Does $||f'||_\infty \leq \sqrt{t_F-t_0}\,||f'||_2$ hold for time-limited continuous functions $f(t)$ with $\sup_t |f'(t)|<\infty$?

Question

Prerequisites for the answer:

I am trying to understand which conditions makes the derivative of non-everywhere-differentiable continuous time-limited functions to be bounded $\sup_t |f'(t)|<\infty$, so the derivative $f'(t)$ could be discontinuous, but cannot diverge to infinity in these discontinuities (because the requested restriction will be instantly violated), so $\mathbf{\sup_t |f'(t)|<\infty}$ is a requirement. Also, if $f(t)$ is differentiable, the derivative will be bounded so $f(t)$ has to have at least one point where is not differentiable (as example, having one finite-size-gap jump-discontinuity).

Also, the function $f(t)$ must have a Fourier Transform, so the function $f(t)$ must be Lebesgue integrable, or equivalently, fulfill that $\int_{t_0}^{t_F} |f(t)|dt < \infty$, and also, the amount of discontinuities must be finite, so nowhere-differentiable functions are not allowed.

With saying that $f(t)$ is "Time-Limited" I am meaning that the function $f(t)$ has a starting point at a time $t_0$ and and ending point at a time $t_F$ with $t_F>t_0$, so, $f(t) = 0,\,t<t_0$ and $f(t) = 0,\,t>t_F$, and can be any function that fulfill the other requirements within the boundaries of its domain $\partial t =\{t_0,\,t_F\}$. Since at the discontinuity that could rise at the edges of the domain could made some problems with the Fourier Transform, if you wish you can assume that $f(t_0)=f(t_F)=0$, but this issue can be easily avoided using the transform $\mathring{\mathbb{F}}\{\cdot\}$ is introduced here, so I will continue assuming that this problem is nonexistent (and using $\mathbb{F}\{\cdot\}$ or $\mathring{\mathbb{F}}\{\cdot\}$ indistinguishable as I need it).

Please note that time-limited functions are compact-supported, so since I am interesting in continuous functions $f(t)$, they will be also bounded functions $\sup_t |f(t)| < \infty$, and since are also Lebesgue integrable, both conditions together imply also that $f(t)$ is of finite energy $\int_{t_0}^{t_F} |f(t)|^2 dt < \infty$.

The problem

Since the traditional upper bound of the derivative $||f'||_\infty \leq ||wF(w)||_1$ will diverge if $f'(t)$ have discontinuities $\Rightarrow ||wF(w)||_1 \to \infty$ (here $F(w)$ is the Fourier transform of $f(t)$), I am looking for alternative upper bounds that could be useful for time-limited continuous functions $f(t)$ with a finite number of discontinuities in $f'(t)$.

On the Wikipedia website of Norms appears some inequalities that fulfills some of the vector norms:

1. $||x||_2 < ||x||_1 < \sqrt{n}\,||x||_2 $
2. $||x||_\infty < ||x||_2 < \sqrt{n}\,||x||_\infty $
3. $||x||_\infty < ||x||_1 < n\,||x||_\infty $

But through some questions I have already done here in SE (1, 2, 3, 4, 5), I know now that these inequalities are not general for functions, which some inequalities could be even inverted for the same function depending on the integration domain.

But experimenting with the following functions $f(t)=1-|t|,\,|t|\leq 1$ and $f(t)=e^{-|t|},\,|t|\leq 1$, that have bounded $||f'(t)|| < \infty$ but makes $||wF(w)|| \to \infty$ since they have bounded-size jump-discontinuities en $f'(t)$ (examples I am taking from the answer to 2), I believe that maybe some of the inequalities could apply for time-limited continuous functions.

The question:

Does any of these inequalities holds in general for time-limited continuous functions $\mathbf{f(t)}$ with bounded derivative $\mathbf{\sup_t |f'(t)|<\infty}$?

1. $\sup_t |f'(t)| \leq \sqrt{(t_F - t_0)\,\int\limits_{t_0}^{t_F}|f'(t)|^2\,dt}$? (this is the main one I am interested in $||f'||_\infty \leq \sqrt{t_F-t_0}\,||f'||_2$)
2. $\sup_t |f'(t)| \leq \sqrt{\frac{\pi}{2}\,\int\limits_{t_0}^{t_F}|f'(t)|^2\,dt}$?
3. $\sup_t |f'(t)| \leq \int\limits_{t_0}^{t_F}|f'(t)|\,dt$?
4. $\sup_t |f'(t)| \leq \int\limits_{t_0}^{t_F}|f(t)|\,dt$?

I am asking explicitly if they holds for general domain $\mathbf{-\infty<t_0<t_F<+\infty}$.

Answer 1

No, none of these bounds hold.

First remark, the derivative is a local object, so the fact to be time limited will not help to bound a derivative. The derivative at a point do not change if you change your function at other points.

Then, defining $g = f'$, you can see that what you are asking has (almost) nothing to do with derivatives and just with bounding the $L^\infty$ norm by the $L^2$ norm. This is in general false: there are a lot of unbounded functions that are square integrable, such as $g(t) = |t|^{-1/4}$ with $t\in[-1,1]$ and $g(t)=0$ on the complement.

The only thing that is special about $g$ being the derivative of a compactly supported function is the fact that you have to ensure that $∫_{t_0}^{t_f} g = 0$, and so you can replace the above example by $g(t) = |t|^{-5/4}t$ (for $t_0 = -1$ and $t_f=1$, but you can translate and dilate the coordinates to get any $t_0$ and $t_f$). In terms of $f$, it leads to $$ f(t) = \frac{4}{3} \,(t^{3/4} - 1) \ \text{ if } t\in[-1,1] $$ and $f(t)=0$ if $|t|>1$ as a counterexample to all your inequalities.