Mistake when using the Cauchy-Schwarz inequality with complex functions lead to working upper bounds (Maybe accidental discovery?)

Question

Mistake when using the Cauchy-Schwarz inequality with complex functions lead to working upper bounds (Maybe accidental discovery?):

This question has been edited since user @LL 3.14 show me where I was wrong. But since the upper bounds I found through a mistake are properly working, even better than known results, a more interesting question arises, and is Why it is working?

Introduction:

I am trying to find upper bounds for $max_t |df(t)/dt|$ for time-limited continuous and differentiable functions for which the following integral converges $\int_{-\infty}^\infty |wF(jw)|dw < \infty$.

To be precise, I am using $j = \sqrt{-1}$ as the imaginary unit, and $F(jw)$ is the Fourier Transform (the $F(jw)$ is just notation, it could be $F(w)$), and the definitions of the Fourier Transforms given in the book "Signals and Systems, 2nd Edition" (Alan V. Oppenheim, Alan S. Willsky, with S. Hamid) [1], $F(j w) = \int_{-\infty}^{\infty} f(t) e^{-j w t} dt$, so the function can be described as $f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(j w) e^{j w t} dw$.

As an example of a function which fulfill the requirement I am using: $$f(t) = \begin{cases} \cos^2\left(\frac{t\pi}{2}\right),\,\,|t|\leq 1 \\ 0,\,\,|t|> 1 \end{cases}$$ Which Fourier transform, following Wolfram-Alpha (W-A) here and here, is: $$ F(jw) = \frac{\pi^2\sin(w)}{\pi^2 w-w^3}$$ Note that since the function is time limited, the integrations limits of the Fourier transform are not $(-\infty, \infty)$ but instead its boundaries $\partial t = \{-1,\,1\}$.

These functions are going to be used as Benchmark in the following sections.

For the maximum rate of change of this $f(t)$, since is compacted-supported, problems will arise at the boundaries $\partial t = \{-1,\,1\}$ since is the same that taking the derivative of $f(t) = \cos^2\left(\frac{t\pi}{2}\right)\cdot\Pi\left( \frac{t}{2}\right)$ with $\Pi(t)$ the standard rectangular function, so the derivative will have terms of the form $\delta(t\pm 1)$: $df(t)/dt = \cos^2\left(\frac{t\pi}{2}\right)(\delta(t-1)+\delta(t+1))+\Pi(t)\cdot d/dt\left(\cos^2\left(\frac{t\pi}{2}\right) \right)$. For avoiding them, I will "conceptually assume" for now that this issue is nonexistent, here partially supported by the fact that for this specific $f(t)$ happens to be true that $\lim_{t \to \partial t} \cos^2\left(\frac{t\pi}{2}\right) = \lim_{t \to \partial t} d/dt(\cos^2\left(\frac{t\pi}{2}\right)) = 0$ and $t\cdot \delta(t)=0\,\forall\,t\in\mathbb{R}$.

With this in mind, as benchmark, I will have that I can obtain $\max_t |d/dt\left[\cos^2\left(\frac{t\pi}{2}\right)\Pi(t/2)\right] | \approx \max_t |d/dt\left[\cos^2\left(\frac{t\pi}{2}\right)\right] |\cdot \Pi(t/2) = \frac{\pi}{2} = 1.5708 $

Given that, following the inverse Fourier transform formula (and the Riemann-Lebesgue Lemma): $$\begin{array}{r c l} \max_t \left| \frac{df(t)}{dt} \right| = \max_t \left| \frac{1}{2\pi} \int\limits_{-\infty}^\infty jw\,F(jw)\,e^{jwt}\,dw \right| & \leq & \max_t \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|w\,F(jw)\right|dw \\ & \overset{\text{indep. of}\, t}{=} & \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|w\,F(jw)\right|dw \,\,\,\texttt{Eq. 1} \end{array}$$

For the benchmark $f(t)$, Eq. 1 on W-A gives as result $2.28547$, but I am not sure about its accuracy, but I know the integral of Eq. 1 converges by the following bounds obtained using Hölder's Inequality as I previously ask here, choosing some "clever" $g(w)\neq 0\,\forall w$: $$ \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|w\,F(jw)\right|dw = \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|w\,g(w)\frac{F(jw)}{g(w)}\right|dw \leq \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|w\,g(w)\right|dw \cdot \sup\limits_w \left|\frac{F(jw)}{g(w)} \right| \,\,\,\texttt{Eq. 2} $$ Which have worked "fine" for the functions $g(w)=e^{-\sqrt{|w|}}$ and $g(w)=\frac{1}{1+|w|^{2.5}}$ (meaning here that the integral and the supremum converges) when used against the benchmark $f(t)$: $$\begin{array}{r c l} \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|w\,e^{-\sqrt{|w|}}\right|dw \cdot \sup\limits_w \left|F(jw)\,e^{\sqrt{|w|}} \right| = & \frac{1}{2\pi}\cdot 24 \cdot \sup\limits_w \left| e^{\sqrt{|w|}}\frac{\pi^2\sin(w)}{\pi^2 w-w^3} \right| & = \frac{1}{2\pi}\cdot 24 \cdot 3.18407 = 12.623 < \infty\,\,\,\texttt{Eq. 3} \\ \frac{1}{2\pi} \int\limits_{-\infty}^\infty \left|\frac{w}{1+|w|^{2.5}} \right|dw \cdot \sup\limits_w \left|F(jw)\,(1+|w|^{2.5}) \right| = & \frac{1}{2\pi}\cdot \frac{4}{5}\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right) \cdot \sup\limits_w \left| \frac{\pi^2\sin(w)(1+|w|^{2.5})}{\pi^2 w-w^3} \right| & = \frac{1}{2\pi}\cdot 4.2758 \cdot 10.1353 = 6.8972 < \infty\,\,\,\texttt{Eq. 4} \end{array}$$

The "Lucky Strike"

Here is where I made my mistake, I was trying to use the Cauchy-Schwartz inequality and I take a mistaken version from internet that said $(\int fg\,dt)^2 \leq \int f^2\,dt \cdot \int g^2\,dt\,$, when, as @LL 3.14 explained in the comments below, the proper inequality is $(\int fg\,dt)^2 \leq \int |f|^2\,dt \cdot \int |g|^2\,dt$, and because of this mistake I was founding bounds lower than the left side for complex functions.

As you will see below, I am keeping "half" the mistake to form some Fourier transforms, by using "as true" the following inequality $(\int fg\,dt)^2 \leq \int |f|^2\,dt \cdot \int g^2\,dt$, which I know now is false in general as user @LL 3.14 shown in the comments with a counterexample. So, I am using something that I know now is conceptually wrong, but by now, please follow the explanation as is where right, so at the end the question will be Why in these particular cases this false-in-general-inequality is showing to be working.

Since has been explained by other person (user @Alex Ortiz) in the answers to this question, there is a lot of possible cancellation happening so the bound of Eq. 1 could be "too loose", so to try to find "better" upper bounds without moving the "absolute value" inside the integrand, I am trying to use the (illegal) Cauchy-Schwarz Inequality in the following way, similarly of how I did in Eq. 2 but now for a "clever" function $\sqrt{g(2w)}$: $$\begin{array}{r c l} \max_t \left| \frac{1}{2\pi} \int\limits_{-\infty}^\infty jw\,F(jw)\,e^{jwt}\,dw \right| & = & \max_t \left| \frac{j}{2\pi} \sqrt{\left( \int\limits_{-\infty}^\infty w\,F(jw)\frac{\sqrt{g(2w)}}{\sqrt{g(2w)}}\,e^{jwt}\,dw \right)^2} \right| \\ & \overset{\text{illegal Cauchy-Schwarz}}{\leq} & \max_t \left| \frac{j}{2\pi} \sqrt{\int\limits_{-\infty}^\infty \left(w\,\sqrt{g(2w)}\,e^{jwt}\right)^2\,dw}\cdot \underbrace{\sqrt{\int\limits_{-\infty}^\infty \left|\frac{F(jw)}{\sqrt{g(2w)}}\right|^2\,dw}}_{\text{indep. of}\,t}\right| \\ & \leq & \frac{1}{2\pi} \cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\int\limits_{-\infty}^\infty \frac{4}{4} w^2\,g(2w)\,e^{j2wt}\,\frac{2}{2} dw} \,\right| \\ & \overset{u = 2w \to du=2dw}{=} & \frac{1}{2\pi} \cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\frac{2\pi}{8}\frac{1}{2\pi} \int\limits_{-\infty}^\infty u^2\,g(u)\,e^{jut}\, du} \,\right| \\ & = & \frac{\sqrt{\pi}}{4\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\mathbb{F}_u^{-1}\left\{u^2 g(u)\right\}(t)} \right|\,\,\,\texttt{Eq. 5} \end{array}$$ where $\mathbb{F}_u^{-1}\left\{u^2 g(u)\right\}(t)$ is the inverse Fourier transform (function of $t$ - be careful of the definition since could change the constants). Here, the "clever" $g(u)$ must be such the integral converges, but also have a inverse Fourier transform in closed form (at least for me, since I don´t know how to work with other kind of functions).

I have tried some functions with known transforms like $e^{|w|}$ and $\text{sech(2w)}$ but the integral diverges, and unfortunately I don't know the transform of $e^{-\sqrt{|w|}}$ (I left it as a question here).

Also, note that the term $w^2$ could be move between the integral and the Fourier transform in the following way: $$\begin{array}{r c l} \max_t \left| \frac{1}{2\pi} \int\limits_{-\infty}^\infty jw\,F(jw)\,e^{jwt}\,dw \right| & = & \max_t \left| \frac{j}{2\pi} \sqrt{\left( \int\limits_{-\infty}^\infty w\,F(jw)\frac{\sqrt{g(2w)}}{\sqrt{g(2w)}}\,e^{jwt}\,dw \right)^2} \right| \\ & \overset{\text{illegal Cauchy-Schwarz}}{\leq} & \max_t \left| \frac{j}{2\pi} \sqrt{\int\limits_{-\infty}^\infty \left(\sqrt{g(2w)}\,e^{jwt}\right)^2\,dw}\cdot \underbrace{\sqrt{\int\limits_{-\infty}^\infty \left|\frac{w\,F(jw)}{\sqrt{g(2w)}}\right|^2\,dw}}_{\text{indep. of}\,t}\right| \\ & \leq & \frac{1}{2\pi} \cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{w^2\,F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\int\limits_{-\infty}^\infty g(2w)\,e^{j2wt}\,\frac{2}{2} dw} \,\right| \\ & \overset{u = 2w \to du=2dw}{=} & \frac{1}{2\pi} \cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{w^2\,F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\frac{2\pi}{2}\frac{1}{2\pi} \int\limits_{-\infty}^\infty g(u)\,e^{jut}\, du} \,\right| \\ & = & \frac{\sqrt{\pi}}{2\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left| \frac{w^2\,F^2(jw)}{g(2w)} \right| \,dw} \right|\cdot \max_t \left|\sqrt{\mathbb{F}_u^{-1}\left\{g(u)\right\}(t)} \right| \,\,\,\texttt{Eq. 6} \end{array}$$

For the upper bound of Eq. 6 I have found that the function $g(2w) = \frac{1}{1+(2w)^2}$ the integrands converges: $$ \begin{array}{r c l} \frac{\sqrt{\pi}}{2\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left| w^2(1+4w^2)F^2(jw)\right|\,dw} \right|\cdot \max_t \left|\sqrt{\mathbb{F}_u^{-1}\left\{\frac{1}{1+u^2}\right\}(t)} \right| & = & \frac{\sqrt{\pi}}{2\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left| w^2(1+4w^2)F^2(jw)\right|\,dw} \right|\cdot \max_t \left|\sqrt{\mathbb{F}_u^{-1}\left\{\frac{1}{1+u^2}\right\}(t)} \right| \\ & = & \frac{\sqrt{\pi}}{2\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|w^2(1+4w^2)F^2(jw)\right|\,dw} \right|\cdot \underbrace{\max_t \left|\sqrt{\frac{\displaystyle{e^{-|t|}}}{\displaystyle{2}}} \right|}_{\displaystyle{\frac{1}{\sqrt{2}}}} \\ & = & \frac{\sqrt{2\pi}}{4\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|w^2(1+4w^2)\left( \frac{\pi^2\sin(w)}{\pi^2 w-w^3} \right)^2\right|\,dw} \right|\\ & = & \frac{\sqrt{2\pi}}{4\pi}\cdot \left| \sqrt{|627.543|} \right| = 0.1994 \cdot 25.0508 = 4.996 \,\,\,\texttt{Eq. 7} \end{array}$$ which is higher that the upper bound of the Eq. 1 (but lower than the Eq. 3 and Eq. 4), so I am still trying to find a "clever" $g(w)$ which could do "the magic" (been lower than Eq. 1).

(Added later in the original question) I believe I have a bound that works better than the bound of Eq. 1, by splitting the $w$ term between both integrands (here another "illegal" thing is done, since I have split the term $w$ as $\sqrt{|w|}\sqrt{w}$ - be aware of it): $$\begin{array}{r c l} \max_t \left| \frac{1}{2\pi} \int\limits_{-\infty}^\infty jw\,F(jw)\,e^{jwt}\,dw \right| & \overset{\text{illegal split}}{=} & \max_t \left| \frac{j}{2\pi} \sqrt{\left( \int\limits_{-\infty}^\infty \sqrt{|w|}\,F(jw)\frac{\sqrt{w\,g(2w)}}{\sqrt{g(2w)}}\,e^{jwt}\,dw \right)^2} \right| \\ & \overset{\text{illegal Cauchy-Schwarz}}{\leq} & \max_t \left| \frac{j}{2\pi} \sqrt{\int\limits_{-\infty}^\infty \left(\sqrt{w\,g(2w)}\,e^{jwt}\right)^2\,dw}\cdot \underbrace{\sqrt{\int\limits_{-\infty}^\infty \left|\frac{\sqrt{|w|}\,F(jw)}{\sqrt{g(2w)}}\right|^2\,dw}}_{\text{indep. of}\,t}\right| \\ & \leq & \frac{1}{2\pi} \cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{|w|\,F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\int\limits_{-\infty}^\infty \frac{2w}{2}g(2w)\,e^{j2wt}\,\frac{2}{2} dw} \,\right| \\ & \overset{u = 2w \to du=2dw}{=} & \frac{1}{2\pi} \cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{|w|\,F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\frac{2\pi}{4}\frac{1}{2\pi} \int\limits_{-\infty}^\infty u\,g(u)\,e^{jut}\, du} \,\right| \\ & = & \frac{\sqrt{2\pi}}{4\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left|\frac{|w|\,F^2(jw)}{g(2w)}\right|\,dw} \right|\cdot \max_t \left|\sqrt{\mathbb{F}_u^{-1}\left\{u\,g(u)\right\}(t)} \right|\,\,\,\texttt{Eq. 8} \end{array}$$

Now, testing with the following $g(2w) = \frac{1}{1+(2w)^2}$, Eq. 8 becomes: $$\begin{array}{r c l} \frac{\sqrt{2\pi}}{4\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left| |w|(1-4w^2)F^2(jw)\right|\,dw} \right|\cdot \max_t \left|\sqrt{\mathbb{F}_u^{-1}\left\{\frac{u}{1+u^2}\right\}(t)} \right| &=& \frac{\sqrt{2\pi}}{4\pi}\cdot \left| \sqrt{\int\limits_{-\infty}^\infty \left| |w|(1-4w^2) \left( \frac{\pi^2\sin(w)}{\pi^2 w-w^3} \right)^2\right| \,dw} \right|\cdot \max_t \left|\sqrt{\frac{j}{2}e^{-|t|}\text{sgn}(t)} \right|\\ & = & \frac{\sqrt{2\pi}}{4\pi} \cdot \left| \sqrt{|182.779|} \right|\cdot \frac{1}{\sqrt{2}} = 1.90690 \,\,\,\texttt{(Eq. 9)} \end{array}$$ which is higher than the maximum rate of change of $\pi/2 = 1.57$ and lower than the bound of Eq. 1 of $2.28547$.

Testing these bounds

I have tested the bounds against some functions $g(2w)$ for the benchmark $f(t)$, and Eq. 6 shows convergence for $g(2w) = \frac{1}{1+2w}\,$, $g(2w) = \frac{1}{2w}\,$, $g(2w) = \frac{1}{1+j2w}\,$, and $g(2w) = \frac{1}{1+j(2w)^2}\,$, and Eq. 8 have shown also finite values for $g(2w) = \frac{1}{1+(2w)^2}\,$, $g(2w) = \frac{1}{1+j(2w)^2}\,$, and $g(2w) = \frac{1}{1-j(2w)^2}\,$, but I will only continue with the results that shows to be lower than the bound of Eq. 1: for Eq. 6 $g(2w) = \frac{1}{1+(2w)}\,$ and for Eq. 8 the functions $g(2w) = \frac{1}{1+(2w)^2}\,$ and $g(2w) = \frac{1}{1+j(2w)^2}\,$ (note here that a sign change in the denominators leads to different values):

$$\begin{array}{r c l} g(2w)=\frac{1}{1+(2w)^2} \rightarrow \texttt{(Eq. 8)} & = & \frac{ \sqrt{\pi} }{4\pi} \left| \sqrt{ \int_{-\infty}^\infty \left| |w| (1+4w^2) F^2(jw) \right| dw } \right| \,\,\,\texttt{(Eq. 10)} \\ g(2w)=\frac{1}{1+j(2w)^2} \rightarrow \texttt{(Eq. 8)} & = & \frac{ \sqrt{\pi} }{4\pi} \left| \sqrt{ \int_{-\infty}^\infty \left| |w| (1+j4 w^2) F^2(jw) \right| dw } \right| \,\,\,\texttt{(Eq. 11)} \\ g(2w)=\frac{1}{1+2w} \rightarrow \texttt{(Eq. 6)} & = & \frac{ \sqrt{2\pi} }{4\pi} \left| \sqrt{ \int_{-\infty}^\infty \left| w^2 (1+2w) F^2(jw) \right| dw } \right| \,\,\,\texttt{(Eq. 12)} \end{array}$$

Using Wolfram-Alpha, I have tested these three bounds with the functions of the table 2 of this question for which Eq. 1 converges, but for accuracy issues I will compare them also with the bound of Eq. 4.

As in the mentioned question, the time-limited Fourier transform is going to be $F(j\omega)=\mathbb{F}_{[a\,;\,b]}\{f(t)\}(\omega) = \int_{t_0 }^{t_F} f(t) e^{-j \omega t} dt$ and the maximum rate of change is obtained as $\max_{t_0 < t < t_F} |f'(t)| = \max_t \left| \frac{d}{dt}\left(f(t)_\text{unbounded}\right)\cdot(\theta(t-t_0)-\theta(t-t_F))\right|$.

$$ \begin{array}{|c:c|c:c|c|c:c:c:c:c|} \hline f(t) & \text{dom}(f) = [a\,;\,b] & \mathbb{F}_{[a\,;\,b]}\{f(t)\}(\omega) & \text{dom}(F(j\omega)) & \max_{a < t < b} |f'(t)| & \frac{1}{2 \pi} \int_{-\infty}^{\infty} |j\omega F(j\omega)|d\omega & Eq. 4 & Eq. 10 & Eq. 11 & Eq. 12 \\ \hline \cos^2(\frac{t\pi}{2}) & [-1; 1] & \frac{\pi^2\sin(\omega)}{(\pi^2\omega-\omega^3)} & (-\infty; \infty) & \frac{\pi}{2} = 1.57079 & 2.28547^* & 6.8973 & 1.9069 & 1.8733 & 1.8708 \\ \hdashline \frac{(1+\cos(t\pi))^2}{4} & [-1; 1] & \frac{3\,\pi^4\sin(\omega)}{(\omega^5-5\pi^2\omega^3+4\pi^4\omega)} & (-\infty; \infty) & \frac{3\sqrt{3}\pi}{8} = 2.0405 & 2.61265^* & 9.9209 & 2.4135 & 2.3876 & 2.3864 \\ \hdashline \sin(\frac{t\pi}{2})\cos^2(\frac{t\pi}{2}) & [-1; 1] & j\frac{16\, \pi^2\, \omega \cos(\omega)}{(16\, \omega^4-40\, \pi^2 \omega^2+9\,\pi^4)} & (-\infty; \infty) & 1.5708 & 1.93647^* & 8.6398 & 1.8215 & 1.8077 & 1.8075 \\ \hdashline \text{sinc}(t\pi)\cos(\frac{t\pi}{2}) & [-1; 1] & \frac{1}{2\pi}\left(\text{Si}(\frac{\pi}{2}-\omega)+\text{Si}(\frac{3\pi}{2}-\omega)+\text{Si}(\frac{\pi}{2}+\omega)+\text{Si}(\frac{3\pi}{2}+\omega)\right) & (-\infty; \infty) & 1.62897 & \infty^* & 7.2904 & 1.9552 & 1.9229 & \textbf{1.9276} \\ \hdashline 1-\sin^4(\frac{t\pi}{2}) & [-1; 1] & \frac{\pi^2(5\pi^2 - 2\,\omega^2) \sin(\omega)}{(\omega^5 - 5\pi^2\omega^3 + 4\pi^4\omega)} & (-\infty; \infty) & \frac{3\sqrt{3}\pi}{8} = 2.0405 & 3.01547^* & 10.8628 & 2.2717 & 2.2362 & 2.2332 \\ \hline \end{array} $$

For all these time-limited functions the proposed bounds show to work and even better than the bound of Eq. 1, in increasing order of $Eq. 10 > Eq. 11 > Eq. 12$, except for the value marked bold (don´t know if a numerical issue or if the order of magnitudes among bounds is just no always true).

The question

Since the bounds are working even better than the traditional bound of Eq. 1, the obvious is Why is that so? Are these bounds "valid" (formally speaking)? Can they be proven to be valid upper bounds?... or if they aren't valid in general, For which kind of functions they will do work right?

The request

Since I can prove these by myself with my knowledge, I hope you can help me to figure out put. I have had another lucky strike for the other question, so if you interested, soon I will update it.

I have not tested these bounds with band-limited functions, but if they work, after proving they are valid, surely it will be a publishable result. If these methods and bounds wasn´t discovered before, please let called them "Herreros' method" and "Herreros' bounds" (this because of ego $\texttt{XD}$), but since I have already write the introduction, I hope you be will friendly enough to include me into the authors (the last one at least $\texttt{XD}$), so you can find my info here (it will help me later if I decide to come back to work on Research).

I believe these bounds could be helpful to find the conditions I am asking in the other question, about the time-limited functions and its maximum rate of change when its bounded, which can be also useful in physics and optimization. This because these bounds can be surely improved with other "cleaver" functions $g(2w)$, as example, for some values $0 < a < 3$ the inverse Fourier transform $g(u) = \frac{1}{1+ju^a}$ on Wolfram-Alpha shows to have closed form results through Meijer G-functions (at least for $a = \{0.25,\,0.2,\,0.4,\,0.8,\,1.6,\,2.4,\,2.5\}$), but they are "too long" to be analyzed through Wolfram-Alpha webpage, and maybe looking for and optimum "$a$" to "make the trick" will tell information about the conditions I am looking for, so please, let me updated with what you find.

Thank you very much.