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Are there any time-limited and continuous one-variable functions $f(t)$ with bounded derivative $||f'(t)||_\infty < \infty$ (not meaning here they are also necessarily differentiable), such as their Fourier transform $F(w)$ makes diverge the following integral $\int\limits_{-\infty}^\infty|iwF(w)+f(t_F)\,e^{-iwt_F}-f(t_0)\,e^{-iwt_0}|\,dw \to \infty$?? Or these kind of functions are an empty set (for each of the following scenarios)?

The different terms from the questions of the tittle are just for avoiding the effects of the discontinuity on the edges of the compact-support $\partial t = \{t_0,\,t_F\}$ (starting and ending times), since they introduce Dirac's Delta functions $\delta(t)$ in the derivative $f'(t)$ ("artificially" in my opinion, since to model time limited phenomena I am interested only in what is happening "within" the compact support).

If you feel uncomfortable with them, just assume also that the functions $f(t)$ begins and finishes at zero $f(t_0)=f(t_F)=0$. From the following, I will use both definitions as equivalent since the problem is avoidable (I explained one way to overcome it here). Please keep it in mind, or it will make harder to find counterexamples since this edges-discontinuities will make the standard $\int_{-\infty}^\infty|wF(w)|dw$ always diverge, since the derivative will be unbounded because of these delta functions, as I will explain now.

I am trying to understand the figure of the integral $\int_{-\infty}^\infty|wF(w)|dw$ which is an upper bound for the maximum rate of change of the function $f(t)$: $$ \sup\limits_t \left| f'(t)\right| \leq \int_{-\infty}^\infty|wF(w)|dw$$ It has an individual name? (as the Dirichlet Energy, as example), this for being able to look for its properties by myself. Any references are welcome.

Directly from the inequality I know that if the derivative is unbounded $||f'(t)||_\infty \to \infty \Rightarrow \int_{-\infty}^\infty|wF(w)|dw \to \infty$ will always diverge, and conversely, if "this" integral is bounded $\int_{-\infty}^\infty|wF(w)|dw < \infty \Rightarrow ||f'(t)||_\infty < \infty$ the maximum rate of change will be bounded (even when time-limited functions has unlimited bandwidth on the frequencies), but I want to know if there exists any cases of functions that lie in-between these two scenarios (I have already looked unsuccessfully for counterexamples by myself).

I am specially interested in these five scenarios (from less to more restrictive - I believe):

  1. General time-limited and continuous one-variable functions $f(t)$, as is already asked
  2. Time-limited continuous one-variable functions which are also absolutely integrable $\int\limits_{t_0}^{t_F}|f(t)|\,dt < \infty$ and energy finite $\int\limits_{t_0}^{t_F}|f(t)|^2 dt < \infty$
  3. Functions that fulfill (1) and (2) and are also have their absolute value of its Fourier Transform bounded $\int\limits_{-\infty}^{\infty} |F(w)| dw < \infty$
  4. Functions that fulfill (1) to (3) and also have finite Dirichlet Energy $\int\limits_{t_0}^{t_F} |f'(t)|^2 dt < \infty$
  5. Functions that fulfill (1) to (4) and there also of bounded total variation $V_{[t_0,\,t_F]}(f(t)) < \infty$

I want to know if any of these intermediate conditions stages makes the integral $\int_{-\infty}^\infty|wF(w)|dw$ becomes bounded, or if are totally unrelated.

Please notice that neither of these conditions are requiring to $f(t)$ to be differentiable. But I am not interested in "bad-behaved" things like nowhere-differentiable functions as Brownian motions, or fractals, or Cantor or Weierstrass functions, and things like that (at least not this time)

Any counterexample will be welcome either. Beforehand, thanks you very much.

Joako
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    Sure, consider the very simple example of the triangle wave, or the unit rectangle convolved with itself $$f(t) = \begin{cases} 1-|t| & |t| < 1 \ 0 & \text{else}\end{cases}$$ – Ninad Munshi Nov 30 '21 at 14:45
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    Consider $f(t)=e^{-|t|}$ on the interval $[t_0,t_F]=[-1,1]$. This is lipschitz continuous everywhere, in particular $|f’|_\infty=1$, but the derivative has a jump in the middle of the interval. The fourier transform (up to constants) is $F(w)=1/(1+w^2)$ which doesn’t have the property $\int wF(w)dw<\infty$. Is this a counterexample you were looking for? – Lorenzo Pompili Nov 30 '21 at 14:47
  • In any case, since you are working on a bounded interval, why don’t you use the Fourier series of $f$ instead of the fourier transform? – Lorenzo Pompili Nov 30 '21 at 14:51
  • @LorenzoPompili I don´t have much knowledge about Fourier Series, Does them required also that $f(t)$ to be periodic?? (If that so, I am thinking in the more general case) – Joako Nov 30 '21 at 15:48
  • @Joako you are correct Fourier series strictly requires functions to be periodic on $\Bbb{R}$ - for time limited functions this means artificially repeating their window forever. But I don't see the point of doing so. – Ninad Munshi Nov 30 '21 at 16:56
  • @NinadMunshi , the point is precisely to obtain the same bound but with the Fourier coefficients instead of the Fourier transform. More precisely, if $f$ is a function on a compact interval, you have the bound $|f’|_\infty\leq\sum |j||\hat f_j|$, where the $\hat f_j$ are the Fourier coefficients of $f$ (always. You don’t need periodicity). Since (as you say) the natural setting of the Fourier series is that of periodic functions, to make a somehow sharp condition you need to have that $f$ can be extended continuously by periodicity on the real line (which is the case, if $f(t_0)=f(t_F)$). – Lorenzo Pompili Nov 30 '21 at 19:24
  • @Joako , you “need” periodicity for the Fourier series in the same way as you “need” a function that is continuous on the whole real line for the Fourier transform: you don’t actually need it :) It’s just that, if you want to have nice Fourier coefficients you have to ensure that $f$ CAN be extended (e.g.) continuously by periodicity, otherwise the fourier coefficients grow too fast; this is in perfect analogy with the reason why you assumed $f(t_0)=f(t_F)=0$: you wanted $f$ to be continuous on the whole real line, otherwise the fourier transform grows too fast (due to the Dirac deltas). – Lorenzo Pompili Nov 30 '21 at 19:35
  • To motivate my point of view even better: extending $f$ as a periodic function is not less natural than taking $f$ on an interval and extend it by $0$ on the rest of the real line. So, in this setting, the Fourier series is not less natural than the Fourier transform. It’s actually much more natural for many reasons. So, @Joako , if you are trying to understand something deeper with this question, I would suggest you to study something about Fourier series as it could be that they work more naturally in the setting you are studying – Lorenzo Pompili Nov 30 '21 at 19:50
  • @LorenzoPompili I did a slight review of the Fourier Series here, please tell me if what you thinking about. I tried first to read about the Fourier Series but there are a full theory, quite abstract sometimes (at least for my limited knowledge), so I get lost. I am really interested about understanding the decay of the coefficients of $|w F(w)|$, so if you can share a reference that review it directly could be awesome. – Joako Nov 30 '21 at 20:54
  • Any reference that talks about the Fourier transform is ok. Just know that $wF(w)$ is essentially the fourier transform of $f’$, up to multiplicative constants. So you don’t have to worry about $w$, and you should start thinking about the Fourier thansform of $f’$ in terms of $f’$ itself. In that way, I think you can manage to find answers more easily. After thinking about what I just wrote, check the link I posted on my answer. – Lorenzo Pompili Nov 30 '21 at 21:06

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By standard properties of the Fourier transform (check in this page the subsections 5.4 and 5.7), the condition $$\int_{-\infty}^{+\infty}| wF(w)|dw<\infty\qquad\text{(1)}$$ implies that $f’$ is continuous and bounded (and of course it has compact support due to the fact that $f$ has compact support). This means that you have counterexamples just picking a Lipschitz continuous function $f$ with discontinuous derivative, e.g., $$f(t)=e^{-|t|}-e\;,\quad t\in [-1,1].$$ What I wrote holds also for the extreme points of the interval, this means that every function that satisfies (1) can’t have non-zero derivatives in the boundary of the interval. For instance, $$f(t)=e^{-t^2}-e\;,\quad t\in [-1,1]$$ is still a counterexample, even though it is smooth inside the interval. (Off topic, this problem can be eliminated using Fourier series. First extend the function oddly around one the two extremes of the interval, then consider the fourier coefficients on this doubled interval… In this case, you can at least hope that the Fourier coefficients with respect to the doubled interval are such that $\sum |j||\hat f_j|<\infty$).

An interesting question would be: what if I assume that $f$ is continuous, compactly supported on $[t_0,t_F]$ and the derivative is also continuous at every point (including the extremes of the interval, in which the derivative has to be $0$)? Here counterexamples are harder to find. I think there still exist counterexamples, but at the moment I can’t exhibit one. I’ll try to find something here on SE.


Edit: The first answer of this post brings a nice example of a continuous compactly supported function whose fourier transform has infinite integral. I think this example could be slightly modified to give a counterexample to your question with a function with continuous derivative in every point, including the extremes.

Lorenzo Pompili
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  • Thanks for the detailed answer. I am not familiarized with functions spaces, so let me see if I follow you right: you are saying that any function with a discontinuity on $f'(t)$ will make diverge the integral $\int |w F(w)| dw$?... I am not really sure about it: as example $f(t) = \sin(\pi | t|),, |t| \leq 1$ has discontinuous but bounded $|f'(t)| \leq \pi$, and $F(w) = \pi e^{-iw}(1+e^{iw})^2/(\pi^2-w^2)$ so, at least following Wolfram-Alpha, $\int |w F(w)| dw < \infty$ (on the next comment the link to Wolfram-Alpha) – Joako Nov 30 '21 at 21:06
  • Not sure if $\sin(\pi |t|)$ is Lipschitz or not, and I have already found mistakes in Wolfram-Alpha, so I am not 100% convinced if is actually convergent. – Joako Nov 30 '21 at 21:08
  • That function is Lipschitz, no question about it. The problem is that the function $F$ that you wrote is not such that $\int wF(w)<\infty$. Wolfram in fact gives back quite a large number – Lorenzo Pompili Nov 30 '21 at 21:17
  • In any case, you understood correctly. Check the first link in my question if you need a proof of what I said. – Lorenzo Pompili Nov 30 '21 at 21:24
  • The mentioned properties of the first link, if I am right, are saying that $\sup_w |wF(w)| \leq \int |f'(t)| dt$, but I don´t understand how this means that $\int |wF(w)| dw \to \infty$ if $f'(t)$ discontinuous, as example, if it has only one "jump" but with finite size, $\int |f'(t)| dt$ could be bounded as the $\sup |w F(w)|$, but still $\int |wF(w)| dw \to \infty$ as in the example of @Ninad Munshi of $f(t) = (1-|t|),,|t|\leq 1$... so How I connect this to the integral convergence? – Joako Nov 30 '21 at 21:35
  • PS: about the edges of the compact-support issue, it can be easily "skipped" by using what I am using here (the $\mathring{\mathbb{F}}{\cdot}$ transform), so is not a problem I think. But understanding that if $f'(t)$ has finite size discontinuities will make diverge $\int |wF(w)| dw$ is not obvious at all at least form me. So please try to follow the insight in this idea. – Joako Nov 30 '21 at 21:43
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    @Joako , the Fourier transform and the inverse fourier transform have essentially the same properties. In particular, $\mathscr F^{-1}=K\mathscr FK$, where $\mathscr F$ is the Fourier transform operator and $K$ is the complex conjugation operator (up to constants, depending on the convention you are using for the Fourier transform). So, for instance, one has $|F|\infty\leq \int|f|$, but it also holds $|f|\infty\leq \int|F|$ (again, possibly up to a multiplicative constant). This should be enough for you to conclude. – Lorenzo Pompili Nov 30 '21 at 22:11
  • Sorry, at first I believe I was understanding but now I realize I am not: I believe it is saying that $||f'(t)||\infty \to \infty \Rightarrow ||wF(w)||_1 \to \infty$ and also if $||w F(w)||_1 < \infty \Rightarrow ||f'(t)||\infty <\infty$, but not in the other direction for which I am trying to understand, How is possible that a discontinuous $f'(t)$ but bounded $||f'(t)|| < \infty$ makes that $||wF(w)||_1 \to \infty$ as in the examples?? (please be explicit, my math background is highly limited/basic - I have been learning through making questions here). – Joako Nov 30 '21 at 23:34
  • Your line of thought is the following? (1) if $||f'(t)||_1 < \infty \Rightarrow iwF(w)$ is uniformly continuous, then (2) if $||iwF(w)||_1 < \infty \Rightarrow f'(t)$ is uniformly continuous, so since having finite $||wF(w)||_1$ requires $f'(t)$ continuous, by contradiction a discontinuous $f'(t)$ will imply an unbounded $||wF(w)||$?? (following point 5.4 of Wikipedia) – Joako Nov 30 '21 at 23:56
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    Just number (2), that’s exactly what I was thinking when writing the answer. We don’t need number (1) (even if it’s true), we already know that the Fourier transform of a compactly supported function is analytic (by Paley-Wiener theorem), but I am not even using that in my argument. I am just using direction (2). – Lorenzo Pompili Dec 01 '21 at 07:42