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When we have say Delta at time $t_0$ on continuous time: $\delta(t-t_0)$.
If we want to move it $t_0/2$, can we scale the time and squeeze by 2 instead of shifting?
How does scaling in time of a shifted function look like? Is it $f(a(t-t_0))$ or $f(at-t_0)$?
Properties or effects of mathematical objects (operators, systems, functions, etc.) generally depend on hypotheses they are supposed to possess (or axioms they obey).
Here, one of the tag the OP used is very important: continuous-signals, and a mention in the question as well: "shifted function". Words are important. The Dirac $\delta$ is not a function. And from the realm of function, it cannot be considered continuous (with standard continuity), in the sense of topological continuity (note to self: ask to rename continuous-signals in continuous-time-signal).
The Dirac $\delta$ is a kind of continuous analogue of the discrete Kronecker delta. Continuous here means that it is defined on continuous time. Its properties are odd, compared to classical functions. For the formula the OP considers, a useful property is: if $g$ is a continuously differentiable function with a real root at $t_0$, and its derivative (well-defined) does not vanish, one could write:
$$\delta(g(t)) = \frac{\delta(t-t_0)}{g'(t_0)}\,.$$
which applies well here. For details, see Dirac delta, Composition with a function.
Despite all the uncensured content here in the comments, I'd like to answer my own question.
I understand that some have difficulties understanding the intuition behind the dilema here. I'll explain this first:
When we have deltas and we do Fourier to them, intuition says since we do scaling to a shifted delta function's time we might maybe somehow exchange a shifting operation on a scaling operation.
$\delta(t-t_0)$ can be scalled in time and since we have y=infinity at $t_0$ and the width of delta is infinitely small, then maybe we could make a little trick in future use?
Well then, wrong.
The thing is that when we do time scalling, the shift moves, that's right. But the time scalles as well. When we have $\delta(a(t-t_0))$ function, it might yield "the same" in time domain, but it's different.
The thing is that while doing fourier to it, since we have an integral during Fourier process, time scalling to delta function will have an impact on the integral and so that:
$\int\delta(a(t-t_0))= {1 \over a}\int\delta(t-t_0)$
Means that we have no way doing time scalling instead of shifting to delta functions despite appealing as much as it can seem to be.
