Would you please compute the behavior of the following composed generalized function?
$g(t) = $ $\delta(e^t)$
Is it even a valid generalized function?
Thank you very much for your time.
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1This old Q&A can perhaps help you answer your question. – Daniele Tampieri Mar 28 '21 at 12:40
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1Thank you so much. – Mohammad Ali Dastgheib Mar 30 '21 at 03:39
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The distribution $\delta(x)$ is concentrated at the point $x=0$. So I guess $\delta(e^t)$ is concentrated at $t=-\infty$. – GEdgar Mar 30 '21 at 11:36
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If you mean the action of $\delta$ on $\exp$, this is of course $1$. If you mean something like the generalisation of change of variables $$\int\delta(e^t)f(t)dt := \int\delta(r) f(\log r)dr/r$$ which if defined for functions vanishing at $-\infty$, is the identically zero distribution. Which makes sense as $e^t$ is never zero.
Calvin Khor
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Change of variables will be $$\int_{-\infty}^{+\infty}\delta(e^t)f(t)dt := \int_0^{+\infty}\frac{\delta(r) f(\log r)}{r},dr$$ so we expect the value to be $$\lim_{t \to -\infty}\frac{f(t)}{e^t}$$ – GEdgar Mar 30 '21 at 11:37
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@GEdgar It depends how you define $\int_0^\infty \delta(r)F(r)dr$. What do you assume on $F$? If $f \in C^\infty_c$ and $Tf(r)=1_{r >0}f(\log r)/r$ then $f\to Tf$ is continuous in the $C^\infty_c$ topology so that $\int_{-\infty}^\infty S(r)Tf(r)dr$ is well-defined and $\int_{-\infty}^\infty S(r)Tf(r)dr=\int_{-\infty}^\infty S(e^t) f(t)dt$ makes sense. With $S=\delta$ it gives $0$. – reuns Mar 30 '21 at 11:55
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Thank you, everyone. I think I am ready now to write the answer to my question. Let me write a synopsis first:
In Distribution Theory (or theory of Generalized Functions), a generalized function is considered not a function in itself but only in relation to how it affects other functions when "integrated" against them. In keeping with this philosophy, to define the delta function properly, it is enough to say what the "integral" of the delta function is against a set of sufficiently "well-behaved" test functions. Hence: A different class of test (or testing) functions will define a different generalized function.
Answer Begins:
from engineering (and applied physics) point of view:
Test functions are usually infinitely differentiable complex-valued (or sometimes real-valued) functions with compact support. Roughly speaking, having compact support means that these test functions vanish (produce only zero output) outside some fixed interval. Due to this property of test functions, the generalized function $g(t)$ = $\delta(e^t)$ maps any test function to the scalar zero (necessary calculations to verify this are straightforward). Indeed, the composed generalized function is mathematically valid (it is the identically zero distribution).from mathematics point of view:
If we would like to analyze any arbitrary vector space of test functions that are equipped with a notion of convergence, then we should refer to the precise definition of "$\delta \circ f$" or "$\delta (f(t))$" which can be found here: [1] (Chapter 1; §1.9 Change of variables in generalized functions; page 22). Simply put, it says:Firstly, we must prepare a sequence of regular distributions converging to the singular delta distribution, e.g. by using "bump functions (Ψ)".
Then we must compose every element of the sequence with $f(t)$, e.g. by computing $Ψ_n(f(t))$ where $n$ is the index of the sequence of distributions.
Finally, we must investigate the convergence of this sequence for our test functions; if it converges, then this converged value is the output of the "$\delta (f(t))$" distribution. If it doesn't converge, then "$\delta (f(t))$" is not a valid distribution over our test function vector space.
[1] Vladimirov, V. S. (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, ISBN 0-415-27356-0