Are there any continuous time-limited Linear and Time-Invariant (LIT) functions with unbounded derivative? Are there any continuous time-limited Linear and Time-Invariant (LIT) functions with unbounded derivative?
Let think about a continuous and time-limited function $q(t)$ that is representing a classical mechanics phenomena, which can be represented as the output of a LIT system with impulse response $h(t)$.
Now, I want to know which is the maximum rate of change that can possible achieve the function $q(t)$, so as a worst case I test this system against a discontinuous jump-alike change by using as input the unitary standard step function $\theta(t)$, so:
$$q(t) = h(t)\circledast\theta(t)$$
related through the convolution operator as every LIT system.
Now, I believe that since $q(t)$ is continuous and time limited, and $\theta(t)$ is not a compacted-supported function, the only alternative to made $q(t)$ as it is, is by an $h(t)$ function that is also continuous and time-limited (please correct me if I am wrong, this is the most important assumption on the presented line-of-thought).
With this, since $q(t)$ and $h(t)$ are both continuous and compact-supported (because they are both time-limited), they are also bounded functions $\sup_t |q(t)| < \infty$ and $\sup_t |h(t)| < \infty$. And also remember, that since $q(t)$ and $h(t)$ are both time-limited, it implies that both are also of unlimited bandwidth (i.e., they aren´t band-limited functions).
Now, following the exercise n° 4.49 of the book "Signals and Systems, 2nd Edition" (Alan V. Oppenheim, Alan S. Willsky, with S. Hamid) [1], and the following properties:

*

*$\frac{d}{dt}\Big(x(t) \circledast y(t) \Big) = \frac{dx(t)}{dt} \circledast y(t) = x(t) \circledast \frac{dy(t)}{dt}$ as is shown on  Wiki: Algebraic Properties --> Relation with differentitation.

*The derivative of the unitary step function is the Dirac's delta function $\theta'(t) = \delta(t)$, as is shown on Wiki.

*and $f(t) = f(t) \circledast \delta(t)$ as is shown on  Wiki: Algebraic Properties --> Multiplicative identity.

So, I will have that $$q'(t) = h(t) \circledast \theta'(t) = h(t) \circledast \delta(t) = h(t)$$
and since $|h(t)|< \infty$ is bounded if my first assumptions are right, I will have that the maximum rate of change of
$$\sup_t |q'(t)| = sup_t |h(t)| < \infty$$
so even they are functions of unlimited bandwidth they will ALWAYS have a bounded derivative:
A) Does this means that every continuous and time-limited LIT functions have a bounded maximum rate of change $\sup_t |q'(t)|< \infty$?
By construction, I already know that if impulse response $h(t)$ is continuous and time-limited then the maximum rate of change of $q(t)$ will be bounded, if the analysis against the unitary step $\theta(t)$ is general enough (Is that so?).
But in the other way, if $q(t)$ is continuous and time-limited, I feel that it could not be necessarily implying that $h(t)$ is also continuous and time-limited, since $q(t) = h(t) \circledast \theta(t) = \int_{-\infty}^t h(t)\,dt$ and I am not sure if the integral of a discontinuous function could lead to a continuous function, hope you can explain this too.

Added later:
I have just notice that any function can be described as:
$$q(t) = \int_{-\infty}^t q'(t)\,dt = q'(t)\circledast \theta(t)$$
so requiring that the impulse response function $h(t) \equiv q'(t)$ to be continuous and time-limited is already requiring that $|q'(t)|<\infty$ been bounded.

Other interesting facts from exercise 4.49:
Since from the properties of the Laplace Transform of $q(t)$ given by $Q(s)$, I will have that:
$$ \lim_{t \to \infty} q(t) = q(\infty) = \lim_{s \to 0} s\,Q(s) = \lim_{s \to 0} \int_{-\infty}^\infty \frac{dq(t)}{dt}e^{-st}dt = \int_{-\infty}^\infty q'(t)\,dt = \int_{-\infty}^\infty h(t)\,dt$$
so I can model the "minimum possible response time to the step function" $\Delta t_{\min}$ as:
$$ \sup_t |q'(t)| = \Bigg| \frac{q(\infty)}{\Delta t_{min}}\Bigg| \Rightarrow \Delta t_{min} = \frac{|q(\infty)|}{\sup_t|q'(t)|} = \frac{|\int_{-\infty}^\infty h(t)\,dt|}{\sup_t |h(t)|} $$
so I could define an "effective Bandwidth" for this system response to the step function using as maximum frequency the quantity $f_{max} = 1/ \Delta t_{min}$:
$$B_W = 2 f_{max} = \frac{2}{\Delta t_{min}} = \frac{2\sup_t |h(t)|}{|\int_{-\infty}^\infty h(t)\,dt|}$$
with Uncertainty relation $$ B_W \cdot \Delta t_{min} > 2$$.
 A: I have realized later that the questions is ill-posed, but since other person bookmarked it, before deleting it I will give what know I think is the answer.
I have found the following paper named "Finite time differential equations" by V. T. Haimo (1985), where continuous time differential equations with finite-duration solutions are studied, an it is stated the following:
"One notices immediately that finite time differential equations cannot be Lipschitz at the origin. As all solutions reach zero in finite time, there is non-uniqueness of solutions through zero in backwards time. This, of course, violates the uniqueness condition for solutions of Lipschitz differential equations."
Since, linear differential equations have solutions that are unique, and finite-duration solutions aren´t, finite-duration phenomena models must be non-linear to show the required behavior (non meaning this, that every non-linear dynamic system support finite-duration solutions).
The paper also show which conditions must fulfill the non-linear differential equation to support finite-duration solutions, at least for first and second order scalar ODEs.
So, if the paper is right (I can't prove or disprove it), no LTI system could have finite-duration solutions, since they are described through linear differential equations, and with this, the question can´t have an answer since is asking for finite-duration LIT system, which can´t exists (or the system is linear, or is of finite-duration, can´t be both).
