Given the finite-duration continuous function $q(t) = \int_{-\infty}^t h(u)\,du$,
- Could $h(t)$ be discontinuous?
- Is mandatory for $h(t)$ to be a function of compact-support? (so, also of finite-duration?)
Also, since $q(t)$ is bounded $\sup_t|q(t)|<\infty$ because is continuous and compact-supported:
- Is mandatory for $h(t)$ to be a bounded function? (here a believe that is not a requirement, since there are unbounded functions with finite area, but if (1) and (2) are both affirmative, it will be bounded).
I have change the term "time-limited" for "finite-duration" since is more accurate and widely used, but it means that the scalar one-variable function $q(t)$ has an starting time $t_s$ for which $q(t) = 0\,\forall t<t_s$, and also an ending time $t_f$ from which $q(t) = 0\,\forall t>t_f$, with $t_s < t_f$
I'll take a stab at answering 1, guessing at 2, and passing on 3. If I have any better ideas at a later time I'll add to my answer.
Yes, by construction, let: $$h(t) = \begin{cases} 0 & x < 0 \\ 1 & 0 \leq x < 1 \\ -1 & 1 \leq x < 2 \\ 0 & x \geq 2 \end{cases}$$ $h(t)$ is discontinuous at 0, 1, and 2. $$q(t) = \int_{-\infty}^{t} h(u) \,du $$ $$q(t) = \begin{cases} 0 & t < 0 \\ t & 0 \leq t < 1 \\ 2-t & 1 \leq t < 2 \\ 0 & x \geq 2 \end{cases}$$ $q(t)$ is a continuous function
I don't have a proof for this, but I suspect this would require a function, that except for finite region is non-zero and yet integrates to zero over any finite interval. This sounds like it would have to be some sort of pathological function. I can't think of a suitable one at the moment, but there are some crazy ones... (maybe the Dirichlet function for the above example except between 0 and 2?)
Is mandatory for $h(t)$ to be a bounded function?