Context: For some function $f\in\mathcal{M}(\mathbb C)$ (meromorphic function on $\mathbb C$), I am interested in linear operators $T_f$ that act on functions of the form $g_a:x\mapsto \exp(ax)$ in the following way: $$T_fg_a(x)=f(a)g_a(x).$$ Here is what I managed to find so far:
- Polynomial: if $f$ is a the polynomial $c_n x^n+\dots+c_0$, then $$T_f:=f\left(\frac{\text{d}}{\text{d}x}\right)$$ satisfies $$T_fg_a(x)=\left(\sum_{k=0}^nc_k\frac{\text{d}^k}{\text{d}x^k}\right)\exp(ax)=\sum_{k=0}^n c_ka^k\exp(ax)=f(a)\exp(ax)$$
- Rational functions: if $f$ is a rational function, then the resolvent formalism allows us to construct a integro-differential operator $T_f$ satisfying $Tg_a=f(a)g_a$.
Now, both the polynomial and the rational cases can be constructed as (pseudo)differential operators. But for general meromorphic functions $f$, it may be trickier, the main difficulty being that $f$ may have poles (for example at $a$, leading to some problem when evaluating $T_fg_a$...). A way to construct such operator may be to consider the Fourier transformation $\mathcal{F}$ and set $$T_f g=\mathcal{F}^{-1}\big[f(t)\mathcal{F}[g](t)\big],$$
Maybe "Pseudo-differential operators with meromorphic symbols and systems of complex differential equations" by Sabir Umarov has a way to construct such an operator (but I can't access this article). So, I'd like to know
- has anyone here worked on such pseudodifferential operator, and know a good reference on this context?
- are there "easier" (in some sense) ways to construct such operators?
Remark: I should state the space $S$ (Schwarz, $L^2$,...) on which $T_f$ must act, but I think the construction of $T_f$ may depend on $S$ (and its topology).
- Edit #1 (an example in the rational case) For $f(x)=1/x$ (which belongs to the section 2. Rational functions), one can observe that $$\int_{-\infty}^x g_a(t)\text{d}t=\int_{-\infty}^x \exp(at)\text{d}t=\frac{\exp(a x)}{a}=f(a)g_a(x)$$ so that $T_f:g\mapsto \int_{-\infty}^x g(t)\text{d}t$ satisfies the condition (except for $a=0$). In general, it is expected that such construction work for all $a$ except for poles of $f$. In full generality, taking powers of this operator yields the solution for $f(x)=x^{-n}$, through the relation $T_{x^{-1}}^n=T_{x^{-n}}$: $$T_{x^{-n}}g(x):=\int_{-\infty}^{t_n}\dots \int_{-\infty}^{t_1}g(t_0)\,\text{d}t_0 \dots \text{d}t_n$$
- Edit #2 (from julio_es_sui_glace's comment) In the first edit, when $\mathfrak{Re}(a)> 0$ the formula doesn't hold. Note that $$\int_{0}^x g_a(t)\text{d}t=\int_0^x \exp(at)\text{d}t=\frac{e^{ax}-1}{a}=f(a)(g_a(x)-g_a(0)).$$ But it is not clear for me how to define a linear operator from the above formula. For example $$T_{x^{-1}}g(x):=\frac{g(0)}{a}+\int_0^xg(t)\text{d}t$$ does not work because the definition should not depend on $a$.
- Edit #3 (some details on the Fourier definition) Define the Fourier transform of $g$ as $$\mathcal{F}[g]:\omega\mapsto\int_{-\infty}^{+\infty} g(t)e^{-i\omega t}\text{d}t$$ so that the inverse Fourier transform is $$\mathcal{F}^{-1}[G]:t\mapsto \frac{1}{2\pi}\int_{-\infty}^{+\infty}G(\omega)e^{i\omega t}\text{d}\omega.$$ It is well known that if $h(t):=g_{ia}(t)=e^{i a t}$ then $\mathcal{F}[h](\omega)=2\pi \delta(\omega-a)$ where $\delta$ is the Dirac delta function (distribution). Hence, $$\mathcal{F}^{-1}[f\mathcal{F}[h]](t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(\omega)\mathcal{F}[h](\omega)e^{i\omega t}\text{d}\omega=f(a)\exp(i a t).$$ Finally, denotation by $D_p$ the dilatation operator defined by $$D_p[g](x)=g(px),$$ and by $M_f$ the multiplication by $f$ operator $$M_f[g](x)=f(x)g(x)$$ (which are both is linear) we can define $$T_f:=D_{-i}\circ \mathcal{F}^{-1}\circ M_f\circ \mathcal{F}\circ D_i.$$ According to the above computations \begin{align} T_fg_a(x)&=(D_{-i}\circ \mathcal{F}^{-1}\circ M_f\circ \mathcal{F}\circ D_i)g_a(x)\\ &= (D_{-i}\circ \mathcal{F}^{-1}\circ M_f\circ \mathcal{F})g_{ia}(x)\\ &=(D_{-i}\circ \mathcal{F}^{-1}\circ M_f)(2\pi \delta(\omega-a))\\ &=(D_{-i}\circ \mathcal{F}^{-1})(2\pi f(\omega) \delta(\omega-a))\\ &=D_{-i}(f(a)\exp(iax))=f(a)\exp(ax). \end{align} Remark: Note that $(\mathcal{F}\circ D_i)^{-1}=D_i^{-1}\circ \mathcal{F}^{-1}=D_{-i}\circ \mathcal{F}^{-1}$ so that $$T_f=(\mathcal{F}\circ D_i)^{-1}\circ M_f\circ (\mathcal{F}\circ D_i)$$ i.e. $T_f$ is the conjugation by $\mathcal{F}\circ D_i$ of the multiplication operator $M_f$. This operator can be thought of as the Laplace transform, but taken on the imaginary axis. The interesting cases are the one where actual poles appear: could this operator be well-defined on some distribution space? Does this generalize integro-differential operator defined through resolvents?
About the bounty: (07/03/25) my question is an open one. I'd be glad to receive partial answers, giving contexts where similar problems appear, equivalent ways to define this kind of operator (in the rational case, for example, as it could be a first step in defining it in a broader context), or even theoretical boundaries that may prevent this operator from existing.
