There exists any real functions $f(t)$ on $[-1; 1]$ with $f(-1)\neq 0$ such $\lim_{w \to \infty} |w\cdot \int_{-1}^1 e^{-iwt}f(t)\,dt| = 0$?

Question

There exists any real functions $f(t)\in \mathbb{R}$ defined on $[-1; 1]$ (so $f(t)=0,\, |t|>1$), with at least $f(-1)\neq 0$ (hopefully also $f(1)\neq 0$), such it Fourier transform on $[-1; 1]$ is such that it will converges when multiplied by the first order polynomial, $\lim_{\omega \to \infty} |\,\omega\cdot \int_{-1}^{1} e^{-i\omega t}\,f(t)\,dt\,| = 0$ ???

Please, if possible, only continuous and differentiable functions "within the open interval" $(-1; 1)$, such the function and its Fourier transform could be simple and in "closed form". And with "functions" I mean that things like the "delta distribution" $\delta(t)$ is not an answer.

As example of what I am asking for, using $\Pi(t)$ as the standard unitary rectangular function [1]:

1. The function here fulfill the requirements (so, it spectrum don't converges to zero when multiplied by $\omega$): $$f_1(t) = \sin\left(\left(t-\frac{1}{4}\right)\frac{\pi}{2}\right)\cdot\cos\left(\left(t-\frac{1}{4}\right)\frac{\pi}{2}\right)\cdot\Pi\left(\frac{t}{2}\right),\, |t|\leq 1,\, (f_1(t)= 0,\,|t|> 1)$$ Were its Fourier transform in $[-1; 1]$ can be seen here, and it's weighted spectrum $|\,\omega\cdot \int_{-1}^{1} e^{-i\omega t}\,f_1(t)\,dt\,|$ did not converges to $0$ as it can be seen here.

2. But this function here did not fulfill the requirements (this one actually have an spectrum that converges to zero when multiplied by $\omega$): $$f_2(t) = \sin\left(\frac{t\pi}{2}\right)\cdot\cos\left(\frac{t\pi}{2}\right)\cdot\Pi\left(\frac{t}{2}\right),\, |t|\leq 1,\, (f_2(t)= 0,\,|t|> 1)$$ Were its Fourier transform in $[-1; 1]$ can be seen here, and it's weighted spectrum $|\,\omega\cdot \int_{-1}^{1} e^{-i\omega t}\,f_1(t)\,dt\,|$ does converges to $0$ as it can be seen here.

In this example, I just added a translation in time, so I will expect to have a multiplication by just a phase on the frequencies, which absolute value is $1$ so nothing must be changing in the spectrum - This is actually wrong really, since I don't move the window defined by $\Pi(t/2)$, but seeing this just as a translation within the domain $[-1;\,1]$, is mysterious for me.

When doing the following question here, I have tried many different examples (actually a few days doing trial-and-error), and I can't find any function whith $f(-1)\neq 0$ for which the mentioned weighted spectrum converges, when even the same "interior" functions were just time translation of the converging ones, so I left it as a conjecture.

I don't have any clue of why this is happening, so if you can explain it it will be awesome.

If they aren't exists in general, please explain why, or let a reference to the explanation.

Or conversely, if there are any proofs that the only way a real function $f(t)\in \mathbb{R}$ defined on $[-1; 1]$ will fulfill $\lim_{\omega \to \infty} |\,\omega\cdot \int_{-1}^{1} e^{-i\omega t}\,f(t)\,dt\,| = 0$ is that $f(-1)=f(1)=0$, please let me know about it.

Also, Is the proof extendable to any finite domain $[a;\,b]$?? Or for unbounded domain $[0;\,\infty)$?? Or $(-\infty;\,\infty)$??

Beforehand, thanks you very much.

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Added Later

I don't know why someone dislike this question, but I believe is interesting, since is related to something I am trying to prove in this other question here (just at the end of the bounded functions examples). I am trying to figure out if is possible to a compacted support function $f(t)$ on $[t_0;\,t_F]$ with $f(t_0) \neq 0$, to have $\int_{-\infty}^{\infty} |w\,F(i w)|\,dw<\infty$, with $F(iw)$ the Fourier transform of $f(t)$ on $[t_0;\,f_F]$.

Answer 1

Here is a partial answer.

In other words, you are asking: if $\tilde f(x) = f(x)\,\mathbb 1_{[-1,1]}(x)$, is it possible that its Fourier transform $g=\mathcal F(\tilde{f})(x) = \widehat{\tilde{f}}(x)$ verifies $|x\,g(x)| \to 0$ when $|x|\to\infty$ and $f(-1)≠ 0$.

Notice that by the Paley-Wiener theorem, $g$ is an analytic function.

Notice that your problem is quite critical since if there is such a function, as $\tilde{f}(-1)≠ 0$, we see that $\tilde f$ is discontinuous and this is the reason why it is more difficult than when $f(-1)=f(1)=0$. In particular, the multiplication by $x$ cannot be replaced by a higher exponent: for any $n>1$, $|x^n\,g(x)|$ is unbounded.

• Proof: By contradiction, if $|x^n\,g(x)| ≤ C$ for any $n> 1$, then since $g$ is analytic, and so bounded by some constant $C_g$ on $[-1,1]$, $$ \int_{\Bbb R} |g| ≤ \int_{-1}^1 |g| + \int_{-\infty}^1 |g| + \int_1^\infty |g| \\ ≤ 2 \,C_g + 2\,C \int_1^\infty x^{-n}\,\mathrm d x < \infty $$ so $g$ would be integrable, hence its Fourier transform $\tilde{f}$ would be continuous, which is false.

More generally you learn that $g$ is a smooth function that is too big at infinity to be integrable (but of course since $x\,g(x)$ converges to $0$ at infinity, it still verifies $|g(x)|≤C/(1+|x|)$)

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Edit: I will prove the following:

Theorem: If the restriction of $f$ to $[-1,1]$ is in $C^2$ and $f(-1)≠0$, then it is not a solution to your problem.

Remark: $C^2$ is just here to simplify, one can more generally take $C^{1,\alpha}$ with $\alpha>0$ instead of $C^2$.

1- A simple case where one can prove that there is no such function. If $f(1)=f(-1)≠ 0$ and the restriction of $f$ to $[-1,1]$ is in $C^2$ (so twice differentiable with continuous second derivative) with $f'(1)=f'(-1)=0$.

• Proof: In this case, the function $\varphi = f - f(1)∈ C^2([-1,1])$ and verifies $\varphi(-1) = \varphi(1) = \varphi'(1) = \varphi'(-1) = 0$. Hence $\varphi ∈ C^1(\Bbb R)$. Since $\varphi ∈C^2([-1,1])$, $\varphi''$ is uniformly bounded on $(-1,1)$, and since $\varphi=0$ out of $[-1,1]$, $\varphi''$ is bounded out of $[-1,1]$. So $\varphi''$ is bounded everywhere (except in $-1$ and $1$ where it might be not defined, but that is not a problem): we deduce that $\varphi ∈ L^\infty$ and $\varphi''∈ L^\infty$. In particular, $$ \|(1+|x|^2)\widehat{\varphi}(x)\|_{L^\infty} = \|\mathcal{F}(\varphi-\varphi'')\|_{L^\infty} ≤ \|\varphi-\varphi''\|_{L^1} $$ so $|x| |\widehat{\varphi}| \underset{x\to\infty}{\to} 0$. But now for $f$ we have $\tilde f = f(1)\,\mathbb 1_{[-1,1]} + \varphi$ and so $$ g = \mathcal{F}(\tilde f) = 2 \frac{\sin(x)}{x} + \widehat{\varphi} $$ Therefore, $|x\,g(x)| = |2\,\sin(x) + x\,\widehat{\varphi}|$. Hence if $|x\,g(x)|\to 0$, then $|2\,\sin(x)| ≤ |x\,g(x)| + |x| |\widehat{\varphi}| \underset{x\to\infty}{\to}$ which is false.

2- The more general case without restrictions on the boundary values.

• Proof: Define $P(x) = a + bx + cx^2 + dx^3$ the polynom such that $P(1)=f(1) = \alpha$, $P(-1)=f(-1) =\beta $, $P'(1)=f'(1)=\gamma$, $P'(-1)=f'(-1)=\delta$ (it is not difficult to find this polynom, I get $a = \frac{\alpha+\beta}{2}+\frac{\gamma+\delta}{4}$, $b=\frac{3(\beta-\alpha)}{4}-\frac{\gamma+\delta}{4}$, $c=\frac{\gamma-\delta}{4}$, $d=\frac{\gamma+\delta}{4}+\frac{\alpha-\beta}{4}$). Then notice that $$ \mathcal F(a\,\mathbb 1_{[-1,1]}) = 2\,a\,\frac{\sin(x)}{x} \\ \mathcal F(b\,x\,\mathbb 1_{[-1,1]}) = 2\,b\,(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}) = 2\,b\,\frac{\cos(x)}{x} + O(x^{-2}) \\ \mathcal F(c\,x^2\,\mathbb 1_{[-1,1]}) = -2\,c\,\frac{\sin(x)}{x} + O(x^{-2}) \\ \mathcal F(b\,x^3\,\mathbb 1_{[-1,1]}) = -2\,d\,\frac{\cos(x)}{x} + O(x^{-2}) $$ Then by the same reasoning as in (1-), $\varphi = f - P$ is a nice function such that $\widehat{\varphi} = O(x^{-2})$ and so $$ x\,g(x) = 2\,(a-c)\,\sin(x) + 2\,(b-d) \,\cos(x) + O(x^{-2}) $$ which converges to $0$ iff $a=c$ and $b=d$. And this implies in particular that $P(x) = (1+x)\,(1-x^2)$, so $f(1)=f(-1)=P(1)=P(-1) = 0$.

Answer 2

I have realized later the following: since any one-variable time-limited function could be described as: $$f(t) = x(t)\cdot (\theta(t-t_0)-\theta(t-t_F))$$ with $\theta(t)$ the unitary step function, given that $\theta'(t) = \delta(t)$ the Dirac's delta function, then: $$ \begin{array}{r c l} \frac{df(t)}{dt} & = & \frac{dx(t)}{dt}\cdot (\theta(t-t_0)-\theta(t-t_F))+x(t)\cdot (\delta(t-t_0)-\delta(t-t_F))\\ & = & \frac{dx(t)}{dt}\cdot (\theta(t-t_0)-\theta(t-t_F))+x(t_0)\delta(t-t_0)-x(t_F)\delta(t-t_F) \end{array}$$ because of the sifting property of the delta function $x(t)\delta(t-a)=x(a)\delta(t-a)$.

Now, given the following definition of the Fourier Transform for a time-limited function $F(w) = \int_{t_0}^{t_F} f(t)\,e^{-iwt}\,dt$ and its inverse as $f(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} F(w)\,e^{iwt}\,dw$ (time-limited functions have unlimited bandwidth), assuming that $f(t)$ is that so it haves a Fourier Transform so it fulfill the Rieman-Lebesgue Lema ($|F(w)|\to 0$ when $|w| \to \infty$), and the use of the triangular inequality, then: $$ \begin{array}{r c l} \max\limits_t \left|\frac{df(t)}{dt} \right| & = & \max\limits_t \left| \frac{1}{2\pi} \int_{-\infty}^{\infty} iw F(w)\,e^{iwt}\,dw\right|\\ & \leq & \max\limits_t \frac{1}{2\pi} \int_{-\infty}^{\infty} \left| i w F(w) \right|dw\\ & = & \frac{1}{2\pi} \int_{-\infty}^{\infty} \left|w F(w) \right|dw \end{array}$$

But the lower bound of this integral will diverge at least for any $x(t_0)\neq 0$ or $x(t_F)\neq 0$ with $x(t_0)\neq x(t_F)$ because $\delta(t-a) = \infty$ and: $$\max\limits_t \left|\frac{df(t)}{dt} \right| = \max\limits_t \left| \frac{dx(t)}{dt}\cdot (\theta(t-t_0)-\theta(t-t_F))+x(t_0)\delta(t-t_0)-x(t_F)\delta(t-t_F)\right|$$

Also noting that $\mathbb{F}\{df(t)/dt\} = iwF(w)$ could be also thought as: $$\begin{array}{r c l} \mathbb{F}\left\{ \frac{df(t)}{dt}\right\} & = & \mathbb{F}\left\{ \frac{dx(t)}{dt}\cdot (\theta(t-t_0)-\theta(t-t_F))+x(t_0)\delta(t-t_0)-x(t_F)\delta(t-t_F)\right\}\\ & = & \mathbb{F}\left\{ \frac{dx(t)}{dt}\cdot (\theta(t-t_0)-\theta(t-t_F))\right\}+\mathbb{F}\left\{x(t_0)\delta(t-t_0)\right\}-\mathbb{F}\left\{x(t_F)\delta(t-t_F)\right\}\\ & = & \int\limits_{t_0}^{t_F}\frac{dx(t)}{dt}\,e^{-iwt}\,dt+ x(t_0)\int\limits_{t_0}^{t_F}\delta(t-t_0)\,e^{-iwt}\,dt-\int\limits_{t_0}^{t_F}x(t_F)\delta(t-t_F)\,e^{-iwt}\,dt\\ & = & iwX(w)+x(t_0)e^{-iwt_0}-x(t_F)e^{-iwt_F} \end{array} $$ With this, the first inequality becomes: $$ \begin{array}{r c l} \max\limits_t \left|\frac{df(t)}{dt} \right| & = & \max\limits_t \left| \frac{1}{2\pi} \int_{-\infty}^{\infty} \left(iw X(w)+x(t_0)e^{-iwt_0}-x(t_F)e^{-iwt_F}\right)\,e^{iwt}\,dw\right|\\ & \leq & \frac{1}{2\pi} \int_{-\infty}^{\infty} \left|w X(w)+ix(t_F)e^{-iwt_F}- ix(t_0)e^{-iwt_0} \right|dw \end{array}$$

So now it can be seen that is equivalent: $$\lim\limits_{w \to \infty} |wF(w)| = \lim\limits_{w \to \infty} |w X(w)+ix(t_F)e^{-iwt_F}- ix(t_0)e^{-iwt_0}| $$

So to have the possibility of having $\lim\limits_{w \to \infty} |wF(w)| =0$ the trigonometric functions have, or been zero so $x(t_0)=x(t_F)=0$, or happen that: $$ w X(w)+ix(t_F)e^{-iwt_F}- ix(t_0)e^{-iwt_0} = 0$$ $$\Rightarrow X(w) = \frac{i}{w}\cdot x(t_0)e^{-iwt_0}- \frac{i}{w}\cdot x(t_F)e^{-iwt_F} $$ which is not a general case.

The only controversial point could be $x(t_0)=x(t_F)\neq 0$ since I could have something weird of the form $(M - M)\cdot\infty$ and $x\delta(x)=0$, but this case have been shown to always be divergent on the already accepted answer gave by user @LL 3.14