For finite-duration continuous $f(t)$ with bounded derivative $\|f'(t)\|_\infty < \infty$: It is true that $\|f'(t)\|_\infty \leq \frac{2\pi \|f'(t)\|_2^2}{\|f'(t)\|_1}$?
I am looking for an upper bound for the derivative of finite-duration continuous functions $f(t)$ that have bounded derivative $\|f'(t)\|_\infty <\infty$ (with unknown upper bound), but with $f'(t)$ discontinuous, but at most in finitely many zero-measure points, where also $f(t)$ has a convergent Fourier Transform $\hat{f}(w)$.
For continuous functions with at most a finite amount of discontinuities in $f'(t)$ the traditional upper bound $\|f'(t)\|_\infty \leq \|w\hat{f}(w)\|_1$ wouldn´t properly work, since it will diverge $\|w\hat{f}(w)\|_1 \to \infty$ for bounded but discontinuous $f'(t)$.
As example, think in $f(t)=\sin(\pi|t|),\,|t|\leq 1$: here $f(t)$ is continuous but has a "pike" at $t=0$, where $f'(t)$ is discontinuous having a jump-discontinuity changing from $-\pi$ to $+\pi$, but the size of the jump is bounded so it stills fulfill $|f'(t)| \leq \pi < \infty$... different from the counterexample $f(t) = |t\,\log(t^2)|(1-t^2+|1-t^2|)$ which also has a "pike" at $t=0$ but $f'(t)$ diverges here.
Since I know that $\|f'(t)\|_\infty \leq \|w\hat{f}(w)\|_1$ doesn´t work for discontinuous but bounded $f'(t)$, I was trying to found another alternative of upper bound for these cases (for other cases, or $f'(t)$ is unbounded, or I already have the mentioned upper bound), and I believe I have found the following upper bound using Parseval's Identity for the derivative: $$\begin{array}{c} \|f'\|_2^2 = \frac{1}{2\pi}\|w\hat{f}\|_2^2 \overset{\text{Hölder}\,p=1}{\leq} \frac{1}{2\pi} \|w\hat{f}\|_\infty \cdot \|w\hat{f}\|_1 \leq \frac{1}{2\pi} \|f'\|_1\cdot \|w\hat{f}\|_1 \\ \overset{\text{norms}\geq 0}{\Rightarrow} \frac{2\pi \|f'(t)\|_2^2}{\|f'(t)\|_1} \leq \|w\hat{f}\|_1 \end{array}$$
So I want to know if is possible to compare both lower bounds of $\|w\hat{f}\|_1$ between each other, hopefully fulfilling that is true that: $\|f'(t)\|_\infty \leq \frac{2\pi \|f'(t)\|_2^2}{\|f'(t)\|_1}$ for functions with already bounded $\mathbf{\|f'(t)\|_\infty <\infty}$ (this is important: I already know is false in general, so there are many examples for which $\|f'(t)\|_\infty \to \infty$ and the bound will be finite $\frac{2\pi \|f'(t)\|_2^2}{\|f'(t)\|_1} < \infty$)...
So, my idea is:
- Test first $\|w\hat{f}(w)\|_1$,
- And then, If the bound (1) diverges, test second (2) $\frac{2\pi \|f'(t)\|_2^2}{\|f'(t)\|_1} $ as an upper bound.
I have already tested it working for these function that rises in the answers to this question: $f(t)=1−|t|,\,|t|\leq 1$ and $f(t)=e^{−|t|},\,|t|\leq1$.
Note 1: I know beforehand that these bounds are not really practical and are sometimes harder than finding directly $\|f'\|_\infty$, but I am trying to understand which characteristics of a finite-duration function will make its derivatives to be unbounded or bounded.
Note 2: Take in consideration that I am asking about continuous functions that are of finite-duration, so by the Extreme value theorem I have already that the function is bounded $\|f(t)\|_\infty < \infty$.
Note 3: I already know that I have $\|f'\|_\infty \leq \|w\hat{f}\|_1$ and $\frac{2\pi \|f'\|_2^2}{\|f'\|_1} \leq \|w\hat{f}\|_1$ does not imply $ \|f'\|_\infty \leq \frac{2\pi \|f'\|_2^2}{\|f'\|_1}$, I am asking If it is possible to prove that only for scenarios where $\|f'\|_\infty < \infty$, then IF $ \|f'\| \leq \frac{2\pi \|f'\|_2^2}{\|f'\|_1}$ could be used as an upper bound, since I have found some examples where it works (maybe is an accident), these to see if there is an alternative upper bound for scenarios of bounded derivative where $\|w\hat{f}\|_1 \to \infty$.
Example:
Lets work with this example (2nd comment by @Lorenzo Pompili), given by the function:
$f(t) = \begin{cases} e^{-|t|}, & \,|t|\leq 1 \\ 0, & \,|t|>1 \end{cases}$
Its derivative is going to be $f'(t) = -\frac{t}{|t|}e^{-|t|},\,|t|\leq 1$ so I will have for this example that the derivative it will be bounded by $\sup_t |f'| = 1 << \infty$, even when its discontinuous.
Now, being proved is bounded, I will take now the Finite Extension Fourier Transform of $f(t)$: $$\hat{f}(w) = \int\limits_{-1}^{1} e^{-|t|}e^{-iwt}dt = \frac{e^{-(1+iw)}\left( 2\,e^{1+iw}+iw-1-(1+iw)e^{2iw}\right)}{1+w^2}$$
So, I will try to compute the classic upper bound: $$\|w\hat{f}(w)\|_\infty = \int\limits_{-\infty}^{\infty} |w\hat{f}(w)|dw \to \infty$$ but since $f'(t)$ is discontinuous it show to diverge, so it don´t going to help as an upper bound (at least as a tight-as-possible one).
Now to see if there is something that can be done using the bound I am asking to prove if it could work for these scenarios where $\|w\hat{f}(w)\|_\infty \to \infty$, lets calculate it by each of its components:
- $$\|f'(t)\|_2^2 = \int\limits_{-1}^1 \left|-\frac{t}{|t|}e^{-|t|} \right|^2 dt = 1 - \frac{1}{e^2} < \infty$$ (here)
- $$\|f'(t)\|_1 = \int\limits_{-1}^1 \left|-\frac{t}{|t|}e^{-|t|} \right| dt = 2 - \frac{2}{e} < \infty $$ (here)
- $$\frac{2\pi\|f'(t)\|_2^2}{\|f'(t)\|_1}=2\pi\frac{\left(1 - \frac{1}{e^2}\right)}{\left(2 - \frac{2}{e}\right)}= \frac{\pi(1+e)}{e} \approx 4.3 << \infty $$ (here)
So at least for this example the "happy accident" of $$ \|f'(t)\|_\infty = 1 \quad < \quad 4.3 = \frac{2\pi\|f'(t)\|_2^2}{\|f'(t)\|_1} \quad << \quad \infty = \|w\hat{f}(w)\|_\infty$$ happen to be true, working as a tightest upper bound for the rate of change than $\|w\hat{f}(w)\|_\infty$.
This is why, knowing beforehand it is false in general, I want to know if there is a way to prove/disprove if this happy accident could be used for the cases where $\|w\hat{f}(w)\|_\infty \to \infty$ as an alternative tighter upper bound for the maximum rate of change.
