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Q&A

Comments on With what classes of functions does the equality $\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)](x) dx$ lead to paradoxes?

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With what classes of functions does the equality $\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_tt f(t) dx$ lead to paradoxes?

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The following operators keep the area under the convergent integrals unchanged:

$$\int_0^\infty f(x)\,dx=\int_0^\infty\mathcal{L}_t[t f(t)](x) \, dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)\,dx$$

But with divergent integrals applying the Laplace-based transform $\mathcal{L}_t[t f(t)](x)$ (I would denote it $\mathcal{T}f(x)$) to a function more than once may lead to a strictly greater or smaller function, which I refer to as an unacceptable paradox.

One such example is the function $f(x)=\frac1{\sqrt{x}}$. Applying the transform $\mathcal{T}$ to it twice, one arrives at function $\frac{\pi }{2 \sqrt{x}}$, which is strictly greater than the original function, and the problematic step seems to be the first one. It seems, one class of the problematic functions is $f(x)=\frac1{x^p}$, where $ 0< p< 1$.

So, I wonder, what is the whole description of the class of the functions that lead to such paradoxes. I outline the two paradoxes I am seeking to avoid:

  1. Functions such that $f(x)>\mathcal{T^n}[f(x)]$ or $f(x)<\mathcal{T^n}[f(x)]$ for all $x>0$, that is the function becomes strictly smaller or greater after the transform.

  2. Functions such that the integral $\int_0^\infty (f(x)-\mathcal{T^n}[f(x)])dx$ is finite but nonzero.

Is it possible to clearly describe the class of functions that lead to such paradoxes so to avoid them?

History

3 comment threads

x-post https://math.stackexchange.com/questions/4768302/with-what-classes-of-functions-the-equality-i... (1 comment)
Why "paradox"? (6 comments)
"applying $\mathcal{T}$ more than once"—what if we apply $\mathcal{T}$ only once? (5 comments)
"applying $\mathcal{T}$ more than once"—what if we apply $\mathcal{T}$ only once?
clemens‭ wrote 4 months ago

You say that $\mathcal{T}^n(f)$ may be strictly greater than $f$. Do you know if this is true only for $n≥2$, or might it also be true for $n=1$?

Anixx‭ wrote 4 months ago · edited 4 months ago

I did not enocounter with such examples. It seems we can apply the transform to $f(x)=\frac1{\sqrt{x}}$ once, and there are no paradoxes, but if we apply it twice we get the original function but changed by a factor $\pi/2$. Maybe, the formula should be chaged to introduce some factor? I do not know, but in other cases the formula works precisely, for instance makes $1/x$ into $1/x$ and $1$ into $1/x^2$ (which is supported by the theory of hyperfunctions).

clemens‭ wrote 4 months ago

@Anixx: Very interesting. It looks to me that $\mathcal{T} f$ where $f(x) := \frac1x$ is $\frac1{x^2}$, though—am I interpreting $\mathcal{T}$ wrong?

Anixx‭ wrote 4 months ago · edited 4 months ago

Maybe, wrongly. This is the Mathematica code: https://cdn.phototourl.com/free/2026-05-11-65dfdc4d-558d-4c0d-ad0c-d9ade4567d35.png First is forward transform, then backward transform, then again forward with another method. See also this post: https://math.stackexchange.com/a/4802878/2513

clemens‭ wrote 4 months ago

You are right, $\mathcal{T}_x \frac1x$ is indeed $\frac1x$. I made a mistake.