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What are the continuous functions $f$ such that on $\mathbb{R}^{+*}$, they satisfy following functional equation:

$$\int_0^\infty f(t) e^{-itx} \, dt =\lambda \frac{1}{x} f\left(\frac{1}{x}\right)$$

$\lambda$ is a constant.

The functions $f(x)=x^{\alpha}$ with $-1<\operatorname{Re}(\alpha)<0$ are solution, but can we find other solutions to this equation ? Any method to solve this problem ? I tried to transform it to find a differential equation but did not succeed ...

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  • $\begingroup$ Since your question seems to be about sort of a "one-sided" Fourier transform $\mathcal{F}_t[f(t)](x)=\int\limits_0^\infty f(t) e^{-i t x}\,dt$ versus the standard Fourier transform $\mathcal{F}_t[f(t)](x)=\int\limits_{-\infty }^\infty f(t) e^{-i t x}\,dt$, perhaps your question is best stated in terms of the Laplace transform. Substituting $s=i x$ into your relationship leads to $\mathcal{L}_t[f(t)](s)=\int\limits_0^\infty f(t) e^{-s t} \, dt=\lambda\frac{i}{s} f\left(\frac{i}{s}\right)$. $\endgroup$
    – Steven Clark
    Commented Jan 29, 2022 at 17:29
  • $\begingroup$ Note that $\mathcal{L}_t\left[t^a\right](s)=\int\limits_0^\infty t^a e^{-s t}\,dt= \Gamma(a+1)\frac{1}{s}\left(\frac{1}{s}\right)^a$ assuming $-1<\Re(a)<0\land \Re(s)>0$. $\endgroup$
    – Steven Clark
    Commented Jan 29, 2022 at 17:29
  • $\begingroup$ Yes, you are right it can be stated with Laplace Transform also. Then do you have any idea on how to treat the generic case and be sure we have all solutions ? $\endgroup$
    – Bertrand
    Commented Feb 2, 2022 at 8:31
  • $\begingroup$ Any reference where we can find functions la equation where Laplace Transform of a function and the function itself in a functional equation ? $\endgroup$
    – Bertrand
    Commented Feb 2, 2022 at 8:34
  • $\begingroup$ Stating in terms of the Laplace transform relationship $\mathcal{L}_t[f(t)](s)=\int_0^\infty f(t)\,e^{-s t}\,dt=\lambda\frac{1}{s}f\left(\frac{1}{s}\right)$ also leads to the inverse Laplace transform relationship $\mathcal{L}_s^{-1}\left[\lambda\frac{1}{s}f\left(\frac{1}{s}\right)\right](t)=\frac{1}{2 \pi i}\int_{\gamma-i \infty}^{\gamma+i \infty}\lambda\frac{1}{s}f\left(\frac{1}{s}\right) e^{s t}\,ds=f(t)$ which perhaps provides some additional insight. $\endgroup$
    – Steven Clark
    Commented Feb 2, 2022 at 18:53

1 Answer 1

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To start with: this functional equation is a "Fredholm integral equation of second kind".

We use the Mellin transform to find solutions.

(See page 657 of "Handbook of integral equations").

Lets make Mellin transform on both side:

$$\int_0^\infty \int_0^\infty f(t) e^{-itx} \, dt \, x^{s-1} dx=\lambda \int_0^\infty \frac{1}{x} f\left(\frac{1}{x}\right) x^{s-1} dx$$

$$\int_0^\infty \int_0^\infty f(t) t^{-s}e^{-ix} \, x^{s-1} \, dt \,dx=\lambda \int_0^\infty f\left(x\right) x^{-s} dx$$

We note $\mathcal{M}(f)(s)$ the Mellin transform of $f$

$$\mathcal{M}(e^{-ix})(s)\cdot \mathcal{M}(f)(1-s) =\lambda \,\mathcal{M}(f)(1-s)$$

$$[\mathcal{M}(e^{-ix})(s) - \lambda] \mathcal{M}(f)(1-s)=0$$

$$\left[\cos\left(\frac{\pi s}{2}\right) + i\sin\left(\frac{\pi s}{2}\right) - \lambda\right] \mathcal{M}(f)(1-s)=0$$

Under this form, we see either $\mathcal{M}(f)(1-s)=0$ either we have $\lambda$ such that there exist $\alpha$ such that $\cos(\frac{\pi \alpha}{2}) + i\sin(\frac{\pi \alpha}{2})=\lambda$ and $\mathcal{M}(f)(1-s)=\delta(s-\alpha)$. So taking Mellin inverse of $\delta(s-\alpha)$ we see the functional equation has only solutions of the form $x^{a}$.

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  • $\begingroup$ I'm not sure I understand your answer as $\mathcal{M}_t\left[t^a\right](s)=2 \pi \delta (i (a+s))$ (see wolframalpha.com/input?i=MellinTransform%5Bt%5Ea%2C+t%2C+s%5D). For a derivation of this result see the last part of the answer I posted at mathoverflow.net/q/294849. $\endgroup$
    – Steven Clark
    Commented Feb 3, 2022 at 22:02
  • $\begingroup$ Yes, you are right, in the sense of distribution, there is a Mellin transform for $X^{\alpha}$. I am not very familiar with distribution and considered only basic definition of Mellin transform. Indeed, considering your remark we can conclude, I will modify my answer. $\endgroup$
    – Bertrand
    Commented Feb 4, 2022 at 12:49

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