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Answering this question, I've tried to alert the OP about the misleading definition of $\delta(t)$ - used in one of the answers - as:

$$ \delta(x) = \left\{\begin{array}{cc} \infty & x = 0 \\ 0 & x\neq 0 \end{array}\right. $$

And gave an example when this definition fails:

I'll show one example that illustrates why:

What is $\mathscr{L}\left[\,\delta(t)\,\right]$? And $\mathscr{L}\left[\,\delta(3t)\,\right]$?

Using the above interpretation, we have: $$ \begin{alignat}{0} \delta(t) = \left\{\begin{array}{cc} \infty & t = 0 \\ 0 & t\neq 0 \end{array}\right. &&\delta(3t) = \left\{\begin{array}{cc} \infty & 3t = 0 \\ 0 & 3t\neq 0 \end{array}\right. = \left\{\begin{array}{cc} \infty & t = 0 \\ 0 & t\neq 0 \end{array}\right. =\delta(t) \end{alignat} $$ So, since we have the same "function"... $\mathscr{L}\left[\,\delta(t)\,\right]$ must be equal to $\mathscr{L}\left[\,\delta(3t)\,\right]$, right? WRONG! $$ \begin{alignat}{1} \mathscr{L}\left[\,\delta(t)\,\right]&=1 \\ \mathscr{L}\left[\,\delta(3t)\,\right]&=\frac 1 3 \end{alignat} $$ But Laplace Transform Uniqueness Theorem states that:

If $f_1$ and $f_2$ are continuous on $[0,∞)$, then $\mathscr{L}[\,f_1\,]\neq\mathscr{L}[\,f_2\,] \iff f_1 \neq f_2$

Then: $$ \mathscr{L}\left[\,\delta(t)\,\right] \neq \mathscr{L}[\,\delta(3t)\,] \iff \delta(t) \neq \delta(3t) $$

I argued that $\delta(t)$ must be different of $\delta(3t)$ because of Laplace Uniqueness Theorem. But now I realized that my argument fails, because Dirac Delta "function" isn't continuous. So, my questions are:

  • Is this theorem valid for Delta "function", even if it isn't continuous in the traditional sense? If yes, why?

  • If no, is there a stronger theorem about uniqueness of Laplace Transform, which allows Dirac Delta "function" to be included?

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Vinicius ACP
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    There isn't a need to go through Laplace transforms. Scaling of a distribution is defined via extending the identity on test functions $\int f(3x) g(x) dx = \frac13\int f(y)g(y/3) dy$, so $\delta(3x)$ has the action $$ \langle \delta(3x) , \phi(x) \rangle := \langle \delta(y) , \frac13 \phi(y/3) \rangle = \phi(0)/3$$ which makes it clear that they aren't the same – Calvin Khor May 02 '19 at 17:06
  • You're right, but I'm trying to look at it using only Laplace transform and uniqueness theorem. – Vinicius ACP May 02 '19 at 17:13
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    Well I don't find this argument convincing because it assumes I know how to interpret $\mathcal L[\delta(3x)]$ without knowing the above basic definition of how to scale a distribution. That said, it seems that Laplace transforms uniquely determine finite measures anyway. (prop 1.8) https://web.math.pmf.unizg.hr/~amimica/lt.pdf – Calvin Khor May 02 '19 at 17:26
  • @CalvinKhor I'll give some context about my question: I've learned Laplace Transform in Electric Circuits class. My teachers defined the Dirac Delta with the mislead definition, and then asked in a test only about Laplace transforms if $\delta(t)=\delta(2t)$. The explanation given in this question was the explanation I could give, given the knowledge I had, and that was the correct answer, according to them. This happened some months ago. I was satisfied with this argument until today, when I realized that it fails. – Vinicius ACP May 02 '19 at 18:09
  • That's why I'm "forcing" the Laplace transform way: if they asked the students this question, considering their knowledge about Delta "function" (the misleading definition), then there must be a way to solve it using only the basic theory related to Laplace Transforms and not to distributions. – Vinicius ACP May 02 '19 at 18:09
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    I guess that the condition of continuity can be dropped as long as we allow the functions to differ on a null set (e.g. a countable set of points). For distributions we don't have that problem and probably $\mathscr{L}[u_1] = \mathscr{L}[u_2] \iff u_1 = u_2.$ – md2perpe May 02 '19 at 18:11
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    Regarding the definition of $\delta$; only saying $$\delta(x) = \begin{cases} \infty & x = 0 \ 0 & x\neq 0 \end{cases}$$ is clearly bogus. But when the condition $\int_{-\infty}^{+\infty}\delta(t),dt=1$ is added, at least the definition becomes somewhat usable. The definition $\int_{-\infty}^{+\infty}f(t),\delta(t),dt=f(0)$ is better and enough, though. – md2perpe May 02 '19 at 18:16
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    @ViniciusACP asking this question of a student who doesn't(shouldn't?) know how to define $\delta(2t)$ is not fair. And as I pointed out, there is still uniqueness for finite measures so the argument does not fail. – Calvin Khor May 02 '19 at 18:17
  • @CalvinKhor I've added the information you provided in my answer. Thanks! – Vinicius ACP May 02 '19 at 19:37

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