papers/books to know whether my various assumptions are valid in their respective contexts.)

Answer 3

I believe an answer could be this:

If a one-variable signal is bandlimited, it means it have a bounded maximum rate of change (as is explained here ), so $$ \sup_{t} \left| \frac{df(t)}{dt} \right| < \infty$$

Now, this make a restriction to $f(t)$, since if it has a discontinuity at a point $t_d$ it will happen that $f(t_d^-) \neq f(t_d^+)$ so at this point its derivative will diverge $f'(t_d) \to \infty$: you can describe this discontinuous function explicitly as another function $x(t)$ such as $$x(t) \cong f(t)\cdot \theta(t_d^--t)+f(t)\cdot\theta(t-t_d^+)$$ with $\theta(t)$ the standard unitary step function, so by applying the differentiation on both sides will leads to: $$\begin{array}{r c l} \frac{dx(t)}{dt} & \cong & \frac{df(t)}{dt}\cdot \theta(t_d^--t)+f(t_d^-)\cdot \delta(t_d^--t)+\frac{df(t)}{dt}\cdot\theta(t-t_d^+)+f(t_d^+)\cdot\delta(t-t_d^+) \\ & = & \frac{df(t)}{dt}\cdot\left( \theta(t_d^--t)+\theta(t-t_d^+)\right)+\delta(t-t_d)\cdot\left(f(t_d^+)-f(t_d^-)\right) \end{array}$$ where $\delta(t)$ is the Dirac's Delta function/distribution, so at the point $t_d$ the function $|dx(t_d)/dt| \to \infty$ since $|\delta(0)| \cong \infty$.

With this, by contradiction, since a bandlimited signal can´t have an unbounded derivative, the function $f(t)$ can´t have a discontinuity of the kind $f(t_d^-) \neq f(t_d^+)$, so since it must happen that $\forall\, t_d\,,\, f(t_d^-) \equiv f(t_d^+)$ then the function must be continuous in this sense... but caution it still could have sharp tips like the absolute value function $|t|$ so $f'(t)$ could be discontinuous, but the step size $|f'(t_d^+)-f'(t_d^-)| < \infty$ must be finite so the maximum rate of change stills be finite... with this, meaning that it $f(t)$ could be "discontinuous" if other definitions of continuity are chosen (I don´t know if the condition $\forall\, t_d\,,\, f(t_d^-) = f(t_d^+)$ implies that $f(t)$ is going to be "uniformly continuous", or "absolutely continuous", etc.).